Largest Square in a Binary Matrix with at most K 1s for multiple Queries

Given a binary matrix M where each element of the matrix will be 0 or 1, the task is to find the largest square that can be formed with center (i, j) and contains atmost K 1’s.

Input: M[][] = {
{1, 0, 1, 0, 0}
{1, 0, 1, 1, 1}
{1, 1, 1, 1, 1}
{1, 0, 0, 1, 0}},
i = 1, j = 2, K = 9
Output: 3
Explanation:
Maximum length square with center at (1, 2)
that can be formed is of 3 length from (0, 1) to (2, 4)

Input: M[][] = {
{ 1, 1, 1 },
{ 1, 1, 1 },
{ 1, 1, 1 }},
i = 1, j = 1, K = 9
Output: 3

Naive Approach:
For each query, consider all squares with center (i, j) and keep incrementing the length of the squares from 1 to its maximum possible value up to which the count of 1s in the square is less than K. The maximum possible value of the length of the square will be 2*MIN_DIST + 1 where MIN_DIST is the minimum distance of the center from the edges of the matrix. So for an R*C matrix with center (i, j),

MIN_DIST = min( i, j, R-i-1, C-j-1 ) 

Below is the implementation of the above approach:



C++

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// C++ implementation to find the 
// largest square in the matrix such
// that it contains atmost K 1's
  
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
  
// Function to calculate the
// largest square with atmost K
// 1s for Q queries
void largestSquare(int matrix[][MAX],
             int R, int C, int q_i[],
            int q_j[], int K, int Q){
                  
    // Loop to solve for each query
    for (int q = 0; q < Q; q++) {
        int i = q_i[q];
        int j = q_j[q];
        int min_dist = min(min(i, j),
          min(R - i - 1, C - j - 1));
        int ans = -1;
        for (int k = 0; k <= min_dist;
                                 k++) {
            int count = 0;
            // Traversing the each sub
            // square and counting total
            for (int row = i - k; 
              row <= i + k; row++)
                for (int col = j - k;
                  col <= j + k; col++)
                    count += matrix[row][col];
              
            // Breaks when exceeds
            // the maximum count
            if (count > K)
                break;
              
            ans = 2 * k + 1;
        }
        cout << ans << "\n";
    }
}
  
// Driver Code
int main()
{
    int matrix[][MAX] = { { 1, 0, 1, 0, 0 },
                        { 1, 0, 1, 1, 1 },
                        { 1, 1, 1, 1, 1 },
                        { 1, 0, 0, 1, 0 } };
    int K = 9, Q = 1;
    int q_i[] = { 1 };
    int q_j[] = { 2 };
    largestSquare(matrix, 4, 5, q_i,
                          q_j, K, Q);
    return 0;
}

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Java

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// Java implementation to find the 
// largest square in the matrix such
// that it contains atmost K 1's
class GFG{
      
static int MAX = 100;
  
// Function to calculate the
// largest square with atmost K
// 1s for Q queries
static void largestSquare(int matrix[][], 
                          int R, int C, 
                          int q_i[], int q_j[], 
                          int K, int Q)
{
      
    // Loop to solve for each query
    for(int q = 0; q < Q; q++) 
    {
       int i = q_i[q];
       int j = q_j[q];
       int min_dist = Math.min(Math.min(i, j),
                      Math.min(R - i - 1
                               C - j - 1));
       int ans = -1;
       for(int k = 0; k <= min_dist; k++)
       {
          int count = 0;
            
          // Traversing the each sub
          // square and counting total
          for(int row = i - k; row <= i + k; row++)
             for(int col = j - k; col <= j + k; col++)
                count += matrix[row][col];
              
            // Breaks when exceeds
            // the maximum count
            if (count > K)
                break;
              
            ans = 2 * k + 1;
        }
          
        System.out.print(ans + "\n");
    }
}
  
// Driver Code
public static void main(String[] args)
{
    int matrix[][] = { { 1, 0, 1, 0, 0 },
                       { 1, 0, 1, 1, 1 },
                       { 1, 1, 1, 1, 1 },
                       { 1, 0, 0, 1, 0 } };
    int K = 9, Q = 1;
    int q_i[] = { 1 };
    int q_j[] = { 2 };
    largestSquare(matrix, 4, 5, q_i, q_j, K, Q);
}
}
  
