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: 3
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: 1
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++ 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 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 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# 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 |
<script> // Javascript 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 function maxCount(mat,k)
{ let n = mat.length;
let m = mat[0].length;
let a = [];
// Loop to find the count of 1's
// in every possible windows
// of rows of matrix
for (let e = 0; e < n; ++e)
{
let s = mat[e];
let q = [];
let c = 0;
// Loop to find the count of
// 1's in the first window
let i;
for (i = 0; i < k; ++i)
{
if (s[i] == 1)
{
c += 1;
}
}
q.push(c);
let p = s[0];
// Loop to find the count of
// 1's in the remaining windows
for (let j = i + 1; j < m; ++j)
{
if (s[j] == 1)
{
c += 1;
}
if (p == 1)
{
c -= 1;
}
q.push(c);
p = s[j - k + 1];
}
a.push(q);
}
let b = [];
let max = 0;
// Loop to find the count of 1's
// in every possible submatrix
for (let i = 0; i < a[0].length; ++i)
{
let c = 0;
let p = a[0][i];
// Loop to find the count of
// 1's in the first window
let j;
for (j = 0; j < k; ++j)
{
c += a[j][i];
}
let q = [];
if (c > max)
{
max = c;
}
q.push(c);
// Loop to find the count of
// 1's in the remaining windows
for (let l = j + 1; j < n; ++j)
{
c += a[l][i];
c -= p;
p = a[l - k + 1][i];
q.push(c);
if (c > max)
{
max = c;
}
}
b.push(q);
}
return max;
} // Driver code let mat=[[1, 0, 1],[1, 1, 0],[0, 1, 0]]; let k = 3; // Function call document.write(maxCount(mat, k)); // This code is contributed by unknown2108 </script> |
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 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 columns
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 |
/*package whatever //do not write package name here */ import java.io.*;
class GFG {
static int findMaxK( 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_;
}
static int [][] buildDPdp( int [][] mat,
int k, int n, int m)
{
// Assign mXn dp list to 0
int [][] dp= new int [n][m];
// 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;
}
static int maxOneInK( int [][] mat, int k)
{
// n is columns
int n = mat.length;
// m is rows
int m = mat[ 0 ].length;
// Build dp list
int [][] dp = buildDPdp(
mat, k, n, m);
// Call the function and return its value
return findMaxK(dp, k, n, m);
}
// Driver code
public static void main (String[] args)
{
// mXn matrix
int [][] mat = { { 1 , 0 , 1 },
{ 1 , 1 , 0 },
{ 0 , 1 , 0 } };
int k = 3 ;
// Calling function
System.out.println( maxOneInK(mat, k));
}
} // This code is contributed by ab2127. |
#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 columns
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 |
// Javascript approach function findMaxK(dp, k, n, m)
{ // Assign first kXk matrix initial
// value as max
let max_ = dp[k - 1][k - 1];
for (let i = k; i < n; i++)
{
let su = dp[i - k][k - 1];
if (max_ < su)
max_ = su;
}
for (let j = k; j < m; j++)
{
let su = dp[k - 1][j - k];
if (max_< su)
max_ = su;
}
for (let i = k; i < n; i++)
{
for (let j = k; j < m; j++)
{
let su = dp[i][j] +
dp[i - k][j - k] -
dp[i - k][j] -
dp[i][j - k];
if ( max_ < su)
max_ = su;
}
}
return max_;
} function buildDPdp( mat,k,n,m)
{ // Assign mXn dp list to 0
let dp= new Array(n);
for (let i=0;i<n;i++)
{
dp[i]= new Array(m);
for (let j=0;j<m;j++)
{
dp[i][j]=0;
}
}
// Assign initial starting value
dp[0][0] = mat[0][0];
for (let i = 1; i < m; i++)
dp[0][i] += (dp[0][i - 1] + mat[0][i]);
for (let i = 1; i < n; i++)
dp[i][0] += (dp[i - 1][0] + mat[i][0]);
for (let i = 1; i < n; i++)
for (let 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;
} function maxOneInK(mat,k)
{ // n is columns
let n = mat.length;
// m is rows
let m = mat[0].length;
// Build dp list
let dp = buildDPdp(
mat, k, n, m);
// Call the function and return its value
return findMaxK(dp, k, n, m);
} // Driver Code // mXn matrix let mat = [[ 1, 0, 1 ],[ 1, 1, 0 ], [ 0, 1, 0 ]];
let k = 3;
// Calling function
console.log(maxOneInK(mat, k));
// This code is contributed by patel2127 |
// C# code to implement the above approach using System;
using System.Collections.Generic;
class Program {
static int FindMaxK(List<List< 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_;
}
static List<List< int > > BuildDPdp(List<List< int > > mat,
int k, int n, int m)
{
// Assign mXn dp list to 0
List<List< int > > dp = new List<List< int > >();
for ( int i = 0; i < n; i++)
dp.Add( new List< int >( new int [m]));
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;
}
// Driver code
static int MaxOneInK(List<List< int > > mat, int k)
{ // n is columns
int n = mat.Count;
// m is rows
int m = mat[0].Count;
// Build dp list
List<List< int > > dp = BuildDPdp(mat, k, n, m);
// Call the function and return its value
return FindMaxK(dp, k, n, m);
}
static void Main( string [] args)
{
// mXn matrix
List<List< int > > mat = new List<List< int > >() {
new List< int >() { 1, 0, 1 }, new List< int >() {
1, 1, 0
}, new List< int >() { 0, 1, 0 }
};
int k = 3;
// Calling function
Console.WriteLine(MaxOneInK(mat, k));
}
} // This code is contributed by Princekumaras |
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).