Find Maximum Length Of A Square Submatrix Having Sum Of Elements At-Most K

Namachila November 19, 2025

0 0 6 minutes read

Given a N x M matrix where N is the number of rows and M is the number of columns in the given matrix and an integer K. The task is to find the maximum length of a square submatrix having the sum of elements less than or equal to K or print 0 if no such square exits.
Examples:

Input: r = 4, c = 4 , k = 6
matrix[][] = { {1, 1, 1, 1},
               {1, 0, 0, 0},
               {1, 0, 0, 0},
               {1, 0, 0, 0},
             } 
Output: 3
Explanation:
Square from (0,0) to (2,2) with 
sum 5 is one of the valid answer.

Input: r = 4, c = 4 , k = 1
matrix[][] = { {2, 2, 2},
               {2, 2, 2},
               {2, 2, 2},
               {2, 2, 2},
             } 
Output: 0
Explanation:
There is no valid answer.

Approach:
The idea is to calculate the prefix sum matrix. Once the prefix sum matrix is calculated, the required sub-matrix sum can be calculated in O(1) time complexity. Use the sliding window technique to calculate the maximum length square submatrix. For every square of length cur_max+1, where cur_max is the currently found maximum length of a square submatrix, we check whether the sum of elements in the current submatrix, having the length of cur_max+1, is less than or equal to K or not. If yes, then increase the result by 1, otherwise, we continue to check.
Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <iostream>
using namespace std;
 
    // Function to return maximum 
    // length of square submatrix
    // having sum of elements at-most K
    int maxLengthSquare(int row, int column, 
                        int arr[][4], int k)
    {
        // Matrix to store prefix sum
        int sum[row + 1][column + 1] ;
     
        for(int i = 1; i <= row; i++)
            for(int j = 0; j <= column; j++)
                sum[i][j] = 0;
                 
        // Current maximum length
        int cur_max = 1;
     
        // Variable for storing 
        // maximum length of square
        int max = 0;
             
        for (int i = 1; i <= row; i++) 
        {
            for (int j = 1; j <= column; j++) 
            {
                // Calculating prefix sum
                sum[i][j] = sum[i - 1][j] + sum[i][j - 1] + 
                            arr[i - 1][j - 1] - sum[i - 1][j - 1];
         
                // Checking whether there 
                // exits square with length 
                // cur_max+1 or not
                if(i >= cur_max && j >= cur_max && 
                    sum[i][j] - sum[i - cur_max][j]
                    - sum[i][j - cur_max] + 
                    sum[i - cur_max][j - cur_max] <= k)
                {
                    max = cur_max++;
                }
            }
        }
     
        // Returning the 
        // maximum length
        return max;
    }
 
    // Driver code
    int main() 
    {
         
        int row = 4, column = 4;
        int matrix[4][4] = { {1, 1, 1, 1},
                        {1, 0, 0, 0},
                        {1, 0, 0, 0},
                        {1, 0, 0, 0}
                        };
     
        int k = 6; 
        int ans = maxLengthSquare(row, column, matrix, k);
        cout << ans;
         
        return 0;
    }
 
// This code is contributed by AnkitRai01

Java

// Java implementation of 
// the above approach
import java.util.*;
 
class GFG
{
    // Function to return maximum 
    // length of square submatrix
    // having sum of elements at-most K
    public static int maxLengthSquare(int row,int column,
                                        int[][] arr, int k)
    {
        // Matrix to store prefix sum
        int sum[][] = new int[row + 1][column + 1];
     
        // Current maximum length
        int cur_max = 1;
     
        // Variable for storing 
        // maximum length of square
        int max = 0;
             
        for (int i = 1; i <= row; i++) 
        {
            for (int j = 1; j <= column; j++) 
            {
                // Calculating prefix sum
                sum[i][j] = sum[i - 1][j] + sum[i][j - 1] + 
                            arr[i - 1][j - 1] - sum[i - 1][j - 1];
         
                // Checking whether there 
                // exits square with length 
                // cur_max+1 or not
                if(i >=cur_max&&j>=cur_max&&sum[i][j]-sum[i - cur_max][j]
                            - sum[i][j - cur_max] 
                            + sum[i - cur_max][j - cur_max] <= k){
                    max = cur_max++;
                }
            }
        }
     
        // Returning the 
        // maximum length
        return max;
    }
 
    // Driver code
    public static void main(String args[]) 
    {
         
        int row = 4 , column = 4;
        int matrix[][] = { {1, 1, 1, 1},
                        {1, 0, 0, 0},
                        {1, 0, 0, 0},
                        {1, 0, 0, 0}
                        };
     
        int k = 6; 
        int ans = maxLengthSquare(row,column,matrix, k);
        System.out.println(ans);
    }
}

Python3

# Python3 implementation of the above approach 
import numpy as np
 
# Function to return maximum 
# length of square submatrix 
# having sum of elements at-most K 
def maxLengthSquare(row, column, arr, k) :
     
