Maximum size of square such that all submatrices of that size have sum less than K

Given an N x M matrix of integers and an integer K, the task is to find the size of the maximum square sub-matrix (S x S), such that all square sub-matrices of the given matrix of that size have a sum less than K.
Examples: 
 

Input: K = 30 
mat[N][M] = {{1, 2, 3, 4, 6}, 
             {5, 3, 8, 1, 2}, 
             {4, 6, 7, 5, 5}, 
             {2, 4, 8, 9, 4} }; 
Output: 2
Explanation:
All Sub-matrices of size 2 x 2 
have sum less than 30

Input : K = 100 
mat[N][M] = { { 1, 2, 3, 4 },
              { 5, 6, 7, 8 }, 
              { 9, 10, 11, 12 }, 
              { 13, 14, 15, 16 } };
Output: 3
Explanation:
All Sub-matrices of size 3 x 3 
have sum less than 100

Naive Approach The basic solution is to choose the size S of the submatrix and find all the submatrices of that size and check that the sum of all sub-matrices is less than the given sum whereas, this can be improved by computing the sum of the matrix using this approach. Therefore, the task will be to choose the maximum size possible and the starting and ending position of the every possible sub-matrices. Due to which the overall time complexity will be O(N3).
Below is the implementation of the above approach:
 

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// C++ implementation to find the
// maximum size square submatrix
// such that their sum is less than K
  
#include <bits/stdc++.h>
  
using namespace std;
  
// Size of matrix
#define N 4
#define M 5
  
// Function to preprocess the matrix
// for computing the sum of every 
// possible matrix of the given size
void preProcess(int mat[N][M], 
                int aux[N][M])
{
    // Loop to copy the first row
    // of the matrix into the aux matrix
    for (int i = 0; i < M; i++)
        aux[0][i] = mat[0][i];
      
    // Computing the sum column-wise
    for (int i = 1; i < N; i++)
        for (int j = 0; j < M; j++)
            aux[i][j] = mat[i][j] + 
                      aux[i - 1][j];
  
    // Computing row wise sum
    for (int i = 0; i < N; i++)
        for (int j = 1; j < M; j++)
            aux[i][j] += aux[i][j - 1];
}
  
// Function to find the sum of a
// submatrix with the given indices
int sumQuery(int aux[N][M], int tli, 
          int tlj, int rbi, int rbj)
{
    // Overall sum from the top to 
    // right corner of matrix
    int res = aux[rbi][rbj];
      
    // Removing the sum from the top
    // corer of the matrix
    if (tli > 0)
        res = res - aux[tli - 1][rbj];
      
    // Remove the overlapping sum
    if (tlj > 0)
        res = res - aux[rbi][tlj - 1];
      
    // Add the sum of top corner 
    // which is substracted twice
    if (tli > 0 && tlj > 0)
        res = res + 
           aux[tli - 1][tlj - 1];
  
    return res;
}
  
// Function to find the maximum
// square size possible with the 
// such that every submatrix have 
// sum less than the given sum
int maximumSquareSize(int mat[N][M], int K)
{
    int aux[N][M];
    preProcess(mat, aux);
      
    // Loop to choose the size of matrix
    for (int i = min(N, M); i >= 1; i--) {
  
        bool satisfies = true;
          
        // Loop to find the sum of the
        // matrix of every possible submatrix
        for (int x = 0; x < N; x++) {
            for (int y = 0; y < M; y++) {
                if (x + i - 1 <= N - 1 && 
                     y + i - 1 <= M - 1) {
                    if (sumQuery(aux, x, y, 
                   x + i - 1, y + i - 1) > K)
                        satisfies = false;
                }
            }
        }
        if (satisfies == true)
            return (i);
    }
    return 0;
}
  
// Driver Code
int main()
{
    int K = 30;
    int mat[N][M] = { { 1, 2, 3, 4, 6 },
                    { 5, 3, 8, 1, 2 },
                    { 4, 6, 7, 5, 5 },
                    { 2, 4, 8, 9, 4 } };
  
    cout << maximumSquareSize(mat, K);
    return 0;
}
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// Java implementation to find the
// maximum size square submatrix
// such that their sum is less than K
class GFG{
   
// Size of matrix
static final int N = 4;
static final int M = 5;
   
// Function to preprocess the matrix
// for computing the sum of every 
// possible matrix of the given size
static void preProcess(int [][]mat, 
                int [][]aux)
{
    // Loop to copy the first row
    // of the matrix into the aux matrix
    for (int i = 0; i < M; i++)
        aux[0][i] = mat[0][i];
       
