Given two sorted arrays `nums1`

and `nums2`

of size `m`

and `n`

respectively, return **the median** of the two sorted arrays.

The overall run time complexity should be `O(log (m+n))`

.

**Example 1:**

`Input: nums1 = [1,3], nums2 = [2]Output: 2.00000Explanation: merged array = [1,2,3] and median is 2.`

**Example 2:**

`Input: nums1 = [1,2], nums2 = [3,4]Output: 2.50000Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.`

**Constraints:**

`nums1.length == m`

`nums2.length == n`

`0 <= m <= 1000`

`0 <= n <= 1000`

`1 <= m + n <= 2000`

`-10<sup>6</sup> <= nums1[i], nums2[i] <= 10<sup>6</sup>`

`func findMedianSortedArrays(nums1 []int, nums2 []int) float64 { A, B := nums1, nums2 total := len(nums1) + len(nums2) half := (total + 1) / 2 var Aleft, Aright float64 var Bleft, Bright float64 if len(B) < len(A) { A, B = B, A } l, r := 0, len(A)-1 for { i := (l + r) >> 1 // A j := half - i - 2 // B if i >= 0 { Aleft = float64(A[i]) } else { Aleft = math.Inf(-1) } if (i + 1) < len(A) { Aright = float64(A[i+1]) } else { Aright = math.Inf(1) } if j >= 0 { Bleft = float64(B[j]) } else { Bleft = math.Inf(-1) } if (j + 1) < len(B) { Bright = float64(B[j+1]) } else { Bright = math.Inf(1) } // partition is correct if Aleft <= Bright && Bleft <= Aright { // odd if total%2 == 1 { return max(Aleft, Bleft) } // even return (max(Aleft, Bleft) + min(Aright, Bright)) / 2 } else if Aleft > Bright { r = i - 1 } else { l = i + 1 } }}func max(a, b float64) float64 { if a > b { return a } return b}func min(a, b float64) float64 { if a < b { return a } return b}`

This Go code implements the median of two sorted arrays algorithm. The goal is to find the median value of the combined sorted arrays `nums1`

and `nums2`

. The algorithm uses a binary search approach to efficiently partition the arrays.

Here's a breakdown of the code:

**Variable Initialization:**`A`

and`B`

are assigned the input arrays`nums1`

and`nums2`

.`total`

represents the total number of elements in both arrays.`half`

is calculated as the middle index of the combined arrays.

**Partitioning and Binary Search:**If the length of array

`B`

is less than`A`

, swap them.Initialize left (

`l`

) and right (`r`

) pointers for array`A`

.Inside the binary search loop:

Calculate indices

`i`

for array`A`

and`j`

for array`B`

.Update variables (

`Aleft`

,`Aright`

,`Bleft`

,`Bright`

) to store values around the partition.Check if the partition is correct. If so, determine the median based on whether the total number of elements is odd or even.

Adjust the pointers based on the values at the current partition.

**Helper Functions:**`max`

returns the maximum of two given float64 values.`min`

returns the minimum of two given float64 values.

The core logic involves adjusting the partition of the arrays (`A`

and `B`

) until the correct position is found where elements on the left side are smaller or equal to elements on the right side. The median is then calculated based on this partition. The use of `math.Inf`

represents positive and negative infinity for comparison purposes.

Design a time-based key-value data structure that can store multiple values for the same key at different time stamps and retrieve the key's value at a certain timestamp.

Implement the `TimeMap`

class:

`TimeMap()`

Initializes the object of the data structure.`void set(String key, String value, int timestamp)`

Stores the key`key`

with the value`value`

at the given time`timestamp`

.`String get(String key, int timestamp)`

Returns a value such that`set`

was called previously, with`timestamp_prev <= timestamp`

. If there are multiple such values, it returns the value associated with the largest`timestamp_prev`

. If there are no values, it returns`""`

.

**Example 1:**

`Input["TimeMap", "set", "get", "get", "set", "get", "get"][[], ["foo", "bar", 1], ["foo", 1], ["foo", 3], ["foo", "bar2", 4], ["foo", 4], ["foo", 5]]Output[null, null, "bar", "bar", null, "bar2", "bar2"]ExplanationTimeMap timeMap = new TimeMap();timeMap.set("foo", "bar", 1); // store the key "foo" and value "bar" along with timestamp = 1.timeMap.get("foo", 1); // return "bar"timeMap.get("foo", 3); // return "bar", since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 is "bar".timeMap.set("foo", "bar2", 4); // store the key "foo" and value "bar2" along with timestamp = 4.timeMap.get("foo", 4); // return "bar2"timeMap.get("foo", 5); // return "bar2"`

**Constraints:**

`1 <= key.length, value.length <= 100`

`key`

and`value`

consist of lowercase English letters and digits.`1 <= timestamp <= 10<sup>7</sup>`

All the timestamps

`timestamp`

of`set`

are strictly increasing.At most

`2 * 10<sup>5</sup>`

calls will be made to`set`

and`get`

.

`type TimeMap struct { store map[string][]ValStamp}type ValStamp struct { Val string Time int}func Constructor() TimeMap { return TimeMap{store: make(map[string][]ValStamp)}}func (this *TimeMap) Set(key string, value string, timestamp int) { if _, ok := this.store[key]; !ok { this.store[key] = make([]ValStamp, 0) } this.store[key] = append(this.store[key], ValStamp{value, timestamp})}func (this *TimeMap) Get(key string, timestamp int) string { var res string var values []ValStamp if _, ok := this.store[key]; ok { values = this.store[key] } l, r := 0, len(values)-1 for l <= r { mid := l + (r-l)/2 if values[mid].Time <= timestamp { res = values[mid].Val l = mid + 1 } else { r = mid - 1 } } return res}`

This code defines a `TimeMap`

struct, which is essentially a key-value store where each key can have multiple values associated with different timestamps. Here's a detailed breakdown:

**TimeMap Struct:**`TimeMap`

is defined with a single field,`store`

, which is a map with string keys and slices of`ValStamp`

values.`ValStamp`

is a struct representing a value and its associated timestamp.

**Constructor:**- The
`Constructor`

function initializes and returns a new`TimeMap`

with an empty map as its store.

- The
**Set Method:**`Set`

adds a new value along with its timestamp to the map for a given key.If the key doesn't exist in the map, it initializes an empty slice for that key.

It then appends a new

`ValStamp`

to the slice with the provided value and timestamp.

**Get Method:**`Get`

retrieves the value associated with a key at or before the specified timestamp.It initializes two pointers,

`l`

(left) and`r`

(right), for a binary search within the stored values.The function iteratively performs a binary search to find the closest timestamp that is less than or equal to the given timestamp.

The result is updated as it searches, and the final result is the value associated with the found timestamp.

In summary, this code implements a time-based key-value store where values are associated with timestamps, and the `Get`

method retrieves the value closest to or at a specified timestamp using binary search on the stored timestamps.

There is an integer array `nums`

sorted in ascending order (with **distinct** values).

Prior to being passed to your function, `nums`

is **possibly rotated** at an unknown pivot index `k`

(`1 <= k < nums.length`

) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]`

(**0-indexed**). For example, `[0,1,2,4,5,6,7]`

might be rotated at pivot index `3`

and become `[4,5,6,7,0,1,2]`

.

Given the array `nums`

**after** the possible rotation and an integer `target`

, return *the index of* `target`

*if it is in* `nums`

*, or* `-1`

*if it is not in* `nums`

.

You must write an algorithm with `O(log n)`

runtime complexity.

**Example 1:**

`Input: nums = [4,5,6,7,0,1,2], target = 0Output: 4`

**Example 2:**

`Input: nums = [4,5,6,7,0,1,2], target = 3Output: -1`

**Example 3:**

`Input: nums = [1], target = 0Output: -1`

**Constraints:**

`1 <= nums.length <= 5000`

`-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup>`

All values of

`nums`

are**unique**.`nums`

is an ascending array that is possibly rotated.`-10<sup>4</sup> <= target <= 10<sup>4</sup>`

`func search(nums []int, target int) int { left, right := 0, len(nums) - 1 for left <= right { mid := (left + right) / 2 if target == nums[mid] { return mid } // left sorted portion if nums[left] <= nums[mid] { if target > nums[mid] || target < nums[left] { left = mid + 1 } else { right = mid - 1 } // Right sorted portion } else { if target < nums[mid] || target > nums[right] { right = mid - 1 } else { left = mid + 1 } } } return -1}`

This is an implementation of a binary search algorithm to find the index of a target element in a rotated sorted array. The function takes two parameters: a sorted array `nums`

and a target element `target`

. It initializes two pointers, `left`

and `right`

, which represent the range of indices to search within.