// This code is contributed by Amit Katiyar

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Python3

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# Python3 implementation to find the 
# largest square in the matrix such 
# that it contains at most K 1's 
  
MAX = 100
  
# Function to calculate the 
# largest square with atmost K 
# 1s for Q queries 
def largestSquare(matrix, R, C, q_i, q_j, K, Q): 
                  
    # Loop to solve for each query 
    for q in range(Q): 
        i = q_i[q] 
        j = q_j[q] 
        min_dist = min(min(i, j), 
                   min(R - i - 1, C - j - 1)) 
        ans = -1
        for k in range(min_dist + 1):
              
            count = 0
              
            # Traversing the each sub 
            # square and counting total 
            for row in range(i - k, i + k + 1): 
                for col in range(j - k, j + k + 1): 
                    count += matrix[row][col] 
              
            # Breaks when exceeds 
            # the maximum count 
            if count > K: 
                break
              
            ans = 2 * k + 1
        print(ans) 
          
# Driver Code 
matrix = [ [ 1, 0, 1, 0, 0 ], 
           [ 1, 0, 1, 1, 1 ], 
           [ 1, 1, 1, 1, 1 ], 
           [ 1, 0, 0, 1, 0 ] ] 
K = 9
Q = 1
q_i = [ 1
q_j = [ 2
largestSquare(matrix, 4, 5, q_i, q_j, K, Q)
                          
# This code is contributed by divyamohan123                     
                     

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C#

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// C# implementation to find the 
// largest square in the matrix such
// that it contains atmost K 1's
using System;
  
class GFG{
      
//static int MAX = 100;
  
// Function to calculate the
// largest square with atmost K
// 1s for Q queries
static void largestSquare(int [,]matrix, 
                          int R, int C, 
                          int []q_i, int []q_j, 
                          int K, int Q)
{
      
    // Loop to solve for each query
    for(int q = 0; q < Q; q++) 
    {
       int i = q_i[q];
       int j = q_j[q];
       int min_dist = Math.Min(Math.Min(i, j), 
                               Math.Min(R - i - 1,
                                        C - j - 1));
       int ans = -1;
       for(int k = 0; k <= min_dist; k++)
       {
          int count = 0;
            
          // Traversing the each sub
          // square and counting total
          for(int row = i - k; row <= i + k; row++)
             for(int col = j - k; col <= j + k; col++)
                count += matrix[row, col];
                  
          // Breaks when exceeds
          // the maximum count
          if (count > K)
              break;
            
          ans = 2 * k + 1;
       }
       Console.Write(ans + "\n");
    }
}
  
// Driver Code
public static void Main(String[] args)
{
    int [,]matrix = { { 1, 0, 1, 0, 0 },
                      { 1, 0, 1, 1, 1 },
                      { 1, 1, 1, 1, 1 },
                      { 1, 0, 0, 1, 0 } };
    int K = 9, Q = 1;
    int []q_i = {1};
    int []q_j = {2};
      
    largestSquare(matrix, 4, 5, q_i, q_j, K, Q);
}
}
  
// This code is contributed by Amit Katiyar

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Output :

3

The worst-case time complexity for the given solution is O(Q*N*N*MIN_DIST) where N is the length of the square(which is maximum 2*MIN_DIST + 1).

Efficient Approach using Dynamic Programming:
The idea is to use Dynamic Programming to count the number of 1s in each square and then increment the length by 1 until the limit and then finally check the count of the 1s is less than the K or not. If yes, then update the answer.

