LeetCode: Remove element




Given an array nums and a value val, remove all instances of that value in-place and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn’t matter what you leave beyond the new length.

Example 1:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

It doesn't matter what you leave beyond the returned length.


Almost same with previous challenge LeetCode: Remove Duplicates from Sorted Array.

Iterate all the elements of input array, with a extra cnt to track the length of result array.

Time complexity: \(O(N)\)

class Solution {
  int removeElement(vector<int>& nums, int val) {
    if(nums.size() == 0) return 0;
    size_t cnt = 0;
    for(size_t i=0; i<nums.size(); i++) {
      if(nums[i] != val)
        nums[cnt++] = nums[i];
    return cnt;

If we use the std::remove, the code will be much shorter.

int removeElement(vector<int>& nums, int val) {
    return std::distance(nums.begin(), std::remove(nums.begin(), nums.end(), val));

Python solution

class Solution:
    def removeElement(self, nums, val):
        count = 0
        for i in range(len(nums)):
            if nums[i] != val :
                nums[count] = nums[i]
                count +=1
        return count

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