The algorithm then enters a while loop, where it calculates the middle index `mid`

of the current range. If the target is equal to the element at the middle index, it returns the index.

The interesting part comes when dealing with the rotated sorted array. It checks whether the left portion or the right portion of the array is sorted. If the left portion is sorted, it checks whether the target lies within that sorted range. If it does, the search continues in the left portion; otherwise, it shifts the search to the right portion. Similarly, if the right portion is sorted, it checks whether the target lies within that range and adjusts the search accordingly.

This approach allows the algorithm to handle rotated sorted arrays efficiently, maintaining the logarithmic time complexity of the standard binary search.

]]>Suppose an array of length `n`

sorted in ascending order is **rotated** between `1`

and `n`

times. For example, the array `nums = [0,1,2,4,5,6,7]`

might become:

`[4,5,6,7,0,1,2]`

if it was rotated`4`

times.`[0,1,2,4,5,6,7]`

if it was rotated`7`

times.

Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]`

1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`

.

Given the sorted rotated array `nums`

of **unique** elements, return *the minimum element of this array*.

You must write an algorithm that runs in `O(log n) time.`

**Example 1:**

`Input: nums = [3,4,5,1,2]Output: 1Explanation: The original array was [1,2,3,4,5] rotated 3 times.`

**Example 2:**

`Input: nums = [4,5,6,7,0,1,2]Output: 0Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.`

**Example 3:**

`Input: nums = [11,13,15,17]Output: 11Explanation: The original array was [11,13,15,17] and it was rotated 4 times.`

**Constraints:**

`n == nums.length`

`1 <= n <= 5000`

`-5000 <= nums[i] <= 5000`

All the integers of

`nums`

are**unique**.`nums`

is sorted and rotated between`1`

and`n`

times.

`func findMin(nums []int) int { res := nums[0] left, right := 0, len(nums)-1 for left <= right { mid := (left + right) / 2 if nums[mid] >= nums[0] { left = mid + 1 } else { if nums[mid] < res { res = nums[mid] } right = mid - 1 } } return res}`

This Go code defines a function `findMin`

that takes a sorted and rotated array of integers `nums`

as input and returns the minimum element in the array. The array is sorted but may have been rotated, meaning elements have been moved to the left or right.

Let's break down the code step by step:

`res := nums[0]`

: Initializes the result variable`res`

with the first element of the array. This assumes that the array is not empty.`left, right := 0, len(nums)-1`

: Initializes two pointers,`left`

and`right`

, which represent the range of elements currently being considered.`left`

starts at the beginning of the array, and`right`

starts at the end.`for left <= right {`

: Initiates a loop that continues as long as the`left`

pointer is less than or equal to the`right`

pointer.`mid := (left + right) / 2`

: Calculates the middle index of the current range.`if nums[mid] >= nums[0] {`

: Checks if the middle element is greater than or equal to the first element of the array. This condition is used to determine whether the rotation point is on the right side of the array.If true, it means the minimum element is on the right side, so the

`left`

pointer is updated to`mid + 1`

.If false, it means the rotation point, and consequently, the minimum element, is on the left side or at the current position. In this case, it goes to the

`else`

block.

`if nums[mid] < res {`

: Checks if the current element at the middle index is less than the current minimum (`res`

).- If true, it updates the minimum (
`res`

) to the value at the middle index.

- If true, it updates the minimum (
`right = mid - 1`

: Adjusts the`right`

pointer to continue searching on the left side of the array.The loop continues until the

`left`

pointer exceeds the`right`

pointer.`return res`

: Returns the minimum element found during the search.

In summary, this code efficiently finds the minimum element in a sorted and rotated array using a binary search approach. The algorithm exploits the sorted nature of the array and adapts the search based on the comparison between the middle element and the first element of the array.

]]>`n`

piles of bananas, the `i<sup>th</sup>`

pile has `piles[i]`

bananas. The guards have gone and will come back in `h`

hours.Koko can decide her bananas-per-hour eating speed of `k`

. Each hour, she chooses some pile of bananas and eats `k`

bananas from that pile. If the pile has less than `k`

bananas, she eats all of them instead and will not eat any more bananas during this hour.

Koko likes to eat slowly but still wants to finish eating all the bananas before the guards return.

Return *the minimum integer* `k`

*such that she can eat all the bananas within* `h`

*hours*.

**Example 1:**

`Input: piles = [3,6,7,11], h = 8Output: 4`

**Example 2:**

`Input: piles = [30,11,23,4,20], h = 5Output: 30`

**Example 3:**

`Input: piles = [30,11,23,4,20], h = 6Output: 23`

**Constraints:**

`1 <= piles.length <= 10<sup>4</sup>`

`piles.length <= h <= 10<sup>9</sup>`

`1 <= piles[i] <= 10<sup>9</sup>`

`func minEatingSpeed(piles []int, h int) int { lo, hi, ans := 1, 1000000000, 1 for lo <= hi { mid := (lo + hi) / 2 if canEat(piles, h, mid){ ans = mid hi = mid -1 }else { lo = mid +1 } } return ans }func canEat(pile []int, timeLimit, speed int )bool { timeNeed := 0 for _, banana := range pile { timeNeed = timeNeed + (banana + speed - 1) / speed if timeNeed > timeLimit { return false } } return true}`

This code is a Go implementation for finding the minimum eating speed required to eat all the bananas within a given time limit. The function `minEatingSpeed`

takes in two parameters: `piles`

, which is an array of integers representing the number of bananas in each pile, and `h`

, which represents the time limit in hours.

The function initializes three variables: `lo`

, `hi`

, and `ans`

. The `lo`

and `hi`

represents the lower and upper bounds for the eating speed, while `ans`

stores the final answer. The function uses a binary search approach to find the minimum eating speed. It starts by setting `lo`

to 1 and `hi`

to a large number, here 1000000000.

The code then enters a while loop that runs until `lo`

is less than or equal to `hi`

. Inside the loop, it calculates the middle value `mid`

using the formula `(lo + hi) / 2`

. Then it calls the `canEat`

function, passing in the `piles`

, the time limit `h`

, and the current eating speed `mid`

.

If the `canEat`

function returns true, it means it's possible to eat all the bananas within the given time limit using the current speed. In this case, the function updates the `ans`

to the current `mid`

value, indicating a possible minimum speed, and adjusts `hi`

to `mid - 1`

to search for smaller values.

If the `canEat`

function returns false, it means it's not possible to eat all the bananas within the time limit at the current speed. In this case, the function updates `lo`

to `mid + 1`

to search for higher values.

The `canEat`

function calculates the total time needed to eat all the bananas using the current speed. It iterates through each pile of bananas, calculates the time required for each pile at the given speed, and adds it to the total time. If the total time exceeds the given time limit, the function returns false. Otherwise, it returns true.

Finally, the `minEatingSpeed`

function returns the final `ans`

, which represents the minimum eating speed required to eat all the bananas within the given time limit.

You are given an `m x n`

integer matrix `matrix`

with the following two properties:

Each row is sorted in non-decreasing order.

The first integer of each row is greater than the last integer of the previous row.

Given an integer `target`

, return `true`

*if* `target`

*is in* `matrix`

*or* `false`

*otherwise*.

You must write a solution in `O(log(m * n))`

time complexity.

`Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3Output: true`

**Example 2:**

`Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13Output: false`

**Constraints:**

`m == matrix.length`

`n == matrix[i].length`

`1 <= m, n <= 100`

`-10<sup>4</sup> <= matrix[i][j], target <= 10<sup>4</sup>`

`func searchMatrix(matrix [][]int, target int) bool { ROWS := len(matrix) COLUMNS := len(matrix[0]) top := 0 bot := ROWS - 1 for top <= bot { row := (top + bot) / 2 if target > matrix[row][len(matrix[row])-1] { top = row + 1 } else if target < matrix[row][0] { bot = row - 1 } else { break } } if !(top <= bot) { return false } row := (top + bot) / 2 left := 0 right := COLUMNS - 1 for left <= right { middle := (left + right) / 2 if matrix[row][middle] == target { return true } else if matrix[row][middle] < target { left = middle + 1 } else { right = middle - 1 } } return false}`

This code defines a Go function `searchMatrix`

that takes a 2D matrix of integers `matrix`

and an integer `target`

as input and returns a boolean value indicating whether the target is present in the matrix. The function uses a binary search approach to efficiently search for the target element.

