Largest square sub-matrix with equal row, column, and diagonal sum

Last Updated : 16 Jul, 2024

Given a matrix mat[][] of dimensions N*M, the task is to find the size of the largest square submatrix such that the sum of all rows, columns, diagonals in that submatrix are equal.

Examples:

Input: N = 3, M = 4, mat[][] = [[5, 1, 3, 1], [9, 3, 3, 1], [1, 3, 3, 8]]
Output: 2
Explanation:
The submatrix which satisfies all the given conditions is shown in bold
5 1 3 1
9 3 3 1
1 3 3 8
Therefore, the size of the submatrix is 2.
Input: N = 4, M = 5, mat[][] = [[7, 1, 4, 5, 6], [2, 5, 1, 6, 4], [1, 5, 4, 3, 2], [1, 2, 7, 3, 4]]
Output: 3
Explanation:
The submatrix which satisfies all the given conditions is shown in bold
7 1 4 5 6
2 5 1 6 4
1 5 4 3 2
1 2 7 3 4
Therefore, the size of the submatrix is 3.

Approach:

The given problem can be solved by finding the Prefix Sum of all the rows and the columns and then iterate for all possible sizes of the square submatrix from each cell of the matrix and if there exists any such square matrix that satisfies the given criteria then print that size of a square matrix. Follow the below steps to solve the problem:
Maintain two prefix sum arrays prefixSumRow[] and prefixSumColumn[] and store the prefix sum of rows and columns of the given matrix respectively.
Perform the following steps to check if any square matrix starting from the cell (i, j) of size K satisfy the given criteria or not:
Find the sum of elements of primary diagonal of the submatrix mat[i][j] to mat[i + K][j + K] and store it in the variable, say sum.
If the value of the sum is the same as the value mentioned below then return true. Otherwise, return false.
The prefix sum of all the rows i.e., the value of prefixSumRow[k][j + K] - prefixSumRow[k][j] for all values of k over then range [i, i + K].
The prefix sum of all the columns i.e., the value of prefixSumColumn[i + K][j] - prefixSumColumn[i][k] for all values of k over then range [j, j + K].
The prefix sum of anti-diagonal elements.
Now, iterate for all possible sizes of the square matrix that can be formed over the range [min(N, M), 1] and if there exist any possible satisfies the given criteria using the steps in the above steps, then print that size of a square matrix.

Below is the implementation of the above approach:

C++

// C++ program for the above approach #include <bits/stdc++.h> using namespace std;  // Define the prefix sum arrays globally int prefix_sum_row[50][51]; int prefix_sum_col[51][50];  bool is_valid(int r, int c, int size, vector<vector<int> >& grid){     int r_end = r + size, c_end = c + size;      // Diagonal sum     int sum = 0;     for (int i = r, j = c; i < r_end; i++, j++) {         sum += grid[i][j];     }      // Check each row     for (int i = r; i < r_end; i++) {         if (prefix_sum_row[i][c_end] - prefix_sum_row[i][c] != sum) {             return false;         }     }      // Check each column     for (int i = c; i < c_end; i++) {         if (prefix_sum_col[r_end][i] - prefix_sum_col[r][i] != sum) {             return false;         }     }      // Check anti-diagonal     int ad_sum = 0;     for (int i = r, j = c_end - 1; i < r_end; i++, j--) {         ad_sum += grid[i][j];     }      return ad_sum == sum; }  int largestSquareValidMatrix(vector<vector<int> >& grid){     // Store the size of the given grid     int m = grid.size(), n = grid[0].size();      // Compute the prefix sum for the rows     for (int i = 0; i < m; i++) {         for (int j = 1; j <= n; j++) {             prefix_sum_row[i][j] = prefix_sum_row[i][j - 1] + grid[i][j - 1];         }     }      // Compute the prefix sum for the columns     for (int i = 1; i <= m; i++) {         for (int j = 0; j < n; j++) {             prefix_sum_col[i][j] = prefix_sum_col[i - 1][j] + grid[i - 1][j];         }     }      // Check for all possible square submatrix     for (int size = min(m, n); size > 1; size--) {         for (int i = 0; i <= m - size; i++) {             for (int j = 0; j <= n - size; j++) {                 if (is_valid(i, j, size, grid)) {                     return size;                 }             }         }     }      return 1; }  // Driver Code int main(){     vector<vector<int> > grid = { { 7, 1, 4, 5, 6 },                                   { 2, 5, 1, 6, 4 },                                   { 1, 5, 4, 3, 2 },                                   { 1, 2, 7, 3, 4 } };     cout << largestSquareValidMatrix(grid);     return 0; }