    # Matrix to store prefix sum 
    sum = np.zeros((row + 1, column + 1)); 
     
    # Current maximum length 
    cur_max = 1; 
     
    # Variable for storing 
    # maximum length of square 
    max = 0; 
             
    for i in range(1, row + 1) :
        for j in range(1, column + 1) :
             
            # Calculating prefix sum
            sum[i][j] = sum[i - 1][j] + sum[i][j - 1] + \
                        arr[i - 1][j - 1] - \
                        sum[i - 1][j - 1];
             
            # Checking whether there
            # exits square with length
            # cur_max+1 or not
            if(i >= cur_max and j >= cur_max and
                 sum[i][j] - sum[i - cur_max][j] - sum[i][j -
                                     cur_max] + sum[i -
                                     cur_max][j - cur_max] <= k) :
                max = cur_max; 
                cur_max += 1;
     
    # Returning the maximum length 
    return max; 
     
# Driver code 
if __name__ == "__main__" :
     
    row = 4 ;
    column = 4;
    matrix = [[1, 1, 1, 1],
              [1, 0, 0, 0],
              [1, 0, 0, 0],
              [1, 0, 0, 0]];
    k = 6;
    ans = maxLengthSquare(row, column, matrix, k);
    print(ans); 
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach 
using System;
 
class GFG 
{ 
    // Function to return maximum 
    // length of square submatrix 
    // having sum of elements at-most K 
    public static int maxLengthSquare(int row,int column, 
                                        int[,] arr, int k) 
    { 
        // Matrix to store prefix sum 
        int [,]sum = new int[row + 1,column + 1]; 
     
        // Current maximum length 
        int cur_max = 1; 
     
        // Variable for storing 
        // maximum length of square 
        int max = 0; 
             
        for (int i = 1; i <= row; i++) 
        { 
            for (int j = 1; j <= column; j++) 
            { 
                // Calculating prefix sum 
                sum[i, j] = sum[i - 1, j] + sum[i, j - 1] + 
                            arr[i - 1, j - 1] - sum[i - 1, j - 1]; 
         
                // Checking whether there 
                // exits square with length 
                // cur_max+1 or not 
                if(i >=cur_max && j>=cur_max && sum[i, j]-sum[i - cur_max, j] 
                            - sum[i, j - cur_max] 
                            + sum[i - cur_max, j - cur_max] <= k)
                { 
                    max = cur_max++; 
                } 
            } 
        } 
     
        // Returning the 
        // maximum length 
        return max; 
    } 
 
    // Driver code 
    public static void Main() 
    { 
         
        int row = 4 , column = 4; 
        int [,]matrix = { {1, 1, 1, 1}, 
                        {1, 0, 0, 0}, 
                        {1, 0, 0, 0}, 
                        {1, 0, 0, 0} 
                        }; 
     
        int k = 6; 
        int ans = maxLengthSquare(row, column, matrix, k); 
        Console.WriteLine(ans); 
    } 
} 
 
// This code is contributed by AnkitRai01

Javascript

<script>
// Javascript implementation of the above approach
// Function to return maximum
// length of square submatrix
// having sum of elements at-most K
function maxLengthSquare(row, column, arr, k) {
    // Matrix to store prefix sum
    let sum = new Array();[row + 1][column + 1];
 
    for (let i = 0; i < row + 1; i++) {
        let temp = new Array();
        for (let j = 0; j < column + 1; j++) {
            temp.push([])
        }
        sum.push(temp)
    }
 
    for (let i = 1; i <= row; i++)
        for (let j = 0; j <= column; j++)
            sum[i][j] = 0;
 
    // Current maximum length
    let cur_max = 1;
 
    // Variable for storing
    // maximum length of square
    let max = 0;
 
    for (let i = 1; i <= row; i++) {
        for (let j = 1; j <= column; j++) {
            // Calculating prefix sum
            sum[i][j] = sum[i - 1][j] + sum[i][j - 1] +
                arr[i - 1][j - 1] - sum[i - 1][j - 1];
 
            // Checking whether there
            // exits square with length
            // cur_max+1 or not
            if (i >= cur_max && j >= cur_max &&
                sum[i][j] - sum[i - cur_max][j]
                - sum[i][j - cur_max] +
                sum[i - cur_max][j - cur_max] <= k) {
                max = cur_max++;
            }
        }
    }
 
    // Returning the
    // maximum length
    return max;
}
 
// Driver code
 
let row = 4, column = 4;
let matrix = [[1, 1, 1, 1],
[1, 0, 0, 0],
[1, 0, 0, 0],
[1, 0, 0, 0]
];
 
let k = 6;
let ans = maxLengthSquare(row, column, matrix, k);
document.write(ans);
 
// This code is contributed by gfgking
</script>

Output:

Time Complexity: O(N x M)
Auxiliary Space: O(N x M), where N and M are the given rows and columns of the matrix.

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