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Submatrix of given size with maximum 1’s
  • Difficulty Level : Expert
  • Last Updated : 26 Jan, 2021

Given a binary matrix mat[][] and an integer K, the task is to find the submatrix of size K*K such that it contains maximum number of 1’s in the matrix.

Examples: 

Input: mat[][] = {{1, 0, 1}, {1, 1, 0}, {1, 0, 0}}, K = 2 
Output:
Explanation: 
In the given matrix, there are 4 sub-matrix of order 2*2, 
|1 0| |0 1| |1 1| |1 0| 
|1 1|, |1 0|, |1 0|, |0 0| 
Out of these sub-matrix, two matrix contains 3, 1’s.

Input: mat[][] = {{1, 0}, {0, 1}}, K = 1 
Output:
Explanation: 
In the given matrix, there are 4 sub-matrix of order 1*1, 
|1|, |0|, |1|, |0| 
Out of these sub-matrix, two matrix contains 1, 1’s. 

Approach: The idea is to use the sliding window technique to solve this problem, In this technique, we generally compute the value of one window and then slide the window one-by-one to compute the solution for every window of size K.
To compute the maximum 1’s submatrix, count the number of 1’s in the row for every possible window of size K using the sliding window technique and store the counts of the 1’s in the form of a matrix. 



For Example: 

Let the matrix be {{1,0,1}, {1, 1, 0}} and K = 2

For Row 1 -
Subarray 1: (1, 0), Count of 1 = 1
Subarray 2: (0, 1), Count of 1 = 1

For Row 2 -
Subarray 1: (1, 1), Count of 1 = 2
Subarray 2: (1, 0), Count of 1 = 1

Then the final matrix for count of 1's will be -
[ 1, 1 ]
[ 2, 1 ]

Similarly, apply the sliding window technique on every column on this matrix, to compute the count of 1’s in every possible sub-matrix and take the maximum out of those counts.

Below is the implementation of the above approach: 

C++




// C++ implementation to find the
// maximum count of 1's in
// submatrix of order K
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum
// count of 1's in the
// submatrix of order K
int maxCount(vector<vector<int>> &mat, int k) {
 
    int n = mat.size();
    int m = mat[0].size();
    vector<vector<int>> a;
 
    // Loop to find the count of 1's
    // in every possible windows
    // of rows of matrix
    for (int e = 0; e < n; ++e){
        vector<int> s = mat[e];
        vector<int> q;
        int    c = 0;
         
        // Loop to find the count of
        // 1's in the first window
        int i;
        for (i = 0; i < k; ++i)
            if(s[i] == 1)
                c += 1;
 
        q.push_back(c);
        int p = s[0];
         
        // Loop to find the count of
        // 1's in the remaining windows
        for (int j = i + 1; j < m; ++j) {
            if(s[j] == 1)
                c+= 1;
            if(p == 1)
                c-= 1;
            q.push_back(c);
            p = s[j-k + 1];
        }
        a.push_back(q);
    }
 
    vector<vector<int>> b;
    int max = 0;
     
    // Loop to find the count of 1's
    // in every possible submatrix
    for (int i = 0; i < a[0].size(); ++i) {
        int c = 0;
        int p = a[0][i];
         
        // Loop to find the count of
        // 1's in the first window
        int j;
        for (j = 0; j < k; ++j) {
            c+= a[j][i];
        }
        vector<int> q;
        if (c>max)
            max = c;
        q.push_back(c);
         
        // Loop to find the count of
        // 1's in the remaining windows
        for (int l = j + 1; j < n; ++j) {
            c+= a[l][i];
            c-= p;
            p = a[l-k + 1][i];
            q.push_back(c);
            if (c > max)
                max = c;
        }
 
        b.push_back(q);
    }
 
    return max;
}
 
// Driver code
int main()
{
    vector<vector<int>> mat = {{1, 0, 1}, {1, 1, 0}, {0, 1, 0}};
    int k = 3;
     
    // Function call
    cout<< maxCount(mat, k);
 
    return 0;
}


Java




// Java implementation to find the
// maximum count of 1's in
// submatrix of order K
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to find the maximum
// count of 1's in the
// submatrix of order K
static int maxCount(ArrayList<ArrayList<Integer> > mat, int k)
{
    int n = mat.size();
    int m = mat.get(0).size();
    ArrayList<ArrayList<Integer>> a = new ArrayList<ArrayList<Integer>>();
     
