Remove Element [Easy] — Read/Write Two Pointers

Problem Statement

Remove all occurrences of val from array in-place. Return the number of elements not equal to val. The relative order of other elements may change.

Input: nums=[3,2,2,3], val=3 → Output: 2, nums=[2,2,_,_]

Intuition

Read pointer scans entire array; write pointer only advances when the current element should be kept. After the scan, write pointer = new length.

Solutions

C

int removeElement(int* nums, int n, int val) {
    int w=0;
    for (int r=0;r<n;r++) if(nums[r]!=val) nums[w++]=nums[r];
    return w;
}

C++

int removeElement(vector<int>& nums, int val) {
    int w=0;
    for (int x:nums) if(x!=val) nums[w++]=x;
    return w;
}

Java

public int removeElement(int[] nums, int val) {
    int w=0;
    for (int x:nums) if(x!=val) nums[w++]=x;
    return w;
}

JavaScript

var removeElement = function(nums, val) {
    let w=0;
    for (const x of nums) if(x!==val) nums[w++]=x;
    return w;
};

Python

def removeElement(nums: list[int], val: int) -> int:
    w = 0
    for x in nums:
        if x != val:
            nums[w] = x
            w += 1
    return w

Complexity

Time: O(n)
Space: O(1)