# Set Matrix Zeroes

Given an `m x n` `matrix`, return *all elements of the* `matrix` *in spiral order*.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg)

<pre><code><strong>Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
</strong><strong>Output: [1,2,3,6,9,8,7,4,5]
</strong></code></pre>

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg)

<pre><code><strong>Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
</strong><strong>Output: [1,2,3,4,8,12,11,10,9,5,6,7]
</strong></code></pre>

&#x20;

**Constraints:**

* `m == matrix.length`
* `n == matrix[i].length`
* `1 <= m, n <= 10`
* `-100 <= matrix[i][j] <= 100`

### Solutions

**Steps**

1. **Initialize two boolean arrays `row` and `col`**: These arrays are used to keep track of which rows and columns have at least one zero. The size of `row` is the number of rows in the matrix, and the size of `col` is the number of columns.
2. **Identify rows and columns that need to be set to zero**: This is done by iterating over each element in the matrix. If an element is zero, the corresponding entries in the `row` and `col` arrays are set to `true`.
3. **Set identified rows to zero**: For each `true` entry in the `row` array, set all elements in that row of the matrix to zero.
4. **Set identified columns to zero**: Similarly, for each `true` entry in the `col` array, set all elements in that column of the matrix to zero.

```csharp
public class Solution
{
    public void SetZeroes(int[][] matrix)
    {
        bool[] row = new bool[matrix.Length];
        bool[] col = new bool[matrix[0].Length];

        // Identify rows and columns that need to be set to zero
        for (int i = 0; i < matrix.Length; i++) //N*M Times
        {
            for (int j = 0; j < matrix[0].Length; j++)
            {
                if (matrix[i][j] == 0)
                {
                    row[i] = true;
                    col[j] = true;
                }
            }
        }

        // Set identified rows to zero
        for (int i = 0; i < row.Length; i++)
        {
            if (row[i])
            {
                for (int j = 0; j < matrix[0].Length; j++)
                {
                    matrix[i][j] = 0;
                }
            }
        }

        // Set identified columns to zero
        for (int i = 0; i < col.Length; i++)
        {
            if (col[i])
            {
                for (int j = 0; j < matrix.Length; j++)
                {
                    matrix[j][i] = 0;
                }
            }
        }
    }

}

```

> Complexity
>
> Time complexity: O(N\*M)
>
> Space complexity: O(N+M)