Here's a step-by-step explanation of the code:

Get the number of rows (

`ROWS`

) and columns (`COLUMNS`

) in the matrix.Initialize two pointers,

`top`

and`bot`

, to keep track of the range of rows to search. Initially,`top`

points to the first row (0), and`bot`

points to the last row (ROWS - 1).Perform a binary search on the rows of the matrix using the

`top`

and`bot`

pointers. This loop continues until`top`

is less than or equal to`bot`

. Inside the loop:Calculate the middle row as

`(top + bot) / 2`

.Check if the target value is greater than the last element of the middle row (

`matrix[row][len(matrix[row])-1]`

). If it is, update`top`

to`row + 1`

because the target must be in a lower row.If the target is less than the first element of the middle row, update

`bot`

to`row - 1`

because the target must be in a higher row.If neither of the above conditions is met, it means the target is within the range of the current row, so break out of the loop.

After the binary search for rows, check if

`top`

is still less than or equal to`bot`

. If not, it means the target is not in the matrix, so return`false`

.If the target is still potentially in the matrix (based on the row search), calculate the middle row again as

`(top + bot) / 2`

. Initialize two more pointers,`left`

and`right`

, to keep track of the columns within the current row.Perform a binary search on the columns of the current row using the

`left`

and`right`

pointers. This loop continues until`left`

is less than or equal to`right`

. Inside the loop:Calculate the middle column as

`(left + right) / 2`

.Check if the element at

`matrix[row][middle]`

is equal to the target. If it is, return`true`

because the target is found.If the element at the middle column is less than the target, update

`left`

to`middle + 1`

because the target must be in the right half of the row.If the element at the middle column is greater than the target, update

`right`

to`middle - 1`

because the target must be in the left half of the row.

If the loop for column search completes without finding the target, return

`false`

.

In summary, this code efficiently searches for a target value in a sorted 2D matrix by performing two binary searches: one to find the appropriate row where the target might exist and another to find the target within that row. If the target is found, the function returns `true`

; otherwise, it returns `false`

.

Given an array of integers `nums`

which is sorted in ascending order, and an integer `target`

, write a function to search `target`

in `nums`

. If `target`

exists, then return its index. Otherwise, return `-1`

.

You must write an algorithm with `O(log n)`

runtime complexity.

**Example 1:**

`Input: nums = [-1,0,3,5,9,12], target = 9Output: 4Explanation: 9 exists in nums and its index is 4`

**Example 2:**

`Input: nums = [-1,0,3,5,9,12], target = 2Output: -1Explanation: 2 does not exist in nums so return -1`

**Constraints:**

`1 <= nums.length <= 10<sup>4</sup>`

`-10<sup>4</sup> < nums[i], target < 10<sup>4</sup>`

All the integers in

`nums`

are**unique**.`nums`

is sorted in ascending order.`30 <= temperatures[i] <= 100`

`func search(nums []int, target int) int { left := 0 right := len(nums)-1 for left<=right{ mid := (left+right)/2 if nums[mid]==target{ return mid }else if nums[mid]<target{ left = mid +1 }else { right = mid -1 } } return -1}`

This is a Go programming language function called `search`

that performs a binary search to find a target integer within a sorted array of integers. Let me break down the code step by step:

`func search(nums []int, target int) int {`

: This line defines a function named`search`

that takes two arguments:`nums`

is a slice of integers, representing the sorted array in which we want to search for the`target`

value.`target`

is the integer we want to find in the`nums`

array.The function returns an integer, which is the index of the

`target`

in the array if found, or -1 if it's not found.

`left := 0`

and`right := len(nums)-1`

: These lines initialize two variables`left`

and`right`

.`left`

represents the left boundary of the current search range, initialized to the beginning of the array (index 0), and`right`

represents the right boundary, initialized to the end of the array (index`len(nums)-1`

).`for left <= right {`

: This starts a loop that continues as long as the`left`

boundary is less than or equal to the`right`

boundary. This loop is the heart of the binary search algorithm.`mid := (left+right)/2`

: Inside the loop, it calculates the middle index`mid`

of the current search range by taking the average of`left`

and`right`

. This is where the binary search gets its name because it repeatedly divides the search range in half.`if nums[mid] == target {`

: This checks if the value at the middle index`mid`

of the array`nums`

is equal to the`target`

. If it is, this means we've found the`target`

, and the function returns`mid`

, which is the index where`target`

is located in the array.`else if nums[mid] < target {`

: If the value at`mid`

is less than the`target`

, which means the`target`

must be in the right half of the current search range. So, it updates`left`

to`mid + 1`

, effectively narrowing the search range to the right half.`else {`

: If the value at`mid`

is greater than the`target`

, which means the`target`

must be in the left half of the current search range. So, it updates`right`

to`mid - 1`

, narrowing the search range to the left half.If the loop exits without finding the

`target`

, i.e.,`left`

becomes greater than`right`

, the function returns`-1`

to indicate that the`target`

is not in the array.

In summary, this code performs an efficient binary search on a sorted array to find the index of a specific target value, and it returns that index or -1 if the target is not in the array. Binary search is efficient because it repeatedly divides the search range in half, reducing the number of comparisons needed to find the target in a sorted array.

]]>Given an array of integers `heights`

representing the histogram's bar height where the width of each bar is `1`

, return *the area of the largest rectangle in the histogram*.

**Example 1:**

Input: heights = [2,1,5,6,2,3]Output: 10Explanation: The above is a histogram where the width of each bar is 1. The largest rectangle is shown in the red area, which has an area of 10 units.

**Example 2:**

Input: heights = [2,4]Output: 4

`func largestRectangleArea(heights []int) int { stack := []StackValue{} maxArea := 0 var start int for i,h := range heights { start = i for len(stack) != 0 && stack[len(stack)-1].height > h { index, height := stack[len(stack)-1].index , stack[len(stack)-1].height stack = stack[0:len(stack)-1] maxArea = max(maxArea, height*(i-index)) start = index } stack = append(stack, StackValue{start,h}) } for _, h := range stack { maxArea = max(maxArea, h.height*(len(heights)-h.index)) } return maxArea}type StackValue struct { index int height int}func max(a,b int) int { if a > b { return a } return b}`

This Go code defines a function `largestRectangleArea`

that calculates the largest area of a rectangle that can be formed within a given histogram represented by an array of integer heights. The code uses a stack data structure to efficiently compute this.

Here's a step-by-step explanation of the code:

`stack := []StackValue{}`

: Initialize an empty stack to keep track of heights and their corresponding indices in the histogram.`maxArea := 0`

: Initialize a variable`maxArea`

to store the maximum rectangle area found so far.`var start int`

: Initialize a variable`start`

to keep track of the starting index of a potential rectangle.Loop through the input

`heights`

array using a for loop with`i`

representing the current index and`h`

representing the current height.Set

`start`

to the current index`i`

.Check if the stack is not empty and the height at the top of the stack (

`stack[len(stack)-1].height`

) is greater than the current height`h`

. If this condition is true, it means we can potentially calculate the area of a rectangle.Inside the loop, pop elements from the stack while the condition in step 6 is met. For each popped element, calculate the area it represents (height times the width, which is the difference between the current index

`i`

and the index stored in the popped element). Update`maxArea`

if this area is greater than the current`maxArea`

.Set

`start`

to the index of the last popped element, as this is the starting index of the potential rectangle that includes the current height`h`

.Push the current index

`start`

and height`h`

onto the stack as a`StackValue`

.After the loop, there might still be elements left in the stack. Loop through these remaining elements and calculate the area for each of them, considering the entire remaining width of the histogram. Update

`maxArea`

if any of these areas is greater than the current`maxArea`

.Finally, return the

`maxArea`

, which represents the largest rectangle that can be formed within the given histogram.

The `StackValue`

struct is used to store the index and height of elements pushed onto the stack.

The `max`

function is a simple utility function used to find the maximum of two integers.

This code essentially uses a stack to keep track of ascending heights in the histogram and calculates the maximum rectangle area that can be formed efficiently.

]]>There are `n`

cars going to the same destination along a one-lane road. The destination is `target`

miles away.

