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Maximum size square Sub-Matrix with sum less than or equals to K

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  • Difficulty Level : Hard
  • Last Updated : 19 Jul, 2021

Given a Matrix arr[][] of size M x N having positive integers and a number K, the task is to find the size of the largest square sub-matrix whose sum of elements is less than or equals to K. 

Example: 

Input: 
arr[][] = { { 1, 1, 3, 2, 4, 3, 2 },
            { 1, 1, 3, 2, 4, 3, 2 },
            { 1, 1, 3, 2, 4, 3, 2 } },
K = 4
Output: 
2
Explanation:
Maximum size square Sub-Matrix 
with sum less than or equals to 4
      1 1
      1 1
Size is 2.

Input: 
arr[][] = { { 1, 1, 3, 2, 4, 3, 2 },
            { 1, 1, 3, 2, 4, 3, 2 },
            { 1, 1, 3, 2, 4, 3, 2 } }, 
K = 22
Output: 
3
Explanation:
Maximum size square Sub-Matrix 
with sum less than or equals to 22
      1 1 3
      1 1 3
      1 1 3
Size is 3.

Approach:  

  1. For the given matrix arr[][] create a prefix sum matrix(say sum[][]) such that sum[i][j] stores the sum of all the elements of the matrix of size i x j.
  2. For each row in prefix sum matrix sum[][] using Binary Search do the following: 
    • Perform Binary search with the lower limit as 0 end the upper limit as to maximum size of square matrix.
    • Find the middle index (say mid).
    • If the sum of elements of all possible square matrix of size mid is less than or equals to K, then update the lower limit as mid + 1 to find the maximum sum with size greater than mid.
    • Else Update the upper limit as mid – 1 to find the maximum sum with size less than mid.
  3. Keep updating the maximum size of square matrix in each iteration for the given valid condition above.

Below is the implementation of the above approach: 
 

Java




// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function to find the maximum size
    // of matrix with sum <= K
    static void findMaxMatrixSize(int[][] arr, int K)
    {
        int i, j;
 
        // N size of rows and M size of cols
        int n = arr.length;
        int m = arr[0].length;
 
        // To store the prefix sum of matrix
        int[][] sum = new int[n + 1][m + 1];
 
        // Create prefix sum
        for (i = 0; i <= n; i++) {
 
            // Traverse each rows
            for (j = 0; j <= m; j++) {
                if (i == 0 || j == 0) {
                    sum[i][j] = 0;
                    continue;
                }
 
                // Update the prefix sum
                // till index i x j
                sum[i][j] = arr[i - 1][j - 1]
                            + sum[i - 1][j] + sum[i][j - 1]
                            - sum[i - 1][j - 1];
            }
        }
 
        // To store the maximum size of
        // matrix with sum <= K
        int ans = 0;
 
        // Traverse the sum matrix
        for (i = 1; i <= n; i++) {
            for (j = 1; j <= m; j++) {
 
                // Index out of bound
                if (i + ans - 1 > n || j + ans - 1 > m)
                    break;
 
                int mid, lo = ans;
 
                // Maximum possible size
                // of matrix
                int hi = Math.min(n - i + 1, m - j + 1);
 
                // Binary Search
                while (lo < hi) {
 
                    // Find middle index
                    mid = (hi + lo + 1) / 2;
 
                    // Check whether sum <= K
                    // or not
                    // If Yes check for other
                    // half of the search
                    if (sum[i + mid - 1][j + mid - 1]
                            + sum[i - 1][j - 1]
                            - sum[i + mid - 1][j - 1]
                            - sum[i - 1][j + mid - 1]
                        <= K) {
                        lo = mid;
                    }
 
                    // Else check it in first
                    // half
                    else {
                        hi = mid - 1;
                    }
                }
 
                // Update the maximum size matrix
                ans = Math.max(ans, lo);
            }
        }
 
        // Print the final answer
        System.out.print(ans + "\n");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[][] arr = { { 1, 1, 3, 2, 4, 3, 2 },
                        { 1, 1, 3, 2, 4, 3, 2 },
                        { 1, 1, 3, 2, 4, 3, 2 } };
 
        // Given target sum
        int K = 4;
 
        // Function Call
        findMaxMatrixSize(arr, K);
    }
}
 
// This code is contributed by 29AjayKumar

Python3



# Python3 program for the above approach
 
# Function to find the maximum size
# of matrix with sum <= K
 
 
def findMaxMatrixSize(arr, K):
 
    # N size of rows and M size of cols
    n = len(arr)
    m = len(arr[0])
 
    # To store the prefix sum of matrix
    sum = [[0 for i in range(m + 1)] for j in range(n + 1)]
 
    # Create Prefix Sum
    for i in range(n + 1):
 
        # Traverse each rows
        for j in range(m+1):
            if (i == 0 or j == 0):
                sum[i][j] = 0
                continue
 