To compute the number of 1s in a submatrix from (x1, y1) to (x2, y2) using:

Number of 1's = sumDP[x2][y2] - sumDP[x2][y1-1] -
                sumDP[x1-1][y2] + sumDP[x1-1][y1-1]

Below is the implementation of the above approach:

C++

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// C++ implementation to find the 
// largest square in the matrix such
// that it contains atmost K 1's
  
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
  
// Function to find the 
// largest square in the matrix such
// that it contains atmost K 1's
void largestSquare(int matrix[][MAX],
            int R, int C, int q_i[],
            int q_j[], int K, int Q){
      
    int countDP[R][C];
    memset(countDP, 0, sizeof(countDP));
  
    // Precomputing the countDP 
    // prefix sum of the matrix
    countDP[0][0] = matrix[0][0];
    for (int i = 1; i < R; i++)
        countDP[i][0] = countDP[i - 1][0] + 
                             matrix[i][0];
    for (int j = 1; j < C; j++)
        countDP[0][j] = countDP[0][j - 1] +
                             matrix[0][j];
    for (int i = 1; i < R; i++)
        for (int j = 1; j < C; j++)
            countDP[i][j] = matrix[i][j] +
                       countDP[i - 1][j] +
                       countDP[i][j - 1] -
                       countDP[i - 1][j - 1];
      
    // Loop to solve Queries
    for (int q = 0; q < Q; q++) {
        int i = q_i[q];
        int j = q_j[q];
        // Calculating the maximum
        // possible distance of the
        // centre from edge
        int min_dist = min(min(i, j), 
          min(R - i - 1, C - j - 1));
        int ans = -1;
        for (int k = 0; k <= min_dist;
                                  k++) {
            int x1 = i - k, x2 = i + k;
            int y1 = i - k, y2 = j + k;
              
            // Calculating the number
            // of 1s in the submatrix
            int count = countDP[x2][y2];
            if (x1 > 0)
                count -= countDP[x1 - 1][y2];
            if (y1 > 0)
                count -= countDP[x2][y1 - 1];
            if (x1 > 0 && y1 > 0)
                count += countDP[x1 - 1][y1 - 1];
  
            if (count > K)
                break;
            ans = 2 * k + 1;
        }
        cout << ans << "\n";
    }
}
  
// Driver Code
int main()
{
    int matrix[][MAX] = { { 1, 0, 1, 0, 0 },
                        { 1, 0, 1, 1, 1 },
                        { 1, 1, 1, 1, 1 },
                        { 1, 0, 0, 1, 0 } };
  
    int K = 9, Q = 1;
    int q_i[] = { 1 };
    int q_j[] = { 2 };
    largestSquare(matrix, 4, 5, q_i,
                          q_j, K, Q);
    return 0;
}

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Java

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// Java implementation to find the 
// largest square in the matrix such
// that it contains atmost K 1's
import java.util.*;
  
class GFG{
      
static int MAX = 100;
  
// Function to find the 
// largest square in the matrix such
// that it contains atmost K 1's
static void largestSquare(int matrix[][], int R,
                          int C, int q_i[], 
                          int q_j[], int K,
                          int Q)
{
    int [][]countDP = new int[R][C];
  
    // Precomputing the countDP 
    // prefix sum of the matrix
    countDP[0][0] = matrix[0][0];
    for(int i = 1; i < R; i++)
        countDP[i][0] = countDP[i - 1][0] + 
                             matrix[i][0];
    for(int j = 1; j < C; j++)
        countDP[0][j] = countDP[0][j - 1] +
                         matrix[0][j];
    for(int i = 1; i < R; i++)
        for(int j = 1; j < C; j++)
            countDP[i][j] = matrix[i][j] +
                           countDP[i - 1][j] +
                           countDP[i][j - 1] -
                           countDP[i - 1][j - 1];
      
    // Loop to solve Queries
    for(int q = 0; q < Q; q++)
    {
        int i = q_i[q];
        int j = q_j[q];
          
        // Calculating the maximum
        // possible distance of the
        // centre from edge
        int min_dist = Math.min(Math.min(i, j), 
                       Math.min(R - i - 1
                                C - j - 1));
                                  
        int ans = -1;
        for(int k = 0; k <= min_dist; k++)
        {
            int x1 = i - k, x2 = i + k;
            int y1 = i - k, y2 = j + k;
              
