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

Given a Matrix arr[][] of size MxN having positive integers and a number K, the task is to find the size of 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:
Explanation:
Maximum size square Sub-Matrix 
with sum less than or equals to 22
      1 1
      1 1
Size is 4.

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 ixj.
  2. For each row in prefix sum matrix sum[][] using Binary Search do the following:
    • Start the Binary search with lower limit as 0 end upper limit as maximum size of square matrix can be formed.
    • 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:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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):
            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 - 1)
  
# 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

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C#

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// 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

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Output:

2

Time Complexity: O(N*N*log(N))
Auxiliary Space: O(M*N)

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