            # Update the prefix sum
            # till index i x j
            sum[i][j] = arr[i - 1][j - 1] + sum[i - 1][j] + \
                sum[i][j - 1]-sum[i - 1][j - 1]
 
    # To store the maximum size of
    # matrix with sum <= K
    ans = 0
 
    # Traverse the sum matrix
    for i in range(1, n + 1):
        for j in range(1, m + 1):
 
            # Index out of bound
            if (i + ans - 1 > n or j + ans - 1 > m):
                break
 
            mid = ans
            lo = ans
 
            # Maximum possible size
            # of matrix
            hi = min(n - i + 1, m - j + 1)
 
            # Binary Search
            while (lo < hi):
 
                # Find middle index
                mid = (hi + lo + 1) // 2
 
                # Check whether sum <= K
                # or not
                # If Yes check for other
                # half of the search
                if (sum[i + mid - 1][j + mid - 1] +
                    sum[i - 1][j - 1] -
                    sum[i + mid - 1][j - 1] -
                        sum[i - 1][j + mid - 1] <= K):
                    lo = mid
 
                # Else check it in first
                # half
                else:
                    hi = mid - 1
 
            # Update the maximum size matrix
            ans = max(ans, lo)
 
    # Print the final answer
    print(ans)
 
 
# Driver Code
if __name__ == '__main__':
    arr = [[1, 1, 3, 2, 4, 3, 2],
           [1, 1, 3, 2, 4, 3, 2],
           [1, 1, 3, 2, 4, 3, 2]]
 
    # Given target sum
    K = 4
 
    # Function Call
    findMaxMatrixSize(arr, K)
 
# This code is contributed by Surendra_Gangwar
C#



// C# program for the above approach
using System;
 
class GFG {
 
    // Function to find the maximum size
    // of matrix with sum <= K
    static void findMaxMatrixSize(int[, ] arr, int K)
    {
        int i, j;
 
        // N size of rows and M size of cols
        int n = arr.GetLength(0);
        int m = arr.GetLength(1);
 
        // To store the prefix sum of matrix
        int[, ] sum = new int[n + 1, m + 1];
 
        // Create prefix sum
        for (i = 0; i <= n; i++) {
 
            // Traverse each rows
            for (j = 0; j <= m; j++) {
                if (i == 0 || j == 0) {
                    sum[i, j] = 0;
                    continue;
                }
 
                // Update the prefix sum
                // till index i x j
                sum[i, j] = arr[i - 1, j - 1]
                            + sum[i - 1, j] + sum[i, j - 1]
                            - sum[i - 1, j - 1];
            }
        }
 
        // To store the maximum size
        // of matrix with sum <= K
        int ans = 0;
 
        // Traverse the sum matrix
        for (i = 1; i <= n; i++) {
            for (j = 1; j <= m; j++) {
 
                // Index out of bound
                if (i + ans - 1 > n || j + ans - 1 > m)
                    break;
 
                int mid, lo = ans;
 
                // Maximum possible size
                // of matrix
                int hi = Math.Min(n - i + 1, m - j + 1);
 
                // Binary Search
                while (lo < hi) {
 
                    // Find middle index
                    mid = (hi + lo + 1) / 2;
 
                    // Check whether sum <= K
                    // or not
                    // If Yes check for other
                    // half of the search
                    if (sum[i + mid - 1, j + mid - 1]
                            + sum[i - 1, j - 1]
                            - sum[i + mid - 1, j - 1]
                            - sum[i - 1, j + mid - 1]
                        <= K) {
                        lo = mid;
                    }
 
                    // Else check it in first
                    // half
                    else {
                        hi = mid - 1;
                    }
                }
 
                // Update the maximum size matrix
                ans = Math.Max(ans, lo);
            }
        }
 
        // Print the readonly answer
        Console.Write(ans + "\n");
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int[, ] arr = { { 1, 1, 3, 2, 4, 3, 2 },
                        { 1, 1, 3, 2, 4, 3, 2 },
                        { 1, 1, 3, 2, 4, 3, 2 } };
 
        // Given target sum
        int K = 4;
 
        // Function Call
        findMaxMatrixSize(arr, K);
    }
}
 
// This code is contributed by Amit Katiyar
Javascript



<script>
// js program for the above approach
 
// Function to find the maximum size
// of matrix with sum <= K
function findMaxMatrixSize(arr, K)
{
 
    let i, j;
 
    // N size of rows and M size of cols
    let n = arr.length;
    let m = arr[0].length;
 
