Count of submatrix with sum X in a given Matrix

Given a matrix of size N x M and an integer X, the task is to find the number of sub-squares in the matrix with sum of elements equal to X.
Examples:

Input: N = 4, M = 5, X = 10, arr[][]={{2, 4, 3, 2, 10}, {3, 1, 1, 1, 5}, {1, 1, 2, 1, 4}, {2, 1, 1, 1, 3}} 
Output: 3 
Explanation:
{10}, {{2, 4}, {3, 1}} and {{1, 1, 1}, {1, 2, 1}, {1, 1, 1}} are subsquares with sum 10.
Input: N = 3, M = 4, X = 8, arr[][]={{3, 1, 5, 3}, {2, 2, 2, 6}, {1, 2, 2, 4}} 
Output: 2 
Explanation:
Sub-squares {{2, 2}, {2, 2}} and {{3, 1}, {2, 2}} have sum 8. 

Naive Approach:
The simplest approach to solve the problem is to generate all possible sub-squares and check sum of all the elements of the sub-square equals to X.

C++

#include <iostream>
#include <vector>

using namespace std;

int count_sub_squares(vector<vector<int> > matrix, int x)
{
    int count = 0;
    int n = matrix.size();
    int m = matrix[0].size();
    int max_size = min(n, m);
    for (int size = 1; size <= max_size; size++) {
        for (int i = 0; i <= n - size; i++) {
            for (int j = 0; j <= m - size; j++) {
                int sum = 0;
                for (int p = i; p < i + size; p++) {
                    for (int q = j; q < j + size; q++) {
                        sum += matrix[p][q];
                    }
                }
                if (sum == x) {
                    count++;
                }
            }
        }
    }
    return count;
}

int main()
{
    vector<vector<int> > matrix = { { 2, 4, 3, 2, 10 },
        { 3, 1, 1, 1, 5 },
        { 1, 1, 2, 1, 4 },
        { 2, 1, 1, 1, 3 }
    };
    int x = 10;
    int sub_squares = count_sub_squares(matrix, x);
    cout << "Number of sub-squares with sum " << x << " is "
         << sub_squares << endl;
    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;

public class SubSquares {

    public static int
    countSubSquares(List<List<Integer> > matrix, int x)
    {
        int count = 0;
        int n = matrix.size();
        int m = matrix.get(0).size();
        int max_size = Math.min(n, m);
        for (int size = 1; size <= max_size; size++) {
            for (int i = 0; i <= n - size; i++) {
                for (int j = 0; j <= m - size; j++) {
                    int sum = 0;
                    for (int p = i; p < i + size; p++) {
                        for (int q = j; q < j + size; q++) {
                            sum += matrix.get(p).get(q);
                        }
                    }
                    if (sum == x) {
                        count++;
                    }
                }
            }
        }
        return count;
    }

    public static void main(String[] args)
    {
        List<List<Integer> > matrix = new ArrayList<>();
        matrix.add(List.of(2, 4, 3, 2, 10));
        matrix.add(List.of(3, 1, 1, 1, 5));
        matrix.add(List.of(1, 1, 2, 1, 4));
        matrix.add(List.of(2, 1, 1, 1, 3));
        int x = 10;
        int subSquares = countSubSquares(matrix, x);
        System.out.println("Number of sub-squares with sum "
                           + x + " is " + subSquares);
    }
}

Python3

def count_sub_squares(matrix, x):
    count = 0
    n = len(matrix)
    m = len(matrix[0])
    max_size = min(n, m)

    # Loop over all possible square sizes from 1 to max_size
    for size in range(1, max_size + 1):
        # Loop over all possible starting points (i, j) of the square
        for i in range(n - size + 1):
            for j in range(m - size + 1):
                # Calculate the sum of elements in the current square
                sum_ = 0
                for p in range(i, i + size):
                    for q in range(j, j + size):
                        sum_ += matrix[p][q]

                # If the sum of elements in the current square is equal to x, increment count
                if sum_ == x:
                    count += 1

    return count

def main():
    matrix = [
        [2, 4, 3, 2, 10],
        [3, 1, 1, 1, 5],
        [1, 1, 2, 1, 4],
        [2, 1, 1, 1, 3]
    ]
    x = 10
    sub_squares = count_sub_squares(matrix, x)
    print("Number of sub-squares with sum", x, "is", sub_squares)

if __name__ == "__main__":
    main()

Output

Number of sub-squares with sum 10 is 3

Time Complexity: O(N² * M²*min(N,M)) 
Auxiliary Space: O(1),  since no extra space has been taken.

