remove_element.md

Demo

Check My Demo Here

Intuition

Traverse nums, if we encounter an element that equals val we skip it but keep a flag on its index for later.

Approach

Using a flag variable n to keep track of the last index in which the numbers from 0 up to n-1 are not equal to val:

• Whenever we encounter an element that is equal to val, we skip it and move forward.
• else (element not equal to val), then we store it at n and move n forward.
• By the end all non val elements will be at nums[0:n-1]

Complexity

• Time complexity: O(k): where k is the size of nums.

• Space complexity: O(1): No extra space is needed.

Code

class Solution {
    public int removeElement(int[] nums, int val) {
        int n = 0;
        for(int i = 0 ; i < nums.length; i++)
            if(nums[i]!=val)
                nums[n++] = nums[i];
        return n;
    }
}