# Count Square Submatrices with All Ones

Given a `m * n` matrix of ones and zeros, return how many square submatrices have all ones.

Example 1:

```Input: matrix =
[
[0,1,1,1],
[1,1,1,1],
[0,1,1,1]
]
Output: 15
Explanation:
There are 10 squares of side 1.
There are 4 squares of side 2.
There is  1 square of side 3.
Total number of squares = 10 + 4 + 1 = 15.
```

Example 2:

```Input: matrix =
[
[1,0,1],
[1,1,0],
[1,1,0]
]
Output: 7
Explanation:
There are 6 squares of side 1.
There is 1 square of side 2.
Total number of squares = 6 + 1 = 7.
```

Constraints:

• `1 <= arr.length <= 300`
• `1 <= arr[0].length <= 300`
• `0 <= arr[i][j] <= 1`

## Solution

### C# Program with DP Approach

In this program, instead of creating of new 2d array, I solved the problem in-place of the array.

```using System;

namespace LeetcodeProgram
{
class Program
{
//Driver
static void Main(string[] args)
{
int[][] arr = new int[2][];

int[,] matrix = new int[3, 3] { { 1, 0, 1 }, { 1, 1, 0 }, { 1, 1, 0 } };
/*

[[1,0,1],
[1,1,0],
[1,1,0]]

*/
int result = CountSquares(matrix);
Console.WriteLine("CountSquares is {0}", result);
}

public static int CountSquares(int[,] matrix)
{
int row = matrix.GetLength(0);
int col = matrix.GetLength(1);
int sum = 0;
for(int i=1;i<row;i++)
{
for(int j=1;j<col;j++)
{
if(matrix[i,j] != 0 && matrix[i-1,j-1] != 0 && matrix[i-1,j] != 0 && matrix[i,j-1] != 0)
{
matrix[i, j] = Math.Min(Math.Min(matrix[i - 1, j], matrix[i, j - 1]), matrix[i - 1, j - 1]) + 1;
}
sum += matrix[i, j];
}
}

//First Row
for(int i = 0; i < col; i++)
{
if (matrix[0, i] != 0)
sum += matrix[0, i];
}

//First Column
for(int i=1;i<row;i++)
{
if (matrix[i, 0] != 0)
sum += matrix[i, 0];
}
return sum;
}
}
}

```

Time Complexity; O(m x n)

Space Complexity: O(1) Since, I never created any new arrays.