Remove Element — DSA · Engineering Playbook

Problem#

Given an integer array nums and an integer val, remove all occurrences of val in place. Return the new length k; the first k elements of nums must hold the surviving values.

Examples#

nums = [3,2,2,3], val = 3 → k = 2, nums = [2,2,_,_]
nums = [0,1,2,2,3,0,4,2], val = 2 → k = 5, nums = [0,1,3,0,4,_,_,_]

Constraints#

0 <= nums.length <= 100.

Approach#

Write-pointer w lags behind the read-pointer i. Copy nums[i] into nums[w] whenever it’s not the target value, incrementing w. Return w.

Solution#

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

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

from typing import List

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

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

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

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

Editorial#

The write-pointer pattern handles all in-place filtering: copy when the predicate holds, otherwise drop. No swapping needed because the order of survivors is preserved and we only ever read positions ≥ w.

Complexity#

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

Concept revision#

Remove Duplicates from Sorted Array

Problem#

Examples#

Constraints#

Approach#

Solution#

Editorial#

Complexity#

Concept revision#

Related#