Given a two dimensional matrix A of zero’s and one’s and an integer K.
In each move, you can choose any row or column and toggle every value in that row or column. That is, change all 0s to 1s, or all 1s to 0s. After making atmost K of moves, every row of this matrix represents a binary number.
The task is to return the maximum possible value of the sum of these numbers.
Examples:
Input : A[][] = { { 0, 0, 1, 1 }, { 1, 0, 1, 0 }, { 1, 1, 0, 0 } }; K = 2 Output : 36 Input : A[][] = { { 0, 1 }, { 1, 0 }, { 1, 1 } }; K = 1 Output : 7
Notice that a 1 in the i-th column from the right, contributes 2i to the score.
Also knowing the fact that,
Now for rows with first element as 0, make a map with value of row as key and index of that row as element. Now we toggle rows with least value so that after updating it contributes maximum to our total score.
Now, for other subsequent columns we count total zeros and ones.
- If ( zeros > ones and K > 0 ) we toggle the column and update our answer to ans = ans + zero * pow( 2, columns – j – 1), for all
and decrements K by one. - Otherwise we update answer to ans = ans + one * pow( 2, columns – j – 1), for all
.
Below is the implementation of above approach:
// C++ program to find the maximum score after // flipping a Binary Matrix atmost K times #include <bits/stdc++.h> using namespace std;
const int n = 3;
const int m = 4;
// Function to find maximum score of matrix int maxMatrixScore( int A[n][m], int K)
{ map< int , int > update;
// find value of rows having first
// column value equal to zero
for ( int i = 0; i < n; ++i) {
if (A[i][0] == 0) {
int ans = 0;
for ( int j = 1; j < m; ++j)
ans = ans + A[i][j] * pow (2, m - j - 1);
update[ans] = i;
}
}
// update those rows which lead to
// maximum score after toggle
map< int , int >::iterator it = update.begin();
while (K > 0 && it != update.end()) {
int idx = it->second;
for ( int j = 0; j < m; ++j)
A[idx][j] = (A[idx][j] + 1) % 2;
it++;
K--;
}
// Calculating answer
int ans = 0;
for ( int j = 0; j < m; ++j) {
int zero = 0, one = 0;
for ( int i = 0; i < n; ++i) {
A[i][j] == 0 ? zero++ : one++;
}
// check if K > 0 we can toggle if necessary.
if (K > 0 && zero > one) {
ans += zero * pow (2, m - j - 1);
K--;
}
else
ans += one * pow (2, m - j - 1);
}
// return max answer possible
return ans;
} // Driver program int main()
{ int A[n][m] = { { 0, 0, 1, 1 },
{ 1, 0, 1, 0 },
{ 1, 1, 0, 0 } };
int K = 2;
// function call to print required answer
cout << maxMatrixScore(A, K);
return 0;
} |
// Java program to find the maximum score after // flipping a Binary Matrix atmost K times import java.util.*;
class GFG
{ static int n = 3 ;
static int m = 4 ;
// Function to find maximum score of matrix static int maxMatrixScore( int A[][], int K)
{ HashMap<Integer,Integer> update =
new HashMap<Integer,Integer>();
// find value of rows having first
// column value equal to zero
for ( int i = 0 ; i < n; ++i)
{
if (A[i][ 0 ] == 0 )
{
int ans = 0 ;
for ( int j = 1 ; j < m; ++j)
ans = ( int ) (ans + A[i][j] *
Math.pow( 2 , m - j - 1 ));
update.put(ans, i);
}
}
// Update those rows which lead to
// maximum score after toggle
for (Map.Entry<Integer,Integer> it : update.entrySet())
if (K > 0 )
{
int idx = it.getValue();
for ( int j = 0 ; j < m; ++j)
A[idx][j] = (A[idx][j] + 1 ) % 2 ;
K--;
}
// Calculating answer
int ans = 0 ;
for ( int j = 0 ; j < m; ++j)
{
int zero = 0 , one = 0 ;
for ( int i = 0 ; i < n; ++i)
{
if (A[i][j] == 0 )
zero++;
else
one++;
}
// Check if K > 0 we can toggle if necessary.