    // Loop to find the count of 1's
    // in every possible windows
    // of rows of matrix
    for(int e = 0; e < n; ++e)
    {
        ArrayList<Integer> s = mat.get(e);
        ArrayList<Integer> q = new ArrayList<Integer>();
        int c = 0;
         
        // Loop to find the count of
        // 1's in the first window
        int i;
        for(i = 0; i < k; ++i)
        {
            if (s.get(i) == 1)
            {
                c += 1;
            }
        }
        q.add(c);
        int p = s.get(0);
         
        // Loop to find the count of
        // 1's in the remaining windows
        for(int j = i + 1; j < m; ++j)
        {
            if (s.get(j) == 1)
            {
                c += 1;
            }
            if (p == 1)
            {
                c -= 1;
            }
            q.add(c);
            p = s.get(j - k + 1);
        }
        a.add(q);
    }
     
    ArrayList<ArrayList<Integer>> b = new ArrayList<ArrayList<Integer>>();
    int max = 0;
     
    // Loop to find the count of 1's
    // in every possible submatrix
    for(int i = 0; i < a.get(0).size(); ++i)
    {
        int c = 0;
        int p = a.get(0).get(i);
         
        // Loop to find the count of
        // 1's in the first window
        int j;
        for(j = 0; j < k; ++j)
        {
            c += a.get(j).get(i);
        }
        ArrayList<Integer> q = new ArrayList<Integer>();
         
        if (c > max)
        {
            max = c;
        }
        q.add(c);
         
        // Loop to find the count of
        // 1's in the remaining windows
        for(int l = j + 1; j < n; ++j)
        {
             
            c += a.get(l).get(i);
            c -= p;
            p = a.get(l - k + 1).get(i);
            q.add(c);
             
            if (c > max)
            {
                max = c;
            }
        }
        b.add(q);
    }
    return max;
}
 
// Driver code
public static void main(String[] args)
{
    ArrayList<ArrayList<Integer>> mat = new ArrayList<ArrayList<Integer>>();
    mat.add(new ArrayList<Integer>(Arrays.asList(1, 0, 1)));
    mat.add(new ArrayList<Integer>(Arrays.asList(1, 1, 0)));
    mat.add(new ArrayList<Integer>(Arrays.asList(0, 1, 0)));
    int k = 3;
     
    // Function call
    System.out.println(maxCount(mat, k));
}
}
 
// This code is contributed by avanitrachhadiya2155


Python3




# Python3 implementation to find the
# maximum count of 1's in
# submatrix of order K
 
# Function to find the maximum
# count of 1's in the
# submatrix of order K
def maxCount(mat, k):
    n, m = len(mat), len(mat[0])
    a =[]
     
    # Loop to find the count of 1's
    # in every possible windows
    # of rows of matrix
    for e in range(n):
        s = mat[e]
        q =[]
        c = 0
         
        # Loop to find the count of
        # 1's in the first window
        for i in range(k):
            if s[i] == 1:
                c += 1
        q.append(c)
        p = s[0]
         
        # Loop to find the count of
        # 1's in the remaining windows
        for j in range(i + 1, m):
            if s[j]==1:
                c+= 1
            if p ==1:
                c-= 1
            q.append(c)
            p = s[j-k + 1]
        a.append(q)
    b =[]
    max = 0
     
    # Loop to find the count of 1's
    # in every possible submatrix
    for i in range(len(a[0])):
        c = 0
        p = a[0][i]
         
        # Loop to find the count of
        # 1's in the first window
        for j in range(k):
            c+= a[j][i]
        q =[]
        if c>max:
            max = c
        q.append(c)
         
        # Loop to find the count of
        # 1's in the remaining windows
        for l in range(j + 1, n):
            c+= a[l][i]
            c-= p
            p = a[l-k + 1][i]
            q.append(c)
            if c > max:
                max = c
        b.append(q)
    return max
     
# Driver Code
if __name__ == "__main__":
    mat = [[1, 0, 1], [1, 1, 0], [0, 1, 0]]
    k = 3
     
    # Function call
    print(maxCount(mat, k))


C#




// C# implementation to find the
// maximum count of 1's in
// submatrix of order K
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the maximum
// count of 1's in the
// submatrix of order K
static int maxCount(List<List<int>> mat, int k)
{
    int n = mat.Count;
    int m = mat[0].Count;
    List<List<int>> a = new List<List<int>>();
     