You are given two integer array `position`

and `speed`

, both of length `n`

, where `position[i]`

is the position of the `i<sup>th</sup>`

car and `speed[i]`

is the speed of the `i<sup>th</sup>`

car (in miles per hour).

A car can never pass another car ahead of it, but it can catch up to it and drive bumper to bumper **at the same speed**. The faster car will **slow down** to match the slower car's speed. The distance between these two cars is ignored (i.e., they are assumed to have the same position).

A **car fleet** is some non-empty set of cars driving at the same position and same speed. Note that a single car is also a car fleet.

If a car catches up to a car fleet right at the destination point, it will still be considered as one car fleet.

Return *the* *number of car fleets**that will arrive at the destination*.

**Example 1:**

`Input: target = 12, position = [10,8,0,5,3], speed = [2,4,1,1,3]Output: 3Explanation:The cars starting at 10 (speed 2) and 8 (speed 4) become a fleet, meeting each other at 12.The car starting at 0 does not catch up to any other car, so it is a fleet by itself.The cars starting at 5 (speed 1) and 3 (speed 3) become a fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches target.Note that no other cars meet these fleets before the destination, so the answer is 3.`

**Example 2:**

`Input: target = 10, position = [3], speed = [3]Output: 1Explanation: There is only one car, hence there is only one fleet.`

**Example 3:**

`Input: target = 100, position = [0,2,4], speed = [4,2,1]Output: 1Explanation:The cars starting at 0 (speed 4) and 2 (speed 2) become a fleet, meeting each other at 4. The fleet moves at speed 2.Then, the fleet (speed 2) and the car starting at 4 (speed 1) become one fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches target.`

**Constraints:**

`n == position.length == speed.length`

`1 <= n <= 10<sup>5</sup>`

`0 < target <= 10<sup>6</sup>`

`0 <= position[i] < target`

All the values of

`position`

are**unique**.`0 < speed[i] <= 10<sup>6</sup>`

`func carFleet(target int, position []int, speed []int) int { pair := make([]carInfo, 0, len(position)) stack := make([]float32, 0, len(position)) for i, _ := range position { pair = append(pair, carInfo{position[i], speed[i]}) } sort.Slice(pair, func(i, j int) bool { return pair[i].pos < pair[j].pos }) for i := len(pair) - 1; i >= 0; i-- { stack = append(stack, float32(target-pair[i].pos)/float32(pair[i].spd)) if len(stack) >= 2 && stack[len(stack)-1] <= stack[len(stack)-2] { stack = stack[:len(stack)-1] } } return len(stack)}type carInfo struct { pos int spd int}`

This Go code defines a function `carFleet`

calculates the number of car fleets that can reach a target destination. Let me break down the code step by step:

`func carFleet(target int, position []int, speed []int) int`

: This is the function definition. It takes three parameters:`target`

, which is the target destination,`position`

, which is a slice of integers representing the initial positions of cars, and`speed`

, which is a slice of integers representing the speeds of cars.`pair := make([]carInfo, 0, len(position))`

: Here, a slice named`pair`

is created to store car information. It's initially empty but has a capacity equal to the length of the`position`

slice.`stack := make([]float32, 0, len(position))`

: Another slice named`stack`

is created, which will be used to store the time it takes for each car to reach the target. It's initially empty but has a capacity equal to the length of the`position`

slice.`for i, _ := range position { pair = append(pair, carInfo{position[i], speed[i]}) }`

: This loop iterates over the`position`

and`speed`

slices and populates the`pair`

slice with`carInfo`

structures. Each structure stores the position and speed of a car.`sort.Slice(pair, func(i, j int) bool { return pair[i].pos < pair[j].pos })`

: The`pair`

slice is sorted based on the car positions in ascending order using the`sort.Slice`

function.The next loop iterates over the sorted

`pair`

slice in reverse order (from the car closest to the target to the car farthest from the target).`stack = append(stack, float32(target-pair[i].pos)/float32(pair[i].spd))`

: For each car, it calculates the time it takes to reach the target using the formula`(target - position) / speed`

and appends this time to the`stack`

slice.`if len(stack) >= 2 && stack[len(stack)-1] <= stack[len(stack)-2] { stack = stack[:len(stack)-1] }`

: If the`stack`

contains at least two elements and the time for the current car is less than or equal to the time for the previous car, it means the current car will catch up to the previous car in a fleet. In this case, the previous car's time is removed from the`stack`

.Finally, the function returns

`len(stack)`

, which represents the number of car fleets that can reach the target without colliding.

The code leverages sorting and a stack-like approach to efficiently calculate the number of car fleets that can reach the target without collisions.

]]>Given an array of integers `temperatures`

represent the daily temperatures, return *an array*`answer`

*such that* `answer[i]`

*is the number of days you have to wait after the* `i<sup>th</sup>`

*day to get a warmer temperature*. If there is no future day for which this is possible, keep `answer[i] == 0`

instead.

**Example 1:**

`Input: temperatures = [73,74,75,71,69,72,76,73]Output: [1,1,4,2,1,1,0,0]`

**Example 2:**

`Input: temperatures = [30,40,50,60]Output: [1,1,1,0]`

**Example 3:**

`Input: temperatures = [30,60,90]Output: [1,1,0]`

**Constraints:**

`1 <= temperatures.length <= 10<sup>5</sup>`

`30 <= temperatures[i] <= 100`

`func dailyTemperatures(temperatures []int) []int { res := make([]int, len(temperatures)) for i := len(temperatures) - 1; i >= 0; i-- { j := i + 1 for j < len(temperatures) && temperatures[j] <= temperatures[i] { if res[j] <= 0 { break } j += res[j] } if j < len(temperatures) && temperatures[j] > temperatures[i] { res[i] = j - i } } return res}`

This code defines a Go function called `dailyTemperatures`

that takes a slice of integers called `temperatures`

as input and returns another slice of integers as output. The purpose of this function is to calculate the number of days you have to wait until a warmer temperature is forecasted for each day in the input.

Here's a detailed explanation of how the code works:

`res := make([]int, len(temperatures))`

: This line initializes an integer slice called`res`

with the same length as the input`temperatures`

. This slice will store the number of days to wait for a warmer temperature for each day in the input.The code then enters a loop that iterates through the

`temperatures`

slice in reverse order, starting from the last day and moving towards the first day:`for i := len(temperatures) - 1; i >= 0; i--`

.Inside this loop, a variable

`j`

is initialized to`i + 1`

.`j`

will be used to traverse through the`temperatures`

array looking for a warmer temperature.Another inner loop is used to find the next warmer temperature. It continues as long as two conditions are met:

`j < len(temperatures)`

: Ensure that`j`

doesn't go out of bounds.`temperatures[j] <= temperatures[i]`

: Check if the temperature at day`j`

is less than or equal to the temperature at day`i`

.

Within the inner loop, there's a conditional check:

`if res[j] <= 0`

. This checks if there's already a result for day`j`

. If`res[j]`

is less than or equal to 0, it means we haven't found a warmer temperature yet, so we break out of the inner loop.If the inner loop is exited without finding a warmer temperature, the code proceeds to the next step.

After exiting the inner loop, there's another conditional check:

`if j < len(temperatures) && temperatures[j] > temperatures[i]`

. This checks if`j`

is still within bounds, and if the temperature at day`j`

is indeed warmer than the temperature at day`i`

.If both conditions are met, it means a warmer temperature has been found, so the difference between

`j`

and`i`

(i.e.,`j - i`

) is assigned to`res[i]`

. This indicates how many days you have to wait until a warmer temperature is forecasted for day`i`

.The outer loop continues until all days in the

`temperatures`

array have been processed, and the`res`

slice is populated with the waiting days for each day.Finally, the function returns the

`res`

slice, which contains the number of days to wait for a warmer temperature for each day in the input.

In summary, this code efficiently calculates the number of days you have to wait for a warmer temperature for each day, taking into account the temperature forecasts for the upcoming days. It uses a nested loop and dynamic programming to optimize the process.

]]>Given `n`

pairs of parentheses, write a function to *generate all combinations of well-formed parentheses*.