    // To store the prefix sum of matrix
    let sum=[];
    for(i =0;i<n+1;i++){
       sum[i] = [];
       for(j =0;j<m+1;j++){
           sum[i][j] = 0;
       }
    }
    // Create Prefix Sum
    for ( i = 0; i <= n; i++) {
 
        // Traverse each rows
        for (j = 0; j <= m; j++) {
 
            if (i == 0 || j == 0) {
                sum[i][j] = 0;
                continue;
            }
 
            // Update the prefix sum
            // till index i x j
            sum[i][j] = arr[i - 1][j - 1] + sum[i - 1][j]
                        + sum[i][j - 1] - sum[i - 1][j - 1];
        }
    }
 
    // To store the maximum size of
    // matrix with sum <= K
    let ans = 0;
 
    // Traverse the sum matrix
    for (i = 1; i <= n; i++) {
 
        for (j = 1; j <= m; j++) {
 
            // Index out of bound
            if (i + ans - 1 > n || j + ans - 1 > m)
                break;
 
            let mid, lo = ans;
 
            // Maximum possible size
            // of matrix
            let hi = Math.min(n - i + 1, m - j + 1);
 
            // Binary Search
            while (lo < hi) {
 
                // Find middle index
                mid = Math.floor((hi + lo + 1) / 2);
 
                // Check whether sum <= K
                // or not
                // If Yes check for other
                // half of the search
                if (sum[i + mid - 1][j + mid - 1]
                        + sum[i - 1][j - 1]
                        - sum[i + mid - 1][j - 1]
                        - sum[i - 1][j + mid - 1]
                    <= K) {
                    lo = mid;
                }
 
                // Else check it in first
                // half
                else {
                    hi = mid - 1;
                }
            }
 
            // Update the maximum size matrix
            ans = Math.max(ans, lo);
        }
    }
 
    // Print the final answer
    document.write(ans ,'<br>');
}
 
// Driver Code
 
    let arr = [[ 1, 1, 3, 2, 4, 3, 2 ],
            [ 1, 1, 3, 2, 4, 3, 2 ],
            [ 1, 1, 3, 2, 4, 3, 2 ]];
 
    // Given target sum
    let K = 4;
 
    // Function Call
    findMaxMatrixSize(arr, K);
</script>
C++



// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum size
// of matrix with sum <= K
void findMaxMatrixSize(vector<vector<int> > arr, int K)
{
 
    int i, j;
 
    // N size of rows and M size of cols
    int n = arr.size();
    int m = arr[0].size();
 
    // To store the prefix sum of matrix
    int sum[n + 1][m + 1];
 
    // Create Prefix Sum
    for (int i = 0; i <= n; i++) {
 
        // Traverse each rows
        for (int j = 0; j <= m; j++) {
 
            if (i == 0 || j == 0) {
                sum[i][j] = 0;
                continue;
            }
 
            // Update the prefix sum
            // till index i x j
            sum[i][j] = arr[i - 1][j - 1] + sum[i - 1][j]
                        + sum[i][j - 1] - sum[i - 1][j - 1];
        }
    }
 
    // To store the maximum size of
    // matrix with sum <= K
    int ans = 0;
 
    // Traverse the sum matrix
    for (i = 1; i <= n; i++) {
 
        for (j = 1; j <= m; j++) {
 
            // Index out of bound
            if (i + ans - 1 > n || j + ans - 1 > m)
                break;
 
            int mid, lo = ans;
 
            // Maximum possible size
            // of matrix
            int hi = min(n - i + 1, m - j + 1);
 
            // Binary Search
            while (lo < hi) {
 
                // Find middle index
                mid = (hi + lo + 1) / 2;
 
                // Check whether sum <= K
                // or not
                // If Yes check for other
                // half of the search
                if (sum[i + mid - 1][j + mid - 1]
                        + sum[i - 1][j - 1]
                        - sum[i + mid - 1][j - 1]
                        - sum[i - 1][j + mid - 1]
                    <= K) {
                    lo = mid;
                }
 
                // Else check it in first
                // half
                else {
                    hi = mid - 1;
                }
            }
 
            // Update the maximum size matrix
            ans = max(ans, lo);
        }
    }
 
    // Print the final answer
    cout << ans << endl;
}
 
// Driver Code
int main()
{
    vector<vector<int> > arr;
 
    arr = { { 1, 1, 3, 2, 4, 3, 2 },
            { 1, 1, 3, 2, 4, 3, 2 },
            { 1, 1, 3, 2, 4, 3, 2 } };
 
    // Given target sum
    int K = 4;
 
    // Function Call
    findMaxMatrixSize(arr, K);
    return 0;
}
Output
2
Time Complexity: O(N*N*log(N)) 
Auxiliary Space: O(M*N)

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Article Contributed By :
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prajwalkp
@prajwalkp
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