    // Loop to find the count of 1's
    // in every possible windows
    // of rows of matrix
    for(int e = 0; e < n; ++e)
    {
        List<int> s = mat[e];
        List<int> q = new List<int>();
        int c = 0;
         
        // Loop to find the count of
        // 1's in the first window
        int i;
         
        for(i = 0; i < k; ++i)
        {
            if (s[i] == 1)
            {
                c++;
            }
        }
        q.Add(c);
        int p = s[0];
         
        // Loop to find the count of
        // 1's in the remaining windows
        for(int j = i + 1; j < m; ++j)
        {
            if (s[j] == 1)
            {
                c++;
            }
            if (p == 1)
            {
                c--;
            }
            q.Add(c);
            p = s[j - k + 1];
        }
        a.Add(q);
    }
    List<List<int>> b = new List<List<int>>();
    int max = 0;
     
    // Loop to find the count of 1's
    // in every possible submatrix
    for(int i = 0; i < a[0].Count; ++i)
    {
        int c = 0;
        int p = a[0][i];
         
        // Loop to find the count of
        // 1's in the first window
        int j;
        for(j = 0; j < k; ++j)
        {
            c += a[j][i];
        }
         
        List<int> q = new List<int>();
         
        if (c > max)
        {
            max = c;
        }
        q.Add(c);
         
        // Loop to find the count of
        // 1's in the remaining windows
        for(int l = j + 1; j < n; ++j)
        {
            c += a[l][i];
            c -= p;
            p = a[l - k + 1][i];
            q.Add(c);
             
            if (c > max)
            {
                max = c;
            }
        }
        b.Add(q);
    }
    return max;
}
 
// Driver code
static public void Main()
{
    List<List<int>> mat = new List<List<int>>();
    mat.Add(new List<int>(){1, 0, 1});
    mat.Add(new List<int>(){1, 1, 0});
    mat.Add(new List<int>(){0, 1, 0});
    int k = 3;
     
    // Function call
    Console.WriteLine(maxCount(mat, k));
}
}
 
// This code is contributed by rag2127


Output: 

5

 

Performance Analysis: 

  • Time Complexity: As in the above approach, there are two loops which takes O(N*M) time, Hence the Time Complexity will be O(N*M).
  • Space Complexity: As in the above approach, there is extra space used, Hence the space complexity will be O(N).

Approach 2: [Dynamic Programming method] In this technique, we compute the dp[][] matrix using given mat[][] array.In dp[][] array we compute number of 1’s till the index (i,j) using previous dp[][] value and store it in dp[i][j] .

Algorithm : 

1) Construct a dp[][] matrix and assign all elements to 0

    initial dp[0][0] = mat[0][0]

    a) compute first row and column of the dp matrix:

        i) for first row:

            dp[0][i] = dp[0][i-1] + mat[0][i]

        ii) for first column:

            dp[i][0] = dp[i-1][0] + mat[i][0]

    b) now compute remaining dp matrix from (1,1) to (n,m):

        dp[i][j] = mat[i][j] + dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1]

2)now, we find the maximum 1's in k X k sub matrix:

    a) initially we assign max = dp[k-1][k-1]

    b) now first we have to check maximum for k-1 row and k-1 column:

        i) for k-1 row:

            if dp[k-1][j] - dp[k-1][j-k] > max:

                max = dp[k-1][j] - dp[k-1][j-k]

        ii) for k-1 column:

            if dp[i][k-1] - dp[i-k][k-1] > max:

                max = dp[i][k-1] - dp[i-k][k-1]

    c) now, we check max for (k to n) row and (k to m) column:

        for i from k to n-1:

            for j from k to m-1:

                if dp[i][j] + dp[i-k][j-k] - dp[i-k][j] - dp[i][j-k] > max:

                    max = dp[i][j] + dp[i-k][j-k] - dp[i-k][j] - dp[i][j-k]

 now just return the max value.