    // Computing the sum column-wise
    for (int i = 1; i < N; i++)
        for (int j = 0; j < M; j++)
            aux[i][j] = mat[i][j] + 
                      aux[i - 1][j];
   
    // Computing row wise sum
    for (int i = 0; i < N; i++)
        for (int j = 1; j < M; j++)
            aux[i][j] += aux[i][j - 1];
}
   
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [][]aux, int tli, 
          int tlj, int rbi, int rbj)
{
    // Overall sum from the top to 
    // right corner of matrix
    int res = aux[rbi][rbj];
       
    // Removing the sum from the top
    // corer of the matrix
    if (tli > 0)
        res = res - aux[tli - 1][rbj];
       
    // Remove the overlapping sum
    if (tlj > 0)
        res = res - aux[rbi][tlj - 1];
       
    // Add the sum of top corner 
    // which is substracted twice
    if (tli > 0 && tlj > 0)
        res = res + 
           aux[tli - 1][tlj - 1];
   
    return res;
}
   
// Function to find the maximum
// square size possible with the 
// such that every submatrix have 
// sum less than the given sum
static int maximumSquareSize(int [][]mat, int K)
{
    int [][]aux = new int[N][M];
    preProcess(mat, aux);
       
    // Loop to choose the size of matrix
    for (int i = Math.min(N, M); i >= 1; i--) {
   
        boolean satisfies = true;
           
        // Loop to find the sum of the
        // matrix of every possible submatrix
        for (int x = 0; x < N; x++) {
            for (int y = 0; y < M; y++) {
                if (x + i - 1 <= N - 1 && 
                     y + i - 1 <= M - 1) {
                    if (sumQuery(aux, x, y, 
                   x + i - 1, y + i - 1) > K)
                        satisfies = false;
                }
            }
        }
        if (satisfies == true)
            return (i);
    }
    return 0;
}
   
// Driver Code
public static void main(String[] args)
{
    int K = 30;
    int mat[][] = { { 1, 2, 3, 4, 6 },
                    { 5, 3, 8, 1, 2 },
                    { 4, 6, 7, 5, 5 },
                    { 2, 4, 8, 9, 4 } };
   
    System.out.print(maximumSquareSize(mat, K));
}
}
  
// This code is contributed by PrinciRaj1992
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# Python3 implementation to find the
# maximum size square submatrix
# such that their sum is less than K
  
# Size of matrix
N = 4
M = 5
  
# Function to preprocess the matrix
# for computing the sum of every 
# possible matrix of the given size
def preProcess(mat, aux):
  
    # Loop to copy the first row
    # of the matrix into the aux matrix
    for i in range (M):
        aux[0][i] = mat[0][i]
      
    # Computing the sum column-wise
    for i in range (1, N):
        for j in range (M):
            aux[i][j] = (mat[i][j] +
                         aux[i - 1][j])
  
    # Computing row wise sum
    for i in range (N):
        for j in range (1, M):
            aux[i][j] += aux[i][j - 1]
  
# Function to find the sum of a
# submatrix with the given indices
def sumQuery(aux, tli, tlj, rbi, rbj):
  
    # Overall sum from the top to 
    # right corner of matrix
    res = aux[rbi][rbj]
      
    # Removing the sum from the top
    # corer of the matrix
    if (tli > 0):
        res = res - aux[tli - 1][rbj]
      
    # Remove the overlapping sum
    if (tlj > 0):
        res = res - aux[rbi][tlj - 1]
      
    # Add the sum of top corner 
    # which is substracted twice
    if (tli > 0 and tlj > 0):
        res = (res +
        aux[tli - 1][tlj - 1])
  
    return res
  
# Function to find the maximum
# square size possible with the 
# such that every submatrix have 
# sum less than the given sum
def maximumSquareSize(mat, K):
  
    aux = [[0 for x in range (M)]
              for y in range (N)]
    preProcess(mat, aux)
      
    # Loop to choose the size of matrix
    for i in range (min(N, M), 0, -1):
  
        satisfies = True
          
        # Loop to find the sum of the
        # matrix of every possible submatrix
        for x in range (N):
            for y in range (M) :
                if (x + i - 1 <= N - 1 and
                    y + i - 1 <= M - 1):
                    if (sumQuery(aux, x, y, 
                                 x + i - 1
                                 y + i - 1) > K):
                        satisfies = False
              
        if (satisfies == True):
            return (i)
    return 0
  
# Driver Code
if __name__ == "__main__":
  
    K = 30
    mat = [[1, 2, 3, 4, 6],
           [5, 3, 8, 1, 2],
           [4, 6, 7, 5, 5],
           [2, 4, 8, 9, 4]]
  
    print( maximumSquareSize(mat, K))
  
