27. Remove Element

Time: O(n)

Space: O(1)

Intuition

Use two pointers: a slow write pointer k and a fast scan pointer i. Copy nums[i] to nums[k] only when nums[i] != val. This effectively filters out all instances of val in-place.

Algorithm

1Initialise k = 0.
2For i from 0 to n-1: if nums[i] != val, set nums[k] = nums[i], k++.
3Return k.

Example Walkthrough

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

1.i=0: 3==val, skip. i=1: 2!=val -> nums[0]=2, k=1.
2.i=2: 2!=val -> nums[1]=2, k=2. i=3: 3==val, skip.

Output: k = 2, nums = [2,2,_,_]

Common Pitfalls

•The remaining elements beyond index k do not matter - only the first k elements are checked.

27.cs

// Approach: Two pointers — swap each non-target element to the write
// position and advance the write pointer.
// Time: O(n) Space: O(1)

public class Solution
{
    public int RemoveElement(int[] nums, int val)
    {
        int n = nums.Length;
        int k = 0;
        for (int i = 0; i < n; i++)
        {
            if (nums[i] != val)
            {
                int temp = nums[k];
                nums[k++] = nums[i];
                nums[i] = temp;
            }
        }

        return k;
    }
}

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