**Example 1:**

`Input: n = 3Output: ["((()))","(()())","(())()","()(())","()()()"]`

**Example 2:**

`Input: n = 1Output: ["()"]`

**Constraints:**

`1 <= n <= 8`

`import "strings"func generateParenthesis(n int) []string { var stack []string var res []string var backtrack func(int, int) backtrack = func(openN int, closedN int){ if openN == n && closedN == n && openN == closedN{ res = append(res, strings.Join(stack, "")) return } if openN < n{ stack = append(stack, "(") backtrack(openN+1, closedN) pop(&stack) } if closedN < openN{ stack = append(stack, ")") backtrack(openN, closedN+1) pop(&stack) } } backtrack(0,0) return res}func pop(list *[]string){ length := len(*list) *list = (*list)[:length-1]}`

This Go code defines a function `generateParenthesis`

that generates all valid combinations of parentheses pairs with a given number `n`

. Here's a detailed explanation of the code:

`import "strings"`

: This line imports the "strings" package, which is used for manipulating strings, particularly to join them together.`func generateParenthesis(n int) []string { ... }`

: This is the main function that takes an integer`n`

as input and returns a slice of strings representing valid combinations of parentheses.Inside this function, three main variables are defined:

`stack []string`

: A slice of strings used to keep track of the currently open and unclosed parentheses.`res []string`

: A slice of strings to store the final result, i.e., valid combinations of parentheses.`backtrack func(int, int)`

: This is a nested function that performs the recursive backtracking to generate the combinations.

`backtrack`

is a recursive function that takes two arguments:`openN`

: The count of open parentheses.`closedN`

: The count of closed parentheses.

The first

`if`

condition checks if we have reached the desired count of open and closed parentheses, both equal to`n`

. If so, it means we have formed a valid combination, and it's added to the`res`

slice by joining the`stack`

using`strings.Join`

.Next, there are two

`if`

conditions:`if openN < n { ... }`

: This checks if the count of open parentheses is less than`n`

. If true, it appends an open parenthesis "(" to the`stack`

, increments the count of open parentheses, and then calls`backtrack`

recursively. After the recursive call, it "pops" the last element from the`stack`

to backtrack and explore other possibilities.`if closedN < openN { ... }`

: This checks if the count of closed parentheses is less than the count of open parentheses. If true, it appends a closed parenthesis ")" to the`stack`

, increments the count of closed parentheses, and then calls`backtrack`

recursively. It also pops the last element from the`stack`

afterward for backtracking.

Finally, the

`backtrack`

function is initially called with`openN`

and`closedN`

both set to 0, starting the recursive process to generate valid combinations.The function returns the

`res`

slice, which contains all the valid combinations of parentheses.`func pop(list *[]string) { ... }`

: This function is used to remove the last element from a slice of strings. It's a utility function called within`backtrack`

to backtrack and explore other possibilities by removing the last element added to the`stack`

.

In summary, this code uses a recursive backtracking approach to generate all valid combinations of parentheses pairs with a given count `n`

. It maintains a `stack`

to keep track of the currently open and unclosed parentheses and appends and pops elements from it while exploring possibilities. The valid combinations are stored in the `res`

slice and returned as the final result.

You are given an array of strings `tokens`

that represent an arithmetic expression in a Reverse Polish Notation.

Evaluate the expression. Return *an integer that represents the value of the expression*.

**Note** that:

The valid operators are

`'+'`

,`'-'`

,`'*'`

, and`'/'`

.Each operand may be an integer or another expression.

The division between two integers always

**truncates toward zero**.There will not be any division by zero.

The input represents a valid arithmetic expression in a reverse polish notation.

The answer and all the intermediate calculations can be represented in a

**32-bit**integer.

**Example 1:**

`Input: tokens = ["2","1","+","3","*"]Output: 9Explanation: ((2 + 1) * 3) = 9`

**Example 2:**

`Input: tokens = ["4","13","5","/","+"]Output: 6Explanation: (4 + (13 / 5)) = 6`

**Example 3:**

`Input: tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]Output: 22Explanation: ((10 * (6 / ((9 + 3) * -11))) + 17) + 5= ((10 * (6 / (12 * -11))) + 17) + 5= ((10 * (6 / -132)) + 17) + 5= ((10 * 0) + 17) + 5= (0 + 17) + 5= 17 + 5= 22`

**Constraints:**

`1 <= tokens.length <= 10<sup>4</sup>`

`tokens[i]`

is either an operator:`"+"`

,`"-"`

,`"*"`

, or`"/"`

, or an integer in the range`[-200, 200]`

.

`func evalRPN(tokens []string) int { var stack []int for _, val := range tokens { switch val { case "+": stack = append(stack, pop(&stack)+pop(&stack)) case "-": a, b := pop(&stack), pop(&stack) stack = append(stack, b-a) case "*": stack = append(stack, pop(&stack)*pop(&stack)) case "/": a, b := pop(&stack), pop(&stack) stack = append(stack, b/a) default: i, _ := strconv.Atoi(val) stack = append(stack, i) } } return stack[0]}func pop(stack *[]int) int { top := (*stack)[len(*stack)-1] *stack = (*stack)[:len(*stack)-1] return top}`

This code defines a Go function called `evalRPN`

that evaluates an expression in Reverse Polish Notation (RPN) using a stack data structure. Reverse Polish Notation is a way of representing mathematical expressions in which operators come after their operands. For example, instead of writing "3 + 4," you would write "3 4 +".

Here's a detailed explanation of the code:

`func evalRPN(tokens []string) int`

: This function takes a slice of strings called`tokens`

as input and returns an integer as the result of evaluating the RPN expression.`var stack []int`

: This line declares an empty integer stack, which will be used to store the operands as the expression is evaluated.The code then iterates through each token in the

`tokens`

slice using a`for`

loop:`for _, val := range tokens {`

: This loop iterates over each element (`val`

) in the`tokens`

slice, where`_`

is used to ignore the index.

Inside the loop, there is a

`switch`

statement that examines the current token`val`

:`case "+":`

: If the token is "+" (addition), it pops the top two values from the stack, adds them, and pushes the result back onto the stack.`case "-":`

: If the token is "-" (subtraction), it pops the top two values from the stack, subtracts the first popped value from the second popped value, and pushes the result back onto the stack.`case "*":`

: If the token is "*" (multiplication), it pops the top two values from the stack, multiplies them, and pushes the result back onto the stack.`case "/":`

: If the token is "/" (division), it pops the top two values from the stack, divides the second popped value by the first popped value, and pushes the result back onto the stack.`default:`

: If the token is not an operator (i.e., it's a number), it converts the token to an integer using`strconv.Atoi(val)`

and pushes it onto the stack.

The loop continues until all tokens have been processed.

Finally, after processing all tokens, the result of the expression is stored at the top of the stack, and it is returned as

`stack[0]`

. The stack should contain only one element, which is the final result of the RPN expression.The

`pop`

function is defined to remove and return the top element from the stack. It takes a pointer to the stack (`stack *[]int`

) and performs the necessary operations to pop the top element. It's used within the`evalRPN`

function to manage the stack.

In summary, this code evaluates an RPN expression by processing each token one by one, using a stack to keep track of operands and intermediate results. It supports basic arithmetic operations like addition, subtraction, multiplication, and division.

]]>Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

Implement the `MinStack`

class:

`MinStack()`

initializes the stack object.`void push(int val)`

pushes the element`val`

onto the stack.`void pop()`

removes the element on the top of the stack.`int top()`

gets the top element of the stack.`int getMin()`

retrieves the minimum element in the stack.

You must implement a solution with `O(1)`

time complexity for each function.

**Example 1:**

`Input["MinStack","push","push","push","getMin","pop","top","getMin"][[],[-2],[0],[-3],[],[],[],[]]Output[null,null,null,null,-3,null,0,-2]ExplanationMinStack minStack = new MinStack();minStack.push(-2);minStack.push(0);minStack.push(-3);minStack.getMin(); // return -3minStack.pop();minStack.top(); // return 0minStack.getMin(); // return -2`

**Constraints:**

`-2<sup>31</sup> <= val <= 2<sup>31</sup> - 1`

Methods

`pop`

,`top`

and`getMin`

operations will always be called on**non-empty**stacks.At most

`3 * 10<sup>4</sup>`

Calls will be made to`push`

,`pop`

,`top`

, and`getMin`

.

