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 = 3

**Output:** 2, nums = [2,2]

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

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 = 2

**Output:** 5, nums = [0,1,4,0,3]

**Explanation:** Your function should return length = **5**, with the first five elements of *nums* containing **0**, **1**, **3**, **0**, and **4**. Note that the order of those five elements can be arbitrary. It doesn't matter what values are set beyond the returned length.

`class Solution {`

public:

int removeElement(vector<int>& nums, int val) {

for(int i=0; i<nums.size(); i++){

if(nums[i] == val){

nums.erase(nums.begin()+i);

i--;

}

}

return nums.size();

}

};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Remove Element.

Memory Usage: 8.7 MB, less than 68.23% of C++ online submissions for Remove Element.