Java

// Java program for the above approach class GFG {     // Define the prefix sum arrays globally     public static int[][] prefix_sum_row = new int[50][51];     public static int[][] prefix_sum_col = new int[51][50];      public static boolean is_valid(int r, int c, int size, int[][] grid) {         int r_end = r + size, c_end = c + size;          // Diagonal sum         int sum = 0;         for (int i = r, j = c; i < r_end; i++, j++) {             sum += grid[i][j];         }          // Check each row         for (int i = r; i < r_end; i++) {             if (prefix_sum_row[i][c_end] - prefix_sum_row[i][c] != sum) {                 return false;             }         }          // Check each column         for (int i = c; i < c_end; i++) {             if (prefix_sum_col[r_end][i] - prefix_sum_col[r][i] != sum) {                 return false;             }         }          // Check anti-diagonal         int ad_sum = 0;         for (int i = r, j = c_end - 1; i < r_end; i++, j--) {             ad_sum += grid[i][j];         }          return ad_sum == sum;     }      public static int largestSquareValidMatrix(int[][] grid) {         // Store the size of the given grid         int m = grid.length, n = grid[0].length;          // Compute the prefix sum for the rows         for (int i = 0; i < m; i++) {             for (int j = 1; j <= n; j++) {                 prefix_sum_row[i][j] = prefix_sum_row[i][j - 1] + grid[i][j - 1];             }         }          // Compute the prefix sum for the columns         for (int i = 1; i <= m; i++) {             for (int j = 0; j < n; j++) {                 prefix_sum_col[i][j] = prefix_sum_col[i - 1][j] + grid[i - 1][j];             }         }          // Check for all possible square submatrix         for (int size = Math.min(m, n); size > 1; size--) {             for (int i = 0; i <= m - size; i++) {                 for (int j = 0; j <= n - size; j++) {                     if (is_valid(i, j, size, grid)) {                         return size;                     }                 }             }         }          return 1;     }      // Driver Code     public static void main(String args[]) {         int[][] grid = { { 7, 1, 4, 5, 6 }, { 2, 5, 1, 6, 4 }, { 1, 5, 4, 3, 2 }, { 1, 2, 7, 3, 4 } };         System.out.println(largestSquareValidMatrix(grid));      }  }  // This code is contributed by saurabh_jaiswal.

Python

## Python program for the above approach:  ## Define the prefix sum arrays globally prefix_sum_row = [[0]*51 for _ in range(50)] prefix_sum_col = [[0]*50 for _ in range(51)]  def is_valid(r, c, size, grid):          r_end = r + size     c_end = c + size      ## Diagonal sum     sum = 0;     j = c     for i in range(r, r_end):         sum += grid[i][j];         j+=1      ## Check each row     for i in range(r, r_end):         if ((prefix_sum_row[i][c_end] - prefix_sum_row[i][c]) != sum):             return False      ## Check each column     for i in range(c, c_end):         if ((prefix_sum_col[r_end][i] - prefix_sum_col[r][i]) != sum):             return False      ## Check anti-diagonal     ad_sum = 0;     j = c_end - 1     for i in range(r, r_end):         ad_sum += grid[i][j]         j-=1      return (ad_sum == sum)  def largestSquareValidMatrix(grid):      ## Store the size of the given grid     m = len(grid)     n = len(grid[0])      ## Compute the prefix sum for the rows     for i in range(0, m):         for j in range(1, n+1):             prefix_sum_row[i][j] = prefix_sum_row[i][j - 1] + grid[i][j - 1];          ## Compute the prefix sum for the columns     for i in range(1, m+1):         for j in range(0, n):             prefix_sum_col[i][j] = prefix_sum_col[i - 1][j] + grid[i - 1][j]       ## Check for all possible square submatrix     for size in range(min(m, n), 1, -1):         for i in range(0, m-size+1):             for j in range(0, n-size+1):                 if (is_valid(i, j, size, grid)):                     return size      return 1  ## Driver code if __name__=='__main__':      grid =  [   [ 7, 1, 4, 5, 6 ],                 [ 2, 5, 1, 6, 4 ],                 [ 1, 5, 4, 3, 2 ],                 [ 1, 2, 7, 3, 4 ]             ]                  print(largestSquareValidMatrix(grid))      # This code is contributed by subhamgoyal2014.