# This code is contributed by Chitranayal
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// C# implementation to find the
// maximum size square submatrix
// such that their sum is less than K
using System;
  
public class GFG{
    
// Size of matrix
static readonly int N = 4;
static readonly int M = 5;
    
// Function to preprocess the matrix
// for computing the sum of every 
// possible matrix of the given size
static void preProcess(int [,]mat, 
                int [,]aux)
{
    // Loop to copy the first row
    // of the matrix into the aux matrix
    for (int i = 0; i < M; i++)
        aux[0,i] = mat[0,i];
        
    // Computing the sum column-wise
    for (int i = 1; i < N; i++)
        for (int j = 0; j < M; j++)
            aux[i,j] = mat[i,j] + 
                      aux[i - 1,j];
    
    // Computing row wise sum
    for (int i = 0; i < N; i++)
        for (int j = 1; j < M; j++)
            aux[i,j] += aux[i,j - 1];
}
    
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [,]aux, int tli, 
          int tlj, int rbi, int rbj)
{
    // Overall sum from the top to 
    // right corner of matrix
    int res = aux[rbi,rbj];
        
    // Removing the sum from the top
    // corer of the matrix
    if (tli > 0)
        res = res - aux[tli - 1,rbj];
        
    // Remove the overlapping sum
    if (tlj > 0)
        res = res - aux[rbi,tlj - 1];
        
    // Add the sum of top corner 
    // which is substracted twice
    if (tli > 0 && tlj > 0)
        res = res + 
           aux[tli - 1,tlj - 1];
    
    return res;
}
    
// Function to find the maximum
// square size possible with the 
// such that every submatrix have 
// sum less than the given sum
static int maximumSquareSize(int [,]mat, int K)
{
    int [,]aux = new int[N,M];
    preProcess(mat, aux);
        
    // Loop to choose the size of matrix
    for (int i = Math.Min(N, M); i >= 1; i--) {
    
        bool satisfies = true;
            
        // Loop to find the sum of the
        // matrix of every possible submatrix
        for (int x = 0; x < N; x++) {
            for (int y = 0; y < M; y++) {
                if (x + i - 1 <= N - 1 && 
                     y + i - 1 <= M - 1) {
                    if (sumQuery(aux, x, y, 
                   x + i - 1, y + i - 1) > K)
                        satisfies = false;
                }
            }
        }
        if (satisfies == true)
            return (i);
    }
    return 0;
}
    
// Driver Code
public static void Main(String[] args)
{
    int K = 30;
    int [,]mat = { { 1, 2, 3, 4, 6 },
                    { 5, 3, 8, 1, 2 },
                    { 4, 6, 7, 5, 5 },
                    { 2, 4, 8, 9, 4 } };
    
    Console.Write(maximumSquareSize(mat, K));
}
}
   
  
// This code contributed by PrinciRaj1992
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Output: 
2

 

Efficient Approach: The key observation is, if a square of side s is the maximum size satisfying the condition, then all sizes smaller than it will satisfy the condition. Using this we can reduce our search space at each step by half which is precisely the idea of Binary Search. Below is the illustration of the steps of the approach: 
 

low = 1
high = min(N, M)

if checkSubmatrix(mat, mid, K):
    low = mid + 1

if not checkSubmatrix(mat, mid, K):
    high = mid - 1

Below is the implementation of the above approach:
 



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// C++ implementation to find the
// maximum size square submatrix
// such that their sum is less than K
  
#include <bits/stdc++.h>
  
using namespace std;
  
// Size of matrix
#define N 4
#define M 5
  
// Function to preprocess the matrix
// for computing the sum of every 
// possible matrix of the given size
void preProcess(int mat[N][M], 
                     int aux[N][M])
{
    // Loop to copy the first row
    // of the matrix into the aux matrix
    for (int i = 0; i < M; i++)
        aux[0][i] = mat[0][i];
  
    // Computing the sum column-wise
    for (int i = 1; i < N; i++)
        for (int j = 0; j < M; j++)
            aux[i][j] = mat[i][j] + 
                       aux[i - 1][j];
  