Efficient Approach: To optimize the above naive approach the sum of all the element of all the matrix till each cell has to be made. Below are the steps:

• Precalculate the sum of all rectangles with its upper left corner at (0, 0) and lower right at (i, j) in O(N * M) computational complexity.
• Now, it can be observed that, for every upper left corner, there can be at most one square with sum X since elements of the matrix are positive.
• Keeping this in mind we can use binary search to check if there exists a square with sum X.
• For every cell (i, j) in the matrix, fix it as the upper left corner of the subsquare. Then, traverse over all possible subsquares with (i, j) as the upper left corner and increase count if sum is equal to X or not. 
  

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Size of a column
#define m 5

// Function to find the count of submatrix
// whose sum is X
int countSubsquare(int arr[][m],
                   int n, int X)
{
    int dp[n + 1][m + 1];

    memset(dp, 0, sizeof(dp));

    // Copying arr to dp and making
    // it indexed 1
    for (int i = 0; i < n; i++) {

        for (int j = 0; j < m; j++) {

            dp[i + 1][j + 1] = arr[i][j];
        }
    }

    // Precalculate and store the sum
    // of all rectangles with upper
    // left corner at (0, 0);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {

            // Calculating sum in
            // a 2d grid
            dp[i][j] += dp[i - 1][j]
                        + dp[i][j - 1]
                        - dp[i - 1][j - 1];
        }
    }

    // Stores the answer
    int cnt = 0;

    for (int i = 1; i <= n; i++) {

        for (int j = 1; j <= m; j++) {

            // Fix upper left corner
            // at {i, j} and perform
            // binary search on all
            // such possible squares

            // Minimum length of square
            int lo = 1;

            // Maximum length of square
            int hi = min(n - i, m - j) + 1;

            // Flag to set if sub-square
            // with sum X is found
            bool found = false;

            while (lo <= hi) {
                int mid = (lo + hi) / 2;

                // Calculate lower
                // right index if upper
                // right corner is at {i, j}
                int ni = i + mid - 1;
                int nj = j + mid - 1;

                // Calculate the sum of
                // elements in the submatrix
                // with upper left column
                // {i, j} and lower right
                // column at {ni, nj};
                int sum = dp[ni][nj]
                          - dp[ni][j - 1]
                          - dp[i - 1][nj]
                          + dp[i - 1][j - 1];

                if (sum >= X) {

                    // If sum X is found
                    if (sum == X) {
                        found = true;
                    }

                    hi = mid - 1;

                    // If sum > X, then size of
                    // the square with sum X
                    // must be less than mid
                }
                else {

                    // If sum < X, then size of
                    // the square with sum X
                    // must be greater than mid
                    lo = mid + 1;
                }
            }

            // If found, increment
            // count by 1;
            if (found == true) {
                cnt++;
            }
        }
    }
    return cnt;
}

// Driver Code
int main()
{
    int N = 4, X = 10;

    // Given Matrix arr[][]
    int arr[N][m] = { { 2, 4, 3, 2, 10 },
                      { 3, 1, 1, 1, 5 },
                      { 1, 1, 2, 1, 4 },
                      { 2, 1, 1, 1, 3 } };

    // Function Call
    cout << countSubsquare(arr, N, X)
         << endl;

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{

// Size of a column
static final int m = 5;

// Function to find the count of submatrix
// whose sum is X
static int countSubsquare(int arr[][],
                          int n, int X)
{
    int [][]dp = new int[n + 1][m + 1];

    // Copying arr to dp and making
    // it indexed 1
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            dp[i + 1][j + 1] = arr[i][j];
        }
    }

    // Precalculate and store the sum
    // of all rectangles with upper
    // left corner at (0, 0);
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {

            // Calculating sum in
            // a 2d grid
            dp[i][j] += dp[i - 1][j] +
                          dp[i][j - 1] -
                          dp[i - 1][j - 1];
        }
    }

    // Stores the answer
    int cnt = 0;

    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {

            // Fix upper left corner
            // at {i, j} and perform
            // binary search on all
            // such possible squares

            // Minimum length of square
            int lo = 1;

            // Maximum length of square
            int hi = Math.min(n - i, m - j) + 1;

            // Flag to set if sub-square
            // with sum X is found
            boolean found = false;

            while (lo <= hi)
            {
                int mid = (lo + hi) / 2;

                // Calculate lower
                // right index if upper
                // right corner is at {i, j}
                int ni = i + mid - 1;
                int nj = j + mid - 1;

