### Problem Statement :

Given a list of integers nums, put all the zeros to the back of the list by modifying the list in-place. The relative ordering of other elements should stay the same.

Can you do it in \mathcal{O}(1)O(1) additional space?

Constraints

0 ≤ n ≤ 100,000 where n is the length of nums

Example 1

Input
nums = [0, 1, 0, 2, 3]

Output
[1, 2, 3, 0, 0]

Explanation
Note that [1, 2, 3] appear in the same order as in the input.

### Solution :

                        Solution in C++ :

vector<int> solve(vector<int>& nums) {
int count = 0;

for (int i = 0; i < nums.size(); ++i) {
if (nums[i] != 0) {
swap(nums[i], nums[count]);
count++;
}
}


                        Solution in Java :

class Solution {
public int[] solve(int[] nums) {
int i = 0, j = 0;
while (j < nums.length) {
if (nums[j] == 0)
j++;
else
swap(nums, i++, j++);
}
return nums;
}
public void swap(int[] a, int i, int j) {
if (i == j)
return;
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
}


                        Solution in Python :

class Solution:
def solve(self, nums):
j = 0
for i in range(len(nums)):
if nums[i] != 0:
if j != i:
nums[j], nums[i] = nums[i], nums[j]
j += 1

return nums


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