`type MinStack struct { stack []int minStack []int}func Constructor() MinStack { return MinStack{}}func (this *MinStack) Push(val int) { this.stack = append(this.stack, val) if len(this.minStack) == 0 || val <= this.minStack[len(this.minStack)-1] { this.minStack = append(this.minStack, val) }}func (this *MinStack) Pop() { if this.isEmpty() { return } popped := this.stack[len(this.stack)-1] this.stack = this.stack[:len(this.stack)-1] if popped == this.minStack[len(this.minStack)-1] { this.minStack = this.minStack[:len(this.minStack)-1] }}func (this *MinStack) Top() int { if this.isEmpty() { return 0 } return this.stack[len(this.stack)-1]}func (this *MinStack) GetMin() int { if this.isEmpty() { return 0 } return this.minStack[len(this.minStack)-1]}func (this *MinStack) Len() int { return len(this.stack)}func (this *MinStack) isEmpty() bool { return this.Len() == 0}`

This code defines a data structure called `MinStack`

is designed to support efficient retrieval of the minimum element in a stack of integers. It uses two stacks, `stack`

and `minStack`

, to achieve this functionality.

Here's a detailed explanation of each part of the code:

`type MinStack struct`

: This line defines a struct named`MinStack`

, which contains two fields:`stack`

: This field is a slice (dynamic array) that will store the elements of the main stack.`minStack`

: This field is another slice that will store the minimum elements encountered so far in the main stack.

`func Constructor() MinStack`

: This is a constructor function that initializes and returns an instance of the`MinStack`

struct. It simply returns an empty`MinStack`

by calling`MinStack{}`

.`func (this *MinStack) Push(val int)`

: This method is used to push an integer value`val`

onto the main stack. It does the following:Appends

`val`

to the`stack`

.Checks if

`minStack`

is empty or if`val`

is less than or equal to the current minimum value stored in`minStack`

. If true, it also appends`val`

to`minStack`

.

`func (this *MinStack) Pop()`

: This method is used to pop the top element from the main stack. It does the following:Check if the main stack is empty (using the

`isEmpty()`

helper method) and returns if it is.Pops the top element from the

`stack`

.Check if the popped element is equal to the current minimum value stored in

`minStack`

. If true, it also pops the top element from`minStack`

.

`func (this *MinStack) Top() int`

: This method returns the top element of the main stack without removing it. It checks if the stack is empty and returns 0 if it is.`func (this *MinStack) GetMin() int`

: This method returns the current minimum element in the main stack (the top element of`minStack`

). It also checks if the stack is empty and returns 0 if it is.`func (this *MinStack) Len() int`

: This method returns the length (number of elements) of the main stack.`func (this *MinStack) isEmpty() bool`

: This is a helper method that checks if the main stack is empty by comparing its length to 0. It returns`true`

if the stack is empty,`false`

otherwise.

In summary, this code defines a specialized stack data structure that maintains two stacks: one for the main stack and another for keeping track of the minimum element. This allows for efficient retrieval of the minimum element in constant time while supporting standard stack operations like push and pop.

]]>Given a string `s`

containing just the characters `'('`

, `')'`

, `'{'`

, `'}'`

, `'['`

and `']'`

, determine if the input string is valid.

An input string is valid if:

Open brackets must be closed by the same type of brackets.

Open brackets must be closed in the correct order.

Every close bracket has a corresponding open bracket of the same type.

**Example 1:**

`Input: s = "()"Output: true`

**Example 2:**

`Input: s = "()[]{}"Output: true`

**Example 3:**

`Input: s = "(]"Output: false`

**Constraints:**

`1 <= s.length <= 10<sup>4</sup>`

`s`

consists of parentheses only`'()[]{}'`

.

`func isValid(s string) bool { stack := make([]byte,0) pairs := map[byte]byte{ '}' : '{', ']' : '[', ')' : '(', } for _, char := range []byte(s){ pair, ok := pairs[char] if !ok { stack = append(stack, char) continue } if len(stack) == 0 { return false } if stack[len(stack)-1] != pair{ return false } stack = stack [:len(stack)-1] } return len(stack) == 0}`

This code defines a function `isValid(s string) bool`

that checks if a given input string `s`

contains valid pairs of parentheses, square brackets, and curly braces. It utilizes a stack data structure to perform this validation. Here's a detailed explanation of the code:

`stack := make([]byte, 0)`

: This line initializes an empty byte slice called`stack`

, which will be used as a stack to keep track of the opening parentheses, square brackets, and curly braces encountered in the input string.`pairs := map[byte]byte{...}`

: This line creates a map called`pairs`

that defines the valid pairs of closing and opening characters. For example, '}' is paired with '{', ']' is paired with '[', and ')' is paired with '('. This map is used to check if the closing character corresponds to the most recent opening character in the stack.The code then iterates through each character

`char`

in the input string`s`

using a`for`

loop.`pair, ok := pairs[char]`

: This line attempts to look up the corresponding opening character for the current`char`

in the`pairs`

map. If`char`

is not one of the closing characters ('}', ']', or ')'),`ok`

will be`false`

, and`pair`

will be the zero value for`byte`

.`if !ok { ... }`

: If`char`

is not a closing character, it means it's an opening character ('{', '[', or '('), so it's appended to the`stack`

. This step is necessary to keep track of the opening characters until their corresponding closing character is encountered.`if len(stack) == 0 { return false }`

: If`char`

is a closing character and the`stack`

is empty, it means there's no corresponding opening character in the stack. In this case, the function returns`false`

immediately because the string is not valid.`if stack[len(stack)-1] != pair { return false }`

: If`char`

is a closing character and the`stack`

is not empty, this line compares the most recent opening character in the stack (i.e.,`stack[len(stack)-1]`

) with the expected opening character (`pair`

) based on the`pairs`

map. If they don't match, the function returns`false`

because the string is not valid.`stack = stack[:len(stack)-1]`

: If the closing character matches the expected opening character, the opening character is removed from the stack, indicating that the pair has been successfully closed.After processing all characters in the input string, the function checks if the

`stack`

is empty. If it is, it means all opening characters have been successfully closed, and the function returns`true`

. Otherwise, if there are unmatched opening characters left in the stack, it returns`false`

.

In summary, this code uses a stack to keep track of opening characters and checks if each closing character matches the most recent opening character encountered. If at the end of processing the input string the stack is empty, it returns `true`

, indicating that the input string contains valid pairs of parentheses, square brackets, and curly braces; otherwise, it returns `false`

.

You are given an array of integers `nums`

, there is a sliding window of size `k`

which is moving from the very left of the array to the very right. You can only see the `k`

numbers in the window. Each time the sliding window moves right by one position.

Return *the max sliding window*.

**Example 1:**

`Input: nums = [1,3,-1,-3,5,3,6,7], k = 3Output: [3,3,5,5,6,7]Explanation: Window position Max--------------- -----[1 3 -1] -3 5 3 6 7 3 1 [3 -1 -3] 5 3 6 7 3 1 3 [-1 -3 5] 3 6 7 5 1 3 -1 [-3 5 3] 6 7 5 1 3 -1 -3 [5 3 6] 7 6 1 3 -1 -3 5 [3 6 7] 7`

**Example 2:**

`Input: nums = [1], k = 1Output: [1]`

**Constraints:**

`1 <= nums.length <= 10<sup>5</sup>`

`-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup>`

`1 <= k <= nums.length`

`func maxSlidingWindow(nums []int, k int) []int { output := []int{} q := make([]int, 0) l, r := 0, 0 for r < len(nums) { for len(q) != 0 && nums[q[len(q)-1]] < nums[r] { q = q[:len(q)-1] } q = append(q, r) if l > q[0] { q = q[1:] } if (r + 1) >= k { output = append(output, nums[q[0]]) l++ } r++ } return output}`

This code defines a function `maxSlidingWindow`

that takes two parameters: a slice of integers `nums`

and an integer `k`

. The goal of this function is to find the maximum element in a sliding window of size `k`

as it moves from left to right through the `nums`

slice and return the maximum values in a new slice.