                // Calculate the sum of
                // elements in the submatrix
                // with upper left column
                // {i, j} and lower right
                // column at {ni, nj};
                int sum = dp[ni][nj] -
                            dp[ni][j - 1] -
                          dp[i - 1][nj] +
                          dp[i - 1][j - 1];

                if (sum >= X)
                {

                    // If sum X is found
                    if (sum == X)
                    {
                        found = true;
                    }

                    hi = mid - 1;

                    // If sum > X, then size of
                    // the square with sum X
                    // must be less than mid
                }
                else
                {

                    // If sum < X, then size of
                    // the square with sum X
                    // must be greater than mid
                    lo = mid + 1;
                }
            }

            // If found, increment
            // count by 1;
            if (found == true)
            {
                cnt++;
            }
        }
    }
    return cnt;
}

// Driver Code
public static void main(String[] args)
{
    int N = 4, X = 10;

    // Given Matrix arr[][]
    int arr[][] = { { 2, 4, 3, 2, 10 },
                    { 3, 1, 1, 1, 5 },
                    { 1, 1, 2, 1, 4 },
                    { 2, 1, 1, 1, 3 } };

    // Function Call
    System.out.print(countSubsquare(arr, N, X) + "\n");
}
}

// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach

# Size of a column
m = 5

# Function to find the count of
# submatrix whose sum is X
def countSubsquare(arr, n, X):

    dp = [[ 0 for x in range(m + 1)]
              for y in range(n + 1)]

    # Copying arr to dp and making
    # it indexed 1
    for i in range(n):
        for j in range(m):
            dp[i + 1][j + 1] = arr[i][j]

    # Precalculate and store the sum
    # of all rectangles with upper
    # left corner at (0, 0);
    for i in range(1, n + 1):
        for j in range(1, m + 1):

            # Calculating sum in
            # a 2d grid
            dp[i][j] += (dp[i - 1][j] +
                         dp[i][j - 1] -
                         dp[i - 1][j - 1])

    # Stores the answer
    cnt = 0

    for i in range(1, n + 1):
        for j in range(1, m + 1):

            # Fix upper left corner
            # at {i, j} and perform
            # binary search on all
            # such possible squares

            # Minimum length of square
            lo = 1

            # Maximum length of square
            hi = min(n - i, m - j) + 1

            # Flag to set if sub-square
            # with sum X is found
            found = False

            while (lo <= hi):
                mid = (lo + hi) // 2

                # Calculate lower right
                # index if upper right
                # corner is at {i, j}
                ni = i + mid - 1
                nj = j + mid - 1

                # Calculate the sum of
                # elements in the submatrix
                # with upper left column
                # {i, j} and lower right
                # column at {ni, nj};
                sum = (dp[ni][nj] -
                       dp[ni][j - 1] -
                       dp[i - 1][nj] +
                       dp[i - 1][j - 1])

                if (sum >= X):

                    # If sum X is found
                    if (sum == X):
                        found = True

                    hi = mid - 1

                    # If sum > X, then size of
                    # the square with sum X
                    # must be less than mid
                else:

                    # If sum < X, then size of
                    # the square with sum X
                    # must be greater than mid
                    lo = mid + 1

            # If found, increment
            # count by 1;
            if (found == True):
                cnt += 1
    return cnt

# Driver Code
if __name__ =="__main__":

    N, X = 4, 10

    # Given matrix arr[][]
    arr = [ [ 2, 4, 3, 2, 10 ],
            [ 3, 1, 1, 1, 5 ],
            [ 1, 1, 2, 1, 4 ],
            [ 2, 1, 1, 1, 3 ] ]

    # Function call
    print(countSubsquare(arr, N, X))

# This code is contributed by chitranayal

C#

// C# program for the above approach
using System;

class GFG{

// Size of a column
static readonly int m = 5;

// Function to find the count of submatrix
// whose sum is X
static int countSubsquare(int [,]arr,
                          int n, int X)
{
    int [,]dp = new int[n + 1, m + 1];

    // Copying arr to dp and making
    // it indexed 1
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            dp[i + 1, j + 1] = arr[i, j];
        }
    }

    // Precalculate and store the sum
    // of all rectangles with upper
    // left corner at (0, 0);
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
        {

            // Calculating sum in
            // a 2d grid
            dp[i, j] += dp[i - 1, j] +
                        dp[i, j - 1] -
                        dp[i - 1, j - 1];
        }
    }

    // Stores the answer
    int cnt = 0;

    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
        {

            // Fix upper left corner
            // at {i, j} and perform
            // binary search on all
            // such possible squares

            // Minimum length of square
            int lo = 1;

            // Maximum length of square
            int hi = Math.Min(n - i, m - j) + 1;

