Remove element

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.

Consider the number of elements in nums which are not equal to val be k, to get accepted, you need to do the following things:

Change the array nums such that the first k elements of nums contain the elements which are not equal to val. The remaining elements of nums are not important as well as the size of nums.
Return k.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int val = ...; // Value to remove
int[] expectedNums = [...]; // The expected answer with correct length.
                             // It is sorted with no values equaling val.
 
int k = removeElement(nums, val); // Calls your implementation

assert k == expectedNums.length;
sort(nums, 0, k); // Sort the first k elements of nums
for (int i = 0; i < actualLength; i++) {
  assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2,,]
Explanation: Your function should return k = 2, with the first two elements of nums being 2.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,1,2,2,3,0,4,2], val = 2
Output: 5, nums = [0,1,4,0,3,,,_]
Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
Note that the five elements can be returned in any order.
It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints:

0 <= nums.length <= 100
0 <= nums[i] <= 50
0 <= val <= 100

Solution

func removeElement(nums []int, val int) int {
    k := 0
    for i := 0; i < len(nums); i++ {
        if nums[i] != val {
            nums[k] = nums[i]
            k++
        }
    }
    return k
}

Complexity

Time complexity : O(N)
Space complexity : O(1)

Explanation

The idea is very simple: we keep only the elements that are NOT equal to val and overwrite the array from the beginning.

We use a variable k to track where the next valid element should go.

Start with k = 0
Go through the array one by one
For each element:
- If it is not equal to val, we:
  - Place it at index k
  - Increase k by 1
- If it is equal to val, we skip it

So basically:

i is reading the array
k is writing the filtered result

By the end:

The first k positions contain all valid elements
Everything after k doesn’t matter

Example (quick intuition)

For nums = [3,2,2,3], val = 3:

Skip 3
Keep 2 → place at index 0
Keep 2 → place at index 1
Skip 3

Result:

nums = [2,2,...]
k = 2

Why this works

We don’t need extra space → we reuse the same array
Order doesn’t matter, so simple overwrite is fine
k naturally becomes the count of valid elements

In short: filter in-place by copying valid elements forward