            // Calculating the number
            // of 1s in the submatrix
            int count = countDP[x2][y2];
            if (x1 > 0)
                count -= countDP[x1 - 1][y2];
            if (y1 > 0)
                count -= countDP[x2][y1 - 1];
            if (x1 > 0 && y1 > 0)
                count += countDP[x1 - 1][y1 - 1];
            if (count > K)
                break;
            ans = 2 * k + 1;
        }
        System.out.print(ans + "\n");
    }
}
  
// Driver Code
public static void main(String[] args)
{
    int matrix[][] = { { 1, 0, 1, 0, 0 },
                       { 1, 0, 1, 1, 1 },
                       { 1, 1, 1, 1, 1 },
                       { 1, 0, 0, 1, 0 } };
  
    int K = 9, Q = 1;
    int q_i[] = { 1 };
    int q_j[] = { 2 };
      
    largestSquare(matrix, 4, 5, q_i,
                        q_j, K, Q);
}
}
  
// This code is contributed by gauravrajput1

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Output :

3

The worst-case time complexity for the given solution is O(R*C + Q*MIN_DIST) where R, C is the dimensions of the initial matrix.

Efficient Approach using Dynamic Programming and Binary Search:
The idea is to use a Binary search to find the largest square instead of incrementing the length of a side iteratively and converge towards the side which gives at most K 1’s.

The search space for the binary search will be –

// Minimum possible answer will be
// the square with side 0
l = 0

// Maximum possible will be to include
// the whole square possible from (i, j)
r = min(min(i, j), 
        min(R - i - 1, C - i - 1))

Below is the implementation of the above approach:

C++

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// C++ implementation to find the 
// largest square in the matrix such
// that it contains atmost K 1's
  
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
  
// Function to find the 
// largest square in the matrix such
// that it contains atmost K 1's
void largestSquare(int matrix[][MAX], 
            int R, int C, int q_i[],
            int q_j[], int K, int Q){
      
    int countDP[R][C];
    memset(countDP, 0, sizeof(countDP));
  
    // Precomputation of the countDP
    // prefix sum of the matrix
    countDP[0][0] = matrix[0][0];
    for (int i = 1; i < R; i++)
        countDP[i][0] = countDP[i - 1][0] +
                             matrix[i][0];
    for (int j = 1; j < C; j++)
        countDP[0][j] = countDP[0][j - 1] +
                             matrix[0][j];
    for (int i = 1; i < R; i++)
        for (int j = 1; j < C; j++)
            countDP[i][j] = matrix[i][j] +
                       countDP[i - 1][j] +
                       countDP[i][j - 1] -
                   countDP[i - 1][j - 1];
      
    // Loop to solve each query
    for (int q = 0; q < Q; q++) {
        int i = q_i[q];
        int j = q_j[q];
        int min_dist = min(min(i, j),
          min(R - i - 1, C - j - 1));
        int ans = -1, l = 0, u = min_dist;
          
        // Binary Search to the side which
        // have atmost in K 1's in square
        while (l <= u) {
            int mid = (l + u) / 2;
            int x1 = i - mid, x2 = i + mid;
            int y1 = i - mid, y2 = j + mid;
            // Count total number of 1s in
            // the sub square considered
            int count = countDP[x2][y2];
            if (x1 > 0)
                count -= countDP[x1 - 1][y2];
            if (y1 > 0)
                count -= countDP[x2][y1 - 1];
            if (x1 > 0 && y1 > 0)
                count += countDP[x1 - 1][y1 - 1];
              
            // If the count is less than or
            // equals to the maximum move
            // to right half
            if (count <= K) {
                ans = 2 * mid + 1;
                l = mid + 1;
            }
            else
                u = mid - 1;
        }
        cout << ans << "\n";
    }
}
  
int main()
{
    int matrix[][MAX] = { { 1, 0, 1, 0, 0 },
                        { 1, 0, 1, 1, 1 },
                        { 1, 1, 1, 1, 1 },
                        { 1, 0, 0, 1, 0 } };
  
    int K = 9, Q = 1;
    int q_i[] = { 1 };
    int q_j[] = { 2 };
    largestSquare(matrix, 4, 5,
                q_i, q_j, K, Q);
    return 0;
}

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Time Complexity: O( R*C + Q*log(MIN_DIST) )

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