# LeetCode: Remove element



## Description

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.


## Solution

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 {
public:
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

Last Updated on