            // Flag to set if sub-square
            // with sum X is found
            bool found = false;

            while (lo <= hi)
            {
                int mid = (lo + hi) / 2;

                // Calculate lower
                // right index if upper
                // right corner is at {i, j}
                int ni = i + mid - 1;
                int nj = j + mid - 1;

                // Calculate the sum of
                // elements in the submatrix
                // with upper left column
                // {i, j} and lower right
                // column at {ni, nj};
                int sum = dp[ni, nj] -
                          dp[ni, j - 1] -
                          dp[i - 1, nj] +
                          dp[i - 1, j - 1];

                if (sum >= X)
                {

                    // If sum X is found
                    if (sum == X)
                    {
                        found = true;
                    }

                    hi = mid - 1;

                    // If sum > X, then size of
                    // the square with sum X
                    // must be less than mid
                }
                else
                {

                    // If sum < X, then size of
                    // the square with sum X
                    // must be greater than mid
                    lo = mid + 1;
                }
            }

            // If found, increment
            // count by 1;
            if (found == true)
            {
                cnt++;
            }
        }
    }
    return cnt;
}

// Driver Code
public static void Main(String[] args)
{
    int N = 4, X = 10;

    // Given Matrix [,]arr
    int [,]arr = { { 2, 4, 3, 2, 10 },
                   { 3, 1, 1, 1, 5 },
                   { 1, 1, 2, 1, 4 },
                   { 2, 1, 1, 1, 3 } };

    // Function call
    Console.Write(countSubsquare(arr, N, X) + "\n");
}
}

// This code is contributed by amal kumar choubey

Javascript

<script>

// Javascript program for the above approach

    // Size of a column
     var m = 5;

    // Function to find the count of submatrix
    // whose sum is X
    function countSubsquare(arr , n , X) {
        var dp = Array(n + 1);
        for(var i =0;i<n+1;i++)
        dp[i] = Array(m + 1).fill(0);

        // Copying arr to dp and making
        // it indexed 1
        for (i = 0; i < n; i++) {
            for (j = 0; j < m; j++) {
                dp[i + 1][j + 1] = arr[i][j];
            }
        }

        // Precalculate and store the sum
        // of all rectangles with upper
        // left corner at (0, 0);
        for (i = 1; i <= n; i++) {
            for (j = 1; j <= m; j++) {

                // Calculating sum in
                // a 2d grid
                dp[i][j] += dp[i - 1][j] +
                dp[i][j - 1] - dp[i - 1][j - 1];
            }
        }

        // Stores the answer
        var cnt = 0;

        for (i = 1; i <= n; i++) {
            for (j = 1; j <= m; j++) {

                // Fix upper left corner
                // at {i, j} and perform
                // binary search on all
                // such possible squares

                // Minimum length of square
                var lo = 1;

                // Maximum length of square
                var hi = Math.min(n - i, m - j) + 1;

                // Flag to set if sub-square
                // with sum X is found
                var found = false;

                while (lo <= hi) {
                    var mid = parseInt((lo + hi) / 2);

                    // Calculate lower
                    // right index if upper
                    // right corner is at {i, j}
                    var ni = i + mid - 1;
                    var nj = j + mid - 1;

                    // Calculate the sum of
                    // elements in the submatrix
                    // with upper left column
                    // {i, j} and lower right
                    // column at {ni, nj];
                    var sum = dp[ni][nj] - dp[ni][j - 1] -
                    dp[i - 1][nj] + dp[i - 1][j - 1];

                    if (sum >= X) {

                        // If sum X is found
                        if (sum == X) {
                            found = true;
                        }

                        hi = mid - 1;

                        // If sum > X, then size of
                        // the square with sum X
                        // must be less than mid
                    } else {

                        // If sum < X, then size of
                        // the square with sum X
                        // must be greater than mid
                        lo = mid + 1;
                    }
                }

                // If found, increment
                // count by 1;
                if (found == true) {
                    cnt++;
                }
            }
        }
        return cnt;
    }

    // Driver Code

        var N = 4, X = 10;

        // Given Matrix arr
        var arr = [ [ 2, 4, 3, 2, 10 ],
                    [ 3, 1, 1, 1, 5 ],
                    [ 1, 1, 2, 1, 4 ],
                    [ 2, 1, 1, 1, 3 ] ];

        // Function Call
        document.write(countSubsquare(arr, N, X) + "<br/>");

// This code contributed by umadevi9616

</script>

Output

3

Time Complexity: O(N * M * log(max(N, M))) 
Auxiliary Space: O(N * M) since n*m space is being used for populating the 2-d array dp.