Problem - Leetcode

Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

Example 1:

Input: nums = [1,2,3,1]
Output: true

Example 2:

Input: nums = [1,2,3,4]
Output: false

Example 3:

Input: nums = [1,1,1,3,3,4,3,2,4,2]
Output: true

Constraints:

• 1 <= nums.length <= 10<sup>5</sup>

• -10<sup>9</sup> <= nums[i] <= 10<sup>9</sup>

Answer-1 Top Runtime in Golang

func containsDuplicate(nums []int) bool {
    nums_map := map[int]int{}
    for _, n := range nums {
        if _, ok := nums_map[n]; !ok {
            nums_map[n] = 1
        } else {
            return true
        }
    }
    return false
}

This code defines a function named containsDuplicate that takes a single argument nums, which is a slice of integers. The purpose of this function is to determine whether the given slice of integers contains any duplicates or not. It returns a boolean value (true if there are duplicates, and false otherwise).

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

1. The function starts by declaring an empty map named nums_map, which is used to keep track of the frequency of each integer in the input slice. The keys of the map are integers, and the values are integers representing the frequency of each integer in the slice.

2. The code then enters a loop that iterates over each element (n) in the input nums slice.

3. Inside the loop, the code checks whether the current integer n is already present in the nums_map by using the line:

    if _, ok := nums_map[n]; !ok {
   

4. Here, ok is a boolean variable that indicates whether the key n is present in the map or not. If n is not present (!ok is true), then the code sets the frequency of n to 1 in the nums_map, indicating that this integer has been encountered once.

5. If the integer n is already present in the nums_map, it means that there's a duplicate. In that case, the code immediately returns true, indicating that duplicates have been found in the input slice.

6. If no duplicate has been found during the iteration, the code returns false, indicating that no duplicates were detected in the input slice.

In essence, this code uses a map to keep track of the frequency of integers in the input slice. If any integer is encountered more than once, it means there's a duplicate, and the function returns true. Otherwise, if no duplicates are found after iterating through the entire input slice, the function returns false.

Answer-2 Top Memory in Golang

import "sort"
func containsDuplicate(nums []int) bool {
    sort.Slice(nums, func(i, j int) bool{
        return nums[i] < nums[j]
    })
    for i := 1; i < len(nums); i++ {
        if nums[i] == nums[i - 1] {
            return true
        }
    }
    return false
}

This code defines a function called containsDuplicate that takes a slice of integers ([]int) named nums as its parameter and returns a boolean value (true or false). The purpose of this function is to determine whether the given slice contains any duplicate integers.

Let's break down the code step by step:

1. The import "sort" statement at the beginning of the code indicates that the built-in sort package from Go's standard library is being imported. This package provides sorting functionalities that will be used later in the code.

2. The func containsDuplicate(nums []int) bool line defines the function containsDuplicate that takes a single parameter, nums, which is a slice of integers.

3. Inside the function, sort.Slice(nums, func(i, j int) bool {...} sorts the nums slice in ascending order using the sort.Slice function. The second argument to sort.Slice is an anonymous function (also known as a closure) that defines how the sorting should be performed. The function takes two indices i and j, and it compares the values at those indices in the nums slice. If the value at index i is less than the value at index j, the function returns true, indicating that the values should be swapped during sorting.

4. After sorting the nums slice, a loop is started using the for statement: for i := 1; i < len(nums); i++ {...}. The loop iterates through the sorted nums slice, starting from index 1 and going up to the length of the slice minus one.

5. Inside the loop, the code checks if the current element at index i is equal to the previous element at index i - 1. If they are equal, it means a duplicate has been found, and the function immediately returns true.

6. If the loop completes without finding any duplicates, the function returns false to indicate that no duplicates were found in the nums slice.

In summary, this code sorts the given slice of integers and then iterates through the sorted slice, checking for consecutive elements that are equal. If any duplicates are found, the function returns true; otherwise, it returns false. This approach leverages the sorting functionality to efficiently identify duplicate elements in the input slice.