if (K > 0 && zero > one)
{
ans += zero * Math.pow( 2 , m - j - 1 );
K--;
}
else
ans += one * Math.pow( 2 , m - j - 1 );
}
// return max answer possible
return ans;
} // Driver code public static void main(String[] args)
{ int A[][] = { { 0 , 0 , 1 , 1 },
{ 1 , 0 , 1 , 0 },
{ 1 , 1 , 0 , 0 } };
int K = 2 ;
// function call to print required answer
System.out.print(maxMatrixScore(A, K));
} } // This code is contributed by PrinciRaj1992 |
# Python3 program to find the maximum # score after flipping a Binary Matrix # atmost K times n = 3
m = 4
# Function to find maximum score of matrix def maxMatrixScore(A, K):
update = {}
# Find value of rows having first
# column value equal to zero
for i in range ( 0 , n):
if A[i][ 0 ] = = 0 :
ans = 0
for j in range ( 1 , m):
ans = ans + A[i][j] * 2 * * (m - j - 1 )
update[ans] = i
# update those rows which lead to
# maximum score after toggle
for idx in update.values():
for j in range ( 0 , m):
A[idx][j] = (A[idx][j] + 1 ) % 2
K - = 1
if K < = 0 :
break
# Calculating answer
ans = 0
for j in range ( 0 , m):
zero, one = 0 , 0
for i in range ( 0 , n):
if A[i][j] = = 0 : zero + = 1
else : one + = 1
# check if K > 0 we can
# toggle if necessary.
if K > 0 and zero > one:
ans + = zero * 2 * * (m - j - 1 )
K - = 1
else :
ans + = one * 2 * * (m - j - 1 )
# return max answer possible
return ans
# Driver Code if __name__ = = "__main__" :
A = [[ 0 , 0 , 1 , 1 ],
[ 1 , 0 , 1 , 0 ],
[ 1 , 1 , 0 , 0 ]]
K = 2
# function call to print required answer
print (maxMatrixScore(A, K))
# This code is contributed by Rituraj Jain |
// C# program to find the maximum score after // flipping a Binary Matrix atmost K times using System;
using System.Collections.Generic;
class GFG
{ static int n = 3;
static int m = 4;
// Function to find maximum score of matrix static int maxMatrixScore( int [,]A, int K)
{ Dictionary< int , int > update =
new Dictionary< int , int >();
// find value of rows having first
// column value equal to zero
int ans=0;
for ( int i = 0; i < n; ++i)
{
if (A[i, 0] == 0)
{
ans = 0;
for ( int j = 1; j < m; ++j)
ans = ( int ) (ans + A[i, j] *
Math.Pow(2, m - j - 1));
update.Add(( int )ans, i);
}
}
// Update those rows which lead to
// maximum score after toggle
foreach (KeyValuePair< int , int > it in update)
if (K > 0 )
{
int idx = it.Value;
for ( int j = 0; j < m; ++j)
A[idx, j] = (A[idx, j] + 1) % 2;
K--;
}
// Calculating answer
ans = 0;
for ( int j = 0; j < m; ++j)
{
int zero = 0, one = 0;
for ( int i = 0; i < n; ++i)
{
if (A[i, j] == 0)
zero++;
else
one++;
}
// Check if K > 0 we can toggle if necessary.
if (K > 0 && zero > one)
{
ans += zero * ( int )Math.Pow(2, m - j - 1);
K--;
}
else
ans += one * ( int )Math.Pow(2, m - j - 1);
}
// return max answer possible
return ans;
} // Driver code public static void Main(String[] args)
{ int [,]A = { { 0, 0, 1, 1 },
{ 1, 0, 1, 0 },
{ 1, 1, 0, 0 } };
int K = 2;
// function call to print required answer
Console.Write(maxMatrixScore(A, K));
} } // This code is contributed by 29AjayKumar |
<script> // Javascript program to find the maximum score after // flipping a Binary Matrix atmost K times var n = 3;
var m = 4;
// Function to find maximum score of matrix function maxMatrixScore(A, K)
{ var update = new Map();
// find value of rows having first
// column value equal to zero
for ( var i = 0; i < n; ++i) {
if (A[i][0] == 0) {
var ans = 0;
for ( var j = 1; j < m; ++j)
ans = ans + A[i][j] * Math.pow(2, m - j - 1);
update.set(ans, i);
}
}
// update those rows which lead to
// maximum score after toggle
update.forEach((value, key) => {
if (K>0)
{
var idx = value;
for ( var j = 0; j < m; ++j)
A[idx][j] = (A[idx][j] + 1) % 2;
K--;
}
});
// Calculating answer
var ans = 0;
for ( var j = 0; j < m; ++j) {
var zero = 0, one = 0;
for ( var i = 0; i < n; ++i) {
A[i][j] == 0 ? zero++ : one++;
}
// check if K > 0 we can toggle if necessary.
if (K > 0 && zero > one) {
ans += zero * Math.pow(2, m - j - 1);
K--;
}
else
ans += one * Math.pow(2, m - j - 1);
}
// return max answer possible
return ans;
} // Driver program var A = [ [ 0, 0, 1, 1 ],
[ 1, 0, 1, 0 ],
[ 1, 1, 0, 0 ] ];
var K = 2;
// function call to print required answer document.write( maxMatrixScore(A, K)); // This code is contributed by noob2000. </script> |
Output:
36
Time Complexity: O(N*M)
Auxiliary Space: O(N)