// C# program for the above approach using System; class GFG {        // Define the prefix sum arrays globally     public static int[,] prefix_sum_row = new int[50,51];     public static int[,] prefix_sum_col = new int[51,50];      public static bool is_valid(int r, int c, int size, int[,] grid) {         int r_end = r + size, c_end = c + size;          // Diagonal sum         int sum = 0;         for (int i = r, j = c; i < r_end; i++, j++) {             sum += grid[i,j];         }          // Check each row         for (int i = r; i < r_end; i++) {             if (prefix_sum_row[i,c_end] - prefix_sum_row[i,c] != sum) {                 return false;             }         }          // Check each column         for (int i = c; i < c_end; i++) {             if (prefix_sum_col[r_end,i] - prefix_sum_col[r,i] != sum) {                 return false;             }         }          // Check anti-diagonal         int ad_sum = 0;         for (int i = r, j = c_end - 1; i < r_end; i++, j--) {             ad_sum += grid[i,j];         }          return ad_sum == sum;     }      public static int largestSquareValidMatrix(int[,] grid) {         // Store the size of the given grid         int m = grid.GetLength(0), n = grid.GetLength(1);          // Compute the prefix sum for the rows         for (int i = 0; i < m; i++) {             for (int j = 1; j <= n; j++) {                 prefix_sum_row[i,j] = prefix_sum_row[i,j - 1] + grid[i,j - 1];             }         }          // Compute the prefix sum for the columns         for (int i = 1; i <= m; i++) {             for (int j = 0; j < n; j++) {                 prefix_sum_col[i,j] = prefix_sum_col[i - 1,j] + grid[i - 1,j];             }         }          // Check for all possible square submatrix         for (int size = Math.Min(m, n); size > 1; size--) {             for (int i = 0; i <= m - size; i++) {                 for (int j = 0; j <= n - size; j++) {                     if (is_valid(i, j, size, grid)) {                         return size;                     }                 }             }         }          return 1;     }      // Driver Code     public static void Main() {         int[,] grid = { { 7, 1, 4, 5, 6 }, { 2, 5, 1, 6, 4 },                        { 1, 5, 4, 3, 2 }, { 1, 2, 7, 3, 4 } };         Console.WriteLine(largestSquareValidMatrix(grid));      }  }  // This code is contributed by ukasp.

JavaScript

// JavaScript Program to implement // the above approach  // Define the prefix sum arrays globally let prefix_sum_row = new Array(50); for (let i = 0; i < prefix_sum_row.length; i++) {     prefix_sum_row[i] = new Array(51).fill(0); } let prefix_sum_col = new Array(50); for (let i = 0; i < prefix_sum_col.length; i++) {     prefix_sum_col[i] = new Array(51).fill(0); } function is_valid(r, c, size, grid) {     let r_end = r + size, c_end = c + size;      // Diagonal sum     let sum = 0;     for (let i = r, j = c; i < r_end; i++, j++) {         sum += grid[i][j];     }     // Check each row     for (let i = r; i < r_end; i++) {         if (prefix_sum_row[i][c_end]             - prefix_sum_row[i][c]             != sum) {             return false;         }     }      // Check each column     for (let i = c; i < c_end; i++) {         if (prefix_sum_col[r_end][i]             - prefix_sum_col[r][i]             != sum) {             return false;         }     }      // Check anti-diagonal     let ad_sum = 0;     for (let i = r, j = c_end - 1; i < r_end;         i++, j--) {         ad_sum += grid[i][j];     }      return ad_sum == sum; }  function largestSquareValidMatrix(grid) {     // Store the size of the given grid     let m = grid.length, n = grid[0].length;      // Compute the prefix sum for the rows     for (let i = 0; i < m; i++) {         for (let j = 1; j <= n; j++) {             prefix_sum_row[i][j]                 = prefix_sum_row[i][j - 1]                 + grid[i][j - 1];         }     }      // Compute the prefix sum for the columns     for (let i = 1; i <= m; i++) {         for (let j = 0; j < n; j++) {             prefix_sum_col[i][j]                 = prefix_sum_col[i - 1][j]                 + grid[i - 1][j];         }     }      // Check for all possible square submatrix     for (let size = Math.min(m, n); size > 1; size--) {         for (let i = 0; i <= m - size; i++) {             for (let j = 0; j <= n - size; j++) {                 if (is_valid(i, j, size, grid)) {                     return size;                 }             }         }     }      return 1; }  // Driver Code  let grid = [[7, 1, 4, 5, 6], [2, 5, 1, 6, 4], [1, 5, 4, 3, 2], [1, 2, 7, 3, 4]]; console.log(largestSquareValidMatrix(grid));  // This code is contributed by Potta Lokesh

Output

Time Complexity: O(N*M*min(N, M)²), where N is the number of
Auxiliary Space: O(N*M)

Find Matrix With Given Row and Column Sums

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Largest square sub-matrix with equal row, column, and diagonal sum

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