LeetCode Problem 27 - Remove Element

Problem

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.
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];
}

Solution

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int k = 0;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] != val) {
                nums[k++] = nums[i];
            }
        }
        return k;
    }
};

Explanation

This algorithm will always shift the numbers based on the counter k which holds the index for all elements without our value. This is initialized to 0. When we encounter our first non-value element, we place it at index k which is 0. By doing this, we ensure that any value within the first k elements do not include our value.