Below is the implementation of the above approach:

C++14




// C++14 approach
#include <bits/stdc++.h>
using namespace std;
 
int findMaxK(vector<vector<int>> dp,
             int k, int n, int m)
{
     
    // Assign first kXk matrix initial
    // value as max
    int max_ = dp[k - 1][k - 1];
     
    for(int i = k; i < n; i++)
    {
        int su = dp[i - k][k - 1];
            if (max_ < su)
                max_ = su;
    }
    for(int j = k; j < m; j++)
    {
        int su = dp[k - 1][j - k];
            if (max_< su)
                max_ = su;
    }
    for(int i = k; i < n; i++)
    {
        for(int j = k; j < m; j++)
        {
            int su = dp[i][j] +
                     dp[i - k][j - k] -
                     dp[i - k][j] -
                     dp[i][j - k];
                      
            if( max_ < su)
                max_ = su;
        }
    }
    return max_;
}
 
vector<vector<int>> buildDPdp(vector<vector<int>> mat,
                              int k, int n, int m)
{
     
    // Assign mXn dp list to 0
    vector<vector<int>> dp(n, vector<int>(m, 0));
 
    // Assign initial starting value
    dp[0][0] = mat[0][0];
 
    for(int i = 1; i < m; i++)
        dp[0][i] += (dp[0][i - 1] + mat[0][i]);
 
    for(int i = 1; i < n; i++)
        dp[i][0] += (dp[i - 1][0] + mat[i][0]);
 
    for(int i = 1; i < n; i++)
        for(int j = 1; j < m; j++)
            dp[i][j] = dp[i - 1][j] +
                       dp[i][j - 1] +
                       mat[i][j] -
                       dp[i - 1][j - 1];
 
    return dp;
}
 
int maxOneInK(vector<vector<int>> mat, int k)
{
     
    // n is colums
    int n = mat.size();
 
    // m is rows
    int m = mat[0].size();
 
    // Build dp list
    vector<vector<int>> dp = buildDPdp(
        mat, k, n, m);
 
    // Call the function and return its value
    return findMaxK(dp, k, n, m);
}
 
// Driver Code
int main()
{
     
    // mXn matrix
    vector<vector<int>> mat = { { 1, 0, 1 },
                                { 1, 1, 0 },
                                { 0, 1, 0 } };
 
    int k = 3;
 
    // Calling function
    cout << maxOneInK(mat, k);
 
    return 0;
}
 
// This code is contributed by mohit kumar 29


Python3




#python3 approach
 
def findMaxK(dp,k,n,m):
     
    # assign first kXk matrix initial value as max
    max_ = dp[k-1][k-1]
     
     
    for i in range(k,n):
        su = dp[i-k][k-1]
        if max_ < su:
            max_ = su
     
    for j in range(k,m):
        su = dp[k-1][i-k]
        if max_< su:
            max_ = su
             
    for i in range(k,n):
        for j in range(k,m):
            su = dp[i][j] + dp[i-k][j-k] - dp[i-k][j] - dp[i][j-k]
            if max_ < su:
                max_ = su
             
    return max_
     
def buildDPdp(mat,k,n,m):
     
    # assign mXn dp list to 0
    dp = [[0 for i in range(m)] for j in range(n)]
     
    # assign initial starting value
    dp[0][0] = mat[0][0]
     
    for i in range(1,m):
        dp[0][i] += (dp[0][i-1]+mat[0][i])
     
    for i in range(1,n):
        dp[i][0] += (dp[i-1][0]+mat[i][0])
     
     
    for i in range(1,n):
        for j in range(1,m):
            dp[i][j] = dp[i-1][j] + dp[i][j-1] + mat[i][j] - dp[i-1][j-1]
 
    return dp
 
def maxOneInK(mat,k):
     
    # n is colums
    n = len(mat)
     
    # m is rows
    m = len(mat[0])
     
    #build dp list
    dp = buildDPdp(mat,k,n,m)
     
    # call the function and return its value
    return findMaxK(dp,k,n,m)
     
     
         
 
def main():
    # mXn matrix
    mat = [[1, 0, 1], [1, 1, 0], [0, 1, 0]]
     
    k = 3
     
    #callind function
    print(maxOneInK(mat,k))
 
#driver code
main()
 
 
#This code is contributed by Tokir Manva


Output: 

5

 

Performance Analysis: 

  • Time Complexity: As in the above Dynamic program approach we have to calculate N X M dp matrix which takes O(N*M) time, Hence the Time Complexity will be O(N*M).
  • Space Complexity: As in the above approach, there is extra space used for making dp N X M matrix, Hence the space complexity will be O(N*M).

 

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