    // Computing row wise sum
    for (int i = 0; i < N; i++)
        for (int j = 1; j < M; j++)
            aux[i][j] += aux[i][j - 1];
}
  
// Function to find the sum of a
// submatrix with the given indices
int sumQuery(int aux[N][M], int tli, 
          int tlj, int rbi, int rbj)
{
    // Overall sum from the top to 
    // right corner of matrix
    int res = aux[rbi][rbj];
  
    // Removing the sum from the top
    // corer of the matrix
    if (tli > 0)
        res = res - aux[tli - 1][rbj];
  
    // Remove the overlapping sum
    if (tlj > 0)
        res = res - aux[rbi][tlj - 1];
  
    // Add the sum of top corner 
    // which is substracted twice
    if (tli > 0 && tlj > 0)
        res = res + aux[tli - 1][tlj - 1];
  
    return res;
}
  
// Function to check whether square
// sub matrices of size mid satisfy 
// the condition or not
bool check(int mid, int aux[N][M], 
                           int K)
{
  
    bool satisfies = true;
      
    // Iterating throught all possible
    // submatrices of given size
    for (int x = 0; x < N; x++) {
        for (int y = 0; y < M; y++) {
            if (x + mid - 1 <= N - 1 && 
                  y + mid - 1 <= M - 1) {
                if (sumQuery(aux, x, y, 
          x + mid - 1, y + mid - 1) > K)
                    satisfies = false;
            }
        }
    }
    return (satisfies == true);
}
// Function to find the maximum
// square size possible with the 
// such that every submatrix have 
// sum less than the given sum
int maximumSquareSize(int mat[N][M], 
                              int K)
{
    int aux[N][M];
  
    preProcess(mat, aux);
      
    // Search space
    int low = 1, high = min(N, M);
    int mid;
      
    // Binary search for size 
    while (high - low > 1) {
        mid = (low + high) / 2;
          
        // Check if the mid satisfies
        // the given condition
        if (check(mid, aux, K)) {
            low = mid;
        }
        else
            high = mid;
    }
    if (check(high, aux, K))
        return high;
    return low;
}
  
// Driver Code
int main()
{
    int K = 30;
    int mat[N][M] = { { 1, 2, 3, 4, 6 },
                    { 5, 3, 8, 1, 2 },
                    { 4, 6, 7, 5, 5 },
                    { 2, 4, 8, 9, 4 } };
  
    cout << maximumSquareSize(mat, K);
    return 0;
}
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// Java implementation to find the
// maximum size square submatrix
// such that their sum is less than K
class GFG{
  
// Size of matrix
static final int N = 4;
static final int M = 5;
  
// Function to preprocess the matrix
// for computing the sum of every 
// possible matrix of the given size
static void preProcess(int [][]mat, 
                       int [][]aux)
{
      
    // Loop to copy the first row of 
    // the matrix into the aux matrix
    for(int i = 0; i < M; i++)
       aux[0][i] = mat[0][i];
  
    // Computing the sum column-wise
    for(int i = 1; i < N; i++)
       for(int j = 0; j < M; j++)
          aux[i][j] = mat[i][j] + 
                      aux[i - 1][j];
  
    // Computing row wise sum
    for(int i = 0; i < N; i++)
       for(int j = 1; j < M; j++)
          aux[i][j] += aux[i][j - 1];
}
  
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [][]aux, int tli, 
                    int tlj, int rbi, int rbj)
{
      
    // Overall sum from the top to 
    // right corner of matrix
    int res = aux[rbi][rbj];
  
    // Removing the sum from the top
    // corer of the matrix
    if (tli > 0)
        res = res - aux[tli - 1][rbj];
  
    // Remove the overlapping sum
    if (tlj > 0)
        res = res - aux[rbi][tlj - 1];
  
    // Add the sum of top corner 
    // which is substracted twice
    if (tli > 0 && tlj > 0)
        res = res + aux[tli - 1][tlj - 1];
  
    return res;
}
  
// Function to check whether square
// sub matrices of size mid satisfy 
// the condition or not
static boolean check(int mid, int [][]aux, 
                     int K)
{
  
    boolean satisfies = true;
      
    // Iterating throught all possible
    // submatrices of given size
    for(int x = 0; x < N; x++)
    {
       for(int y = 0; y < M; y++)
       {
          if (x + mid - 1 <= N - 1 && 
              y + mid - 1 <= M - 1)
          {
              if (sumQuery(aux, x, y, 
                           x + mid - 1,
                           y + mid - 1) > K)
                  satisfies = false;
          }
       }
    }
    return (satisfies == true);
}
  
