### Problem

In a provided vector `vec`

we need to move all `0`

elements to the end of the vector. A relative order of other elements should not change.

#### Example 1:

*Input:* `vec = [5, 0, 0, 2, 3, 0, 4, 0, 1]`

*Output:* `[5, 2, 3, 4, 1, 0, 0, 0, 0]`

#### Example 2:

*Input:* `vec = [0, 0, 8, 6, 0, 0]`

*Output:* `[8, 6, 0, 0, 0, 0]`

#### Example 3:

*Input:* `vec = [1, 2, 3]`

*Output:* `[1, 2, 3]`

### Solution

```
fn move_zeros_to_end(vec: &mut Vec<i32>) -> &Vec<i32> {
let mut zero_count = 0;
(0..vec.len()).for_each(|index| {
if vec[index] == 0 {
zero_count += 1;
} else if zero_count > 0 {
vec[index - zero_count] = vec[index];
vec[index] = 0;
}
});
return vec;
}
```

### Explanation

We can solve the puzzle in a single pass over the provided vector `vec`

. As we go through the vector `vec`

, we maintain `zero_count`

of `0`

elements we come across. At the same time, we shift non-zero elements to the left to `index - zero_count`

position, and fill their place with `0`

.

### Complexity

The time complexity of this solution is `O(n)`

(`n`

is a size of the input vector).