Here's a step-by-step explanation of how the code works:

Initialize

`output`

as an empty slice of integers to store the maximum values found in the sliding window.Create an empty slice

`q`

to act as a deque (double-ended queue) for storing indices of elements in the`nums`

slice.Initialize two pointers

`l`

and`r`

to 0.`l`

represents the left end of the sliding window, and`r`

represents the right end of the sliding window.Start a

`for`

loop that continues until the right pointer`r`

reaches the end of the`nums`

slice.Inside the loop, there is another

`for`

loop that runs while the deque`q`

is not empty and the element at the last index in`q`

(i.e.,`nums[q[len(q)-1]]`

) is less than the current element`nums[r]`

. This inner loop removes elements from the back of the deque`q`

until the condition is met, effectively maintaining a deque of decreasing elements.After the inner loop, the current index

`r`

is appended to the deque`q`

. This is done because it is possible that the maximum element for the current window might be the element at index`r`

.Check if the left pointer

`l`

is greater than the index stored at the front of the deque`q`

. If it is, this means that the front element in`q`

is outside the current window, so we remove it from the front of the deque.Check if the size of the current sliding window (i.e.,

`r+1`

) is greater than or equal to`k`

. If it is, this means that the window has reached the desired size of`k`

, and we can add the maximum element in the window (which is`nums[q[0]]`

, where`q[0]`

stores the index of the maximum element) to the`output`

slice. Then, increment the left pointer`l`

.Increment the right pointer

`r`

to move the sliding window one step to the right.Repeat steps 4 to 9 until the right pointer

`r`

reaches the end of the`nums`

slice.Finally, return the

`output`

slice, which contains the maximum values for each sliding window of size`k`

.

In summary, this code efficiently finds the maximum values in a sliding window of size `k`

as it moves through the input slice `nums`

using a deque data structure to optimize the process.

Given two strings `s`

and `t`

of lengths `m`

and `n`

respectively, return *the* *minimum window*

*substring*

*of* `s`

*such that every character in*`t`

*(including duplicates) is included in the window*. If there is no such substring, return the

`""`

.The test cases will be generated such that the answer is **unique**.

**Example 1:**

`Input: s = "ADOBECODEBANC", t = "ABC"Output: "BANC"Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.`

**Example 2:**

`Input: s = "a", t = "a"Output: "a"Explanation: The entire string s is the minimum window.`

**Example 3:**

`Input: s = "a", t = "aa"Output: ""Explanation: Both 'a's from t must be included in the window.Since the largest window of s only has one 'a', return empty string.`

**Constraints:**

`m == s.length`

`n == t.length`

`1 <= m, n <= 10<sup>5</sup>`

`s`

and`t`

consists of uppercase and lowercase English letters.

**Follow-up:** Could you find an algorithm that runs in `O(m + n)`

time?

`func minWindow(s string, t string) string { start, end := 0, 0 targetCharacterFrequency := make(map[uint8]int) currentCharacterFrequency := make(map[uint8]int) distinctCharacterCount := 0 minSubstring := "" for index := range t { targetCharacterFrequency[t[index]]++ } for end < len(s) { currentCharacterFrequency[s[end]]++ if targetCharacterFrequency[s[end]] != 0 && targetCharacterFrequency[s[end]] == currentCharacterFrequency[s[end]] { distinctCharacterCount++ } for distinctCharacterCount == len(targetCharacterFrequency) { if minSubstring == "" { minSubstring = s[start:end+1] } if end - start + 1 < len(minSubstring) { minSubstring = s[start:end+1] } currentCharacterFrequency[s[start]]-- if currentCharacterFrequency[s[start]] < targetCharacterFrequency[s[start]] { distinctCharacterCount-- } start++ } end++ } return minSubstring}`

This code is an implementation of the sliding window technique to find the minimum window in string 's' which contains all characters from string 't'. The idea is to maintain two pointers, 'start' and 'end', and slide the window to find the smallest substring in 's' that contains all the characters from 't'.

Here's a detailed breakdown of the code:

Initialize variables:

`start`

and`end`

are pointers to maintain the current window.`targetCharacterFrequency`

is a map to store the frequency of characters in string 't'.`currentCharacterFrequency`

is a map to store the frequency of characters in the current window.`distinctCharacterCount`

keeps track of the number of distinct characters in the current window.`minSubstring`

is initially an empty string and will be updated with the minimum substring containing all characters from 't'.

Loop through string 't' to populate

`targetCharacterFrequency`

with character frequencies.Start a while loop with the 'end' pointer moving through the string 's':

Increment the frequency of the character at 'end' in

`currentCharacterFrequency`

.Check if the character at 'end' is one of the target characters (from 't') and if its frequency in the current window matches the target frequency. If it does, increment

`distinctCharacterCount`

.

Inside the while loop, another nested while loop:

Check if

`distinctCharacterCount`

is equal to the total number of distinct characters in 't'. If it is, it means the current window contains all characters from 't'.Update

`minSubstring`

if it's empty or if the current window is smaller than the previously found minimum.Decrement the frequency of the character at 'start' in

`currentCharacterFrequency`

.If the frequency of the character at 'start' becomes less than the target frequency, decrement

`distinctCharacterCount`

.Move the 'start' pointer to the right to try to find a smaller window containing all characters.

Move the 'end' pointer to the right to expand the window and continue the process.

Once the while loop finishes, return

`minSubstring`

, which will be the minimum window containing all characters from 't'.

In summary, this code efficiently finds the minimum window in string 's' that contains all characters from string 't' by maintaining a sliding window and tracking character frequencies. It's a common algorithmic technique used for substring search problems.

]]>Given two strings `s1`

and `s2`

, return `true`

*if* `s2`

*contains a permutation of* `s1`

*, or* `false`

*otherwise*.

In other words, return `true`

if one of `s1`

's permutations is the substring of `s2`

.

**Example 1:**

`Input: s1 = "ab", s2 = "eidbaooo"Output: trueExplanation: s2 contains one permutation of s1 ("ba").`

**Example 2:**

`Input: s1 = "ab", s2 = "eidboaoo"Output: false`

**Constraints:**

`1 <= s1.length, s2.length <= 10<sup>4</sup>`

`s1`

and`s2`

consists of lowercase English letters.

`func checkInclusion(s1 string, s2 string) bool { if len(s1) > len(s2) { return false } s1Count, s2Count := [26]int{}, [26]int{} for i, _ := range s1 { s1Count[s1[i]-'a']++ s2Count[s2[i]-'a']++ } matches := 0 for i := 0; i < 26; i++ { if s1Count[i] == s2Count[i] { matches += 1 } else { matches += 0 } } l := 0 for r := len(s1); r < len(s2); r++ { if matches == 26 { return true } index := s2[r] - 'a' s2Count[index]++ if s1Count[index] == s2Count[index] { matches++ } else if s1Count[index]+1 == s2Count[index] { matches-- } index = s2[l] - 'a' s2Count[index]-- if s1Count[index] == s2Count[index] { matches++ } else if s1Count[index]-1 == s2Count[index] { matches-- } l++ } return matches == 26}`

This code defines a Go function called `checkInclusion`

that checks if one string, `s1`

, is a permutation of any substring in another string, `s2`

. In other words, it determines if all the characters in `s1`

can be found in any order within `s2`

. Here's a detailed explanation of how the code works:

The function starts by comparing the lengths of

`s1`

and`s2`

. If`s1`

is longer than`s2`

, it immediately returns`false`

, because it's not possible for`s1`

to be a permutation of a longer string.Two arrays,

`s1Count`

and`s2Count`

, of size 26 (representing the lowercase English alphabet) are initialized to store the frequency of characters in`s1`

and`s2`

, respectively.The code iterates through the characters in

`s1`

and increments the corresponding index in`s1Count`

and`s2Count`

arrays based on the ASCII value difference between the character and`'a'`

.The variable

`matches`

is initialized to track the number of characters that have matching frequencies between`s1`

and the current substring of`s2`

.A loop from 0 to 25 iterates through each index representing a character in the arrays. If the character frequency in

`s1`

and`s2`

match,`matches`

is incremented.The code then enters a sliding window approach to compare the substrings of

`s2`

with the length of`s1`

. The`l`

variable represents the left boundary of the sliding window, and the`r`

variable represents the right boundary.At each step, the code checks if

`matches`

is equal to 26 (representing all characters of the alphabet). If yes, it means`s1`

is a permutation of the current substring of`s2`

, and it immediately returns`true`

.The character at index

`r`

of`s2`

is considered. Its frequency in`s2Count`

is incremented. If this increment makes the frequency match that in`s1Count`

,`matches`

is incremented. If the frequency in`s2Count`

exceeds`s1Count`

by one,`matches`

is decremented.Similarly, the character at index

`l`

of the previous substring is considered. Its frequency in`s2Count`

is decremented. If this decrement makes the frequency match that in`s1Count`

,`matches`

is incremented. If the frequency in`s2Count`

becomes one less than`s1Count`

,`matches`

is decremented.The

`l`

index is incremented to slide the window one step to the right.The loop continues until the end of

`s2`

is reached. Finally, the function returns`true`

if`matches`

equals 26, indicating that all characters of`s1`

are permuted within some substring of`s2`

.