// Function to find the maximum
// square size possible with the 
// such that every submatrix have 
// sum less than the given sum
static int maximumSquareSize(int [][]mat, 
                             int K)
{
    int [][]aux = new int[N][M];
  
    preProcess(mat, aux);
      
    // Search space
    int low = 1, high = Math.min(N, M);
    int mid;
      
    // Binary search for size 
    while (high - low > 1)
    {
        mid = (low + high) / 2;
          
        // Check if the mid satisfies
        // the given condition
        if (check(mid, aux, K)) 
        {
            low = mid;
        }
        else
            high = mid;
    }
    if (check(high, aux, K))
        return high;
    return low;
}
  
// Driver Code
public static void main(String[] args)
{
    int K = 30;
    int [][]mat = { { 1, 2, 3, 4, 6 },
                    { 5, 3, 8, 1, 2 },
                    { 4, 6, 7, 5, 5 },
                    { 2, 4, 8, 9, 4 } };
  
    System.out.print(maximumSquareSize(mat, K));
}
}
  
// This code is contributed by Rajput-Ji
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// C# implementation to find the
// maximum size square submatrix
// such that their sum is less than K
using System;
  
class GFG{
  
// Size of matrix
static readonly int N = 4;
static readonly int M = 5;
  
// Function to preprocess the matrix
// for computing the sum of every 
// possible matrix of the given size
static void preProcess(int [,]mat, 
                       int [,]aux)
{
      
    // Loop to copy the first row of 
    // the matrix into the aux matrix
    for(int i = 0; i < M; i++)
       aux[0, i] = mat[0, i];
  
    // Computing the sum column-wise
    for(int i = 1; i < N; i++)
       for(int j = 0; j < M; j++)
          aux[i, j] = mat[i, j] + 
                      aux[i - 1, j];
  
    // Computing row wise sum
    for(int i = 0; i < N; i++)
       for(int j = 1; j < M; j++)
          aux[i, j] += aux[i, j - 1];
}
  
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [,]aux, int tli, 
                    int tlj, int rbi, int rbj)
{
      
    // Overall sum from the top to 
    // right corner of matrix
    int res = aux[rbi, rbj];
  
    // Removing the sum from the top
    // corer of the matrix
    if (tli > 0)
        res = res - aux[tli - 1, rbj];
  
    // Remove the overlapping sum
    if (tlj > 0)
        res = res - aux[rbi, tlj - 1];
  
    // Add the sum of top corner 
    // which is substracted twice
    if (tli > 0 && tlj > 0)
        res = res + aux[tli - 1, tlj - 1];
  
    return res;
}
  
// Function to check whether square
// sub matrices of size mid satisfy 
// the condition or not
static bool check(int mid, int [,]aux, 
                  int K)
{
  
    bool satisfies = true;
      
    // Iterating throught all possible
    // submatrices of given size
    for(int x = 0; x < N; x++)
    {
       for(int y = 0; y < M; y++)
       {
          if (x + mid - 1 <= N - 1 && 
              y + mid - 1 <= M - 1)
          {
              if (sumQuery(aux, x, y, 
                           x + mid - 1,
                           y + mid - 1) > K)
                  satisfies = false;
          }
       }
    }
    return (satisfies == true);
}
  
// Function to find the maximum
// square size possible with the 
// such that every submatrix have 
// sum less than the given sum
static int maximumSquareSize(int [,]mat, 
                             int K)
{
    int [,]aux = new int[N, M];
  
    preProcess(mat, aux);
      
    // Search space
    int low = 1, high = Math.Min(N, M);
    int mid;
      
    // Binary search for size 
    while (high - low > 1)
    {
        mid = (low + high) / 2;
          
        // Check if the mid satisfies
        // the given condition
        if (check(mid, aux, K)) 
        {
            low = mid;
        }
        else
            high = mid;
    }
    if (check(high, aux, K))
        return high;
    return low;
}
  
// Driver Code
public static void Main(String[] args)
{
    int K = 30;
    int [,]mat = { { 1, 2, 3, 4, 6 },
                   { 5, 3, 8, 1, 2 },
                   { 4, 6, 7, 5, 5 },
                   { 2, 4, 8, 9, 4 } };
  
    Console.Write(maximumSquareSize(mat, K));
}
}
  
// This code is contributed by Rajput-Ji
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Output: 
2

 

Performance Analysis: 
 

 





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