Approach: Overwrite pointer

def removeElement(nums, target):
    k = 0
    for x in nums:
        if x != target:
            nums[k] = x; k += 1
    return k

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class Solution:
    def removeElement(self, nums, target):
        k = 0
        for x in nums:
            if x != target:
                nums[k] = x
                k += 1
        return k

import java.util.*;
class Solution {
    public int removeElement(int[] nums, int target) {
        int k = 0;
        for (int x : nums) {
            if (x != target) {
                nums[k] = x;
                k++;
            }
        }
        return k;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int removeElement(vector<int>& nums, int target) {
        int k = 0;
        for (int x : nums) {
            if (x != target) {
                nums[k] = x;
                k++;
            }
        }
        return k;
    }
};

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */ var removeElement = function(nums, target) {
    let k = 0;
    for (const x of nums) {
        if (x !== target) {
            nums[k] = x;
            k++;
        }
    }
    return k;
};

Remove Element

Example 1

Example 2

Constraints