This code essentially uses the sliding window technique along with character frequency tracking to efficiently check for the permutation condition between two strings.

]]>You are given a string `s`

and an integer `k`

. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k`

times.

Return *the length of the longest substring containing the same letter you can get after performing the above operations*.

**Example 1:**

`Input: s = "ABAB", k = 2Output: 4Explanation: Replace the two 'A's with two 'B's or vice versa.`

**Example 2:**

`Input: s = "AABABBA", k = 1Output: 4Explanation: Replace the one 'A' in the middle with 'B' and form "AABBBBA".The substring "BBBB" has the longest repeating letters, which is 4.There may exists other ways to achive this answer too.`

**Constraints:**

`1 <= s.length <= 10<sup>5</sup>`

`s`

consists of only uppercase English letters.`0 <= k <= s.length`

`func characterReplacement(s string, k int) int { count := make(map[byte]int) maxF := 0 left := 0 result := 0 for right, _ := range s { count[s[right]] = 1 + count[s[right]] maxF = max(maxF, count[s[right]]) if (right-left+1)-maxF > k { count[s[left]] -= 1 left++ } result = max(result, right-left+1) } return result}func max(a, b int) int { if a > b { return a } return b}`

This code defines a function called `characterReplacement`

which takes two parameters: a string `s`

and an integer `k`

. The goal of the function is to find the length of the longest substring in which you can replace at most `k`

characters to make all characters in the substring the same. Let's break down the code step by step:

`count := make(map[byte]int)`

: This line initializes a map called`count`

that will store the frequency of each character in the current substring.`maxF := 0`

:`maxF`

keeps track of the maximum frequency of any character in the current substring.`left := 0`

:`left`

is the left pointer of the sliding window that will be used to traverse the string.`result := 0`

:`result`

will store the length of the longest valid substring found.The code enters a loop that goes through each character in the input string

`s`

.a.

`count[s[right]] = 1 + count[s[right]]`

: This line increments the count of the character at the current`right`

index in the`count`

map.b.

`maxF = max(maxF, count[s[right]])`

: Update`maxF`

with the maximum frequency seen so far.c. The code checks if the number of characters that need to be replaced in the current substring

`(right - left + 1) - maxF`

exceeds the given limit`k`

.d. If the limit is exceeded, it means the substring needs more replacements than allowed. So, we move the

`left`

pointer one step to the right and decrease the frequency count of the character at that position.e.

`result = max(result, right - left + 1)`

: Update`result`

with the maximum length of valid substrings encountered so far.After the loop, the function returns the calculated

`result`

, which represents the length of the longest substring with at most`k`

replacements allowed to make all characters the same.There is also a helper function

`max(a, b int) int`

that returns the maximum of two integers.

In summary, this code uses a sliding window approach to find the length of the longest substring where you can replace at most `k`

characters to make all characters the same. The `count`

map keeps track of character frequencies within the current window, and the `maxF`

variable tracks the most frequent character. By adjusting the window using the `left`

pointer, the code keeps track of the maximum valid substring length encountered.

Given a string `s`

, find the length of the **longest substring** without repeating characters.

**Example 1:**

`Input: s = "abcabcbb"Output: 3Explanation: The answer is "abc", with the length of 3.`

**Example 2:**

`Input: s = "bbbbb"Output: 1Explanation: The answer is "b", with the length of 1.`

**Example 3:**

`Input: s = "pwwkew"Output: 3Explanation: The answer is "wke", with the length of 3.Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.`

**Constraints:**

`0 <= s.length <= 5 * 10<sup>4</sup>`

`s`

consists of English letters, digits, symbols and spaces.

`func lengthOfLongestSubstring(s string) int { charSet := make(map[byte]bool) l := 0 res := 0 for r, _ := range s { for charSet[s[r]] { delete(charSet,s[l]) l++ } charSet[s[r]] = true res = max(res, r-l+1) } return res}func max(a,b int) int { if a > b{ return a } return b}`

This code defines a function `lengthOfLongestSubstring`

that takes a string `s`

as input and calculates the length of the longest substring within that string, where all characters in the substring are unique. Let's break down the code step by step:

`charSet := make(map[byte]bool)`

: This line initializes an empty map called`charSet`

, which will be used to keep track of characters encountered in the current substring.`l := 0`

: This initializes a variable`l`

(short for "left") to 0. It represents the left pointer of the current substring.`res := 0`

: This initializes a variable`res`

(short for "result") to 0. It will store the length of the longest substring found.The first

`for`

loop iterates over the characters in the input string`s`

. It uses two variables,`r`

and`_`

, where`r`

represents the current position (right pointer) in the string.Inside the first loop, there's another loop:

`for charSet[s[r]]`

. This loop checks whether the current character`s[r]`

is already present in the`charSet`

map. If it is, that means the current character is a repeating character, so the algorithm needs to shrink the substring by moving the left pointer (`l`

) to the right until the repeated character is removed from the substring.Inside the inner loop,

`delete(charSet, s[l])`

removes the character at the`l`

position from the`charSet`

map, effectively shifting the left pointer to the right.After the inner loop,

`charSet[s[r]] = true`

adds the current character to the`charSet`

map, indicating its presence in the current substring.`res = max(res, r-l+1)`

: This calculates the length of the current substring (from`l`

to`r`

) and compares it to the current maximum length stored in`res`

. Whichever is greater becomes the new value of`res`

.The function

`max(a, b int)`

is defined to return the maximum of two integers`a`

and`b`

.Finally, the function returns the value of

`res`

, which represents the length of the longest substring with unique characters.

In summary, this code uses a sliding window approach to find the longest substring without repeating characters within the given input string `s`

. The `charSet`

map keeps track of characters in the current substring, and the left and right pointers (`l`

and `r`

) control the window boundaries. The result is stored in the `res`

variable and returned at the end.

To install SQLite, you generally don't need to perform a separate installation step because SQLite is often included with many programming languages and platforms. Here are instructions for installing SQLite on various platforms:

**Linux/macOS**: Most Linux distributions and macOS come with SQLite pre-installed. You can access it from the command line using the`sqlite3`

command. Open a terminal and type`sqlite3`

to start the SQLite shell.**Windows**: Windows doesn't have SQLite pre-installed, but you can easily download the command-line shell for SQLite from the official SQLite website:Visit the SQLite download page:

**https://www.sqlite.org/download.html**Scroll down to the "Precompiled Binaries for Windows" section.

Download the "sqlite-tools-win32-x86-*.zip" file.

Extract the contents of the downloaded ZIP file to a directory of your choice.

Open a Command Prompt and navigate to the directory where you extracted the files.

Run

`sqlite3`

to start the SQLite shell.

**Programming Languages and Frameworks**: If you're planning to use SQLite with a specific programming language or framework (like Go, Python, Node.js, etc.), you might not need to install SQLite separately. Many programming languages provide libraries or packages that include SQLite bindings. For example, in Go, you can use the "github.com/mattn/go-sqlite3" package to work with SQLite databases.

Here's how you can execute some basic SQL queries using the SQLite3 command-line tool:

You can open an existing SQLite database or create a new one using the following command:

`$ sqlite3 mydatabase.db`

To create a new table, you can use the `CREATE TABLE`

statement. For example, to create a simple "users" table:

`CREATE TABLE users ( id INTEGER PRIMARY KEY, username TEXT, email TEXT);`

You can use the `INSERT INTO`

statement to add data to your table:

`INSERT INTO users (username, email) VALUES ('john_doe', 'john@example.com');`

To retrieve data from a table, you can use the `SELECT`

statement:

`SELECT * FROM users;`

SELECT * FROM users WHERE username = 'john_doe';

`SELECT * FROM users WHERE username = 'john_doe';`

Use the `UPDATE`

statement to modify existing records:

`UPDATE users SET email = 'new_email@example.com' WHERE id = 1;`

To delete records, use the `DELETE FROM`

statement:

`DELETE FROM users WHERE id = 1;`

After you're done working with the database, you can close it by typing `.exit`

or press Ctrl+D.

Remember that these are just basic examples. SQL supports a wide range of operations, including joining tables, grouping data, using aggregate functions, and more. Make sure to refer to the SQLite documentation or other SQL resources for more advanced queries and concepts.

If you're working with SQLite in a programming language, you'll typically use libraries that provide an interface to SQLite (e.g., SQLite3 library in Python) to execute SQL queries programmatically.

]]>