# Leetcode: Remove Element

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

Given an array and a value, 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:

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

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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/remove-element
## Basic Ideas:
##        index point to the last element which have been processed
## Complexity: Time O(n), Space O(1)
class Solution(object):
def removeElement(self, nums, val):
"""
:type nums: List[int]
:type val: int
:rtype: int
"""
index = 0
for n in nums:
if n != val:
nums[index] = n
index += 1
return index
```

Share It, If You Like It.