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# Remove Element

Two Pointer

## Problem

Given an array *nums* and a value *val*, remove all instances of that value [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) and return the new length.

{% hint style="info" %}
For example:

```
Input: nums = [0,1,2,2,3,0,4,2], val = 2
Output: 5, nums = [0,1,4,0,3]
```

```
Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2]
```

{% endhint %}

### Thought Process

* We'll use two pointer approach: one pointer to iterate through the array and the other pointer to keep track of the position of the elements that aren't key. We will start from index 0.

## Solution

```
class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        i = 0
        j = 0
        
        while i < len(nums):
            if nums[i] != val:
                nums[j] = nums[i]
                j+=1
            i+=1
        return j
```