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=2, 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.

**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:

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

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