Number of square matrices with all 1s

Given an N*M matrix containing only 0s and 1s, the task is to count the number of square submatrices containing all 1s.

Examples:

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



Input: arr[][] = {{1, 0, 1},
{1, 1, 0},
{1, 1, 0}}
Output: 7

Approach: This problem can be solved using Dynamic Programming.

  1. Let the array arr[i][j] store the number of square matrices ending at (i, j)
  2. The recurrence relation to find the number of squares ending at (i, j) can be given by:
    • If arr[i][j] is 1:
      • arr[i][j] = min( min(arr[i-1][j], arr[i][j-1]), arr[i-1][j-1]) + 1
    • Else if arr[i][j] is 0:
      • arr[i][j] = 0
  3. Calculate the sum of the array which is equal to the number of square submatrices with all 1s.

Below is the implementation of the above approach:

CPP

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// C++ program to return the number of
// square submatrices with all 1s
#include <bits/stdc++.h>
using namespace std;
  
#define n 3
#define m 3
  
// Function to return the number of
// square submatrices with all 1s
int countSquareMatrices(int a[][m],
                        int N, int M)
{
    // Initialize count variable
    int count = 0;
  
    for (int i = 1; i < N; i++) {
        for (int j = 1; j < M; j++) {
            // If a[i][j] is equal to 0
            if (a[i][j] == 0)
                continue;
  
            // Calculate number of
            // square submatrices
            // ending at (i, j)
            a[i][j] = min(min(a[i - 1][j],
                              a[i][j - 1]),
                          a[i - 1][j - 1])
                      + 1;
        }
    }
  
    // Calculate the sum of the array
    for (int i = 0; i < N; i++)
        for (int j = 0; j < M; j++)
            count += a[i][j];
  
    return count;
}
  
// Driver code
int main()
{
    int arr[][m] = { { 1, 0, 1 },
                     { 1, 1, 0 },
                     { 1, 1, 0 } };
  
    cout << countSquareMatrices(arr, n, m);
  
    return 0;
}

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Java














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// Java program to return the number of  


// square submatrices with all 1s 


class GFG  


{ 


      


    final static int n = 3;  


    final static int m = 3;  


      


    // Function to return the number of  


    // square submatrices with all 1s  


    static int countSquareMatrices(int a[][], int N, int M)  


    {  


        // Initialize count variable  


        int count = 0;  


      


        for (int i = 1; i < N; i++) 


        {  


            for (int j = 1; j < M; j++)  


            {  


                // If a[i][j] is equal to 0  


                if (a[i][j] == 0)  


                    continue;  


      


                // Calculate number of  


                // square submatrices  


                // ending at (i, j)  


                a[i][j] = Math.min(Math.min(a[i - 1][j], a[i][j - 1]),  


                            a[i - 1][j - 1]) + 1;  


            }  


        }  


      


        // Calculate the sum of the array  


        for (int i = 0; i < N; i++)  


            for (int j = 0; j < M; j++)  


                count += a[i][j];  


      


        return count;  


    }  


      


    // Driver code  


    public static void main (String[] args) 


    {  


        int arr[][] = { { 1, 0, 1 },  


                        { 1, 1, 0 },  


                        { 1, 1, 0 } };  


      


        System.out.println(countSquareMatrices(arr, n, m));  


    }  


} 


  


// This code is contributed by AnkitRai01 



















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Python














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# Python3 program to return the number of 


# square submatrices with all 1s 


n = 3


m = 3


  


# Function to return the number of 


# square submatrices with all 1s 


def countSquareMatrices(a, N, M): 


      


    # Initialize count variable 


    count = 0


  


    for i in range(1, N): 


        for j in range(1, M): 


              


            # If a[i][j] is equal to 0 


            if (a[i][j] == 0): 


                continue


  


            # Calculate number of 


            # square submatrices 


            # ending at (i, j) 


            a[i][j] = min([a[i - 1][j],  


                      a[i][j - 1], a[i - 1][j - 1]])+1


  


    # Calculate the sum of the array 


    for i in range(N): 


        for j in range(M): 


            count += a[i][j] 


  


    return count 


  


# Driver code 


  


arr = [ [ 1, 0, 1], 


    [ 1, 1, 0 ], 


    [ 1, 1, 0 ] ] 


  


print(countSquareMatrices(arr, n, m)) 


  


# This code is contributed by mohit kumar 29 



















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C#














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// C# program to return the number of  


// square submatrices with all 1s 


using System; 


  


class GFG  


{ 


      


    static int n = 3;  


    static int m = 3;  


      


    // Function to return the number of  


    // square submatrices with all 1s  


    static int countSquareMatrices(int [,]a, int N, int M)  


    {  


        // Initialize count variable  


        int count = 0;  


      


        for (int i = 1; i < N; i++) 


        {  


            for (int j = 1; j < M; j++)  


            {  


                // If a[i][j] is equal to 0  


                if (a[i, j] == 0)  


                    continue;  


      


                // Calculate number of  


                // square submatrices  


                // ending at (i, j)  


                a[i, j] = Math.Min(Math.Min(a[i - 1, j], a[i, j - 1]),  


                            a[i - 1, j - 1]) + 1;  


            }  


        }  


      


        // Calculate the sum of the array  


        for (int i = 0; i < N; i++)  


            for (int j = 0; j < M; j++)  


                count += a[i, j];  


      


        return count;  


    }  


      


    // Driver code  


    public static void Main() 


    {  


        int [,]arr = { { 1, 0, 1 },  


                        { 1, 1, 0 },  


                        { 1, 1, 0 } };  


      


        Console.WriteLine(countSquareMatrices(arr, n, m));  


    }  


} 


  


// This code is contributed by AnkitRai01 



















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Output :


7


Time Complexity: O(N*M)



          
          
          

          

    

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