## Description

https://leetcode.com/problems/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.

**Clarification:**

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by **reference**, which means a modification to the input array will be known to the caller as well.

Internally you can think of this:

//numsis passed in by reference. (i.e., without making a copy) int len = removeElement(nums, val); // any modification tonumsin your function would be known by the caller. // using the length returned by your function, it prints the firstlenelements. for (int i = 0; i < len; i++) { print(nums[i]); }

**Example 1:**

Input:nums = [3,2,2,3], val = 3Output:2, nums = [2,2]Explanation:Your function should return length =2, with the first two elements ofnumsbeing2. It doesn't matter what you leave beyond the returned length. For example if you return 2 with nums = [2,2,3,3] or nums = [2,2,0,0], your answer will be accepted.

**Example 2:**

Input:nums = [0,1,2,2,3,0,4,2], val = 2Output:5, nums = [0,1,4,0,3]Explanation:Your function should return length =, with the first five elements of`5`

containing`nums`

,`0`

,`1`

,`3`

, and`0`

4. Note that the order of those five elements can be arbitrary. It doesn't matter what values are set beyond the returned length.

**Constraints:**

`0 <= nums.length <= 100`

`0 <= nums[i] <= 50`

`0 <= val <= 100`

## Explanation

Use two pointers to achieve in-place removing. One pointer is used to iterate the list, the other one is used to track the current first element position equals to the value. If the element is not equal to the value, swap the element with the current first element position which equals to the value.

## Python Solution

```
class Solution:
def removeElement(self, nums: List[int], val: int) -> int:
i = 0
for j, num in enumerate(nums):
if num != val:
nums[i] = nums[j]
nums[j] = nums[i]
i += 1
return i
```

- Time Complexity: O(N).
- Space Complexity: O(1).