## Problem Statement

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:**

Givennums=[3,2,2,3],val=3, Your function should return length =2, with the first two elements ofnumsbeing2. It doesn't matter what you leave beyond the returned length.

**Example 2:**

Givennums=[0,1,2,2,3,0,4,2],val...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.

**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 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]); }

**Solution:**

class Solution:
def removeElement(self, nums: List[int], val: int) -> int:
if not nums: return 0
i=0
j=len(nums)
while i<j:
if nums[i]==val:
nums[i] = nums[j-1]
j-=1
else:
i+=1
return j