# Maximum score after flipping a Binary Matrix atmost K times

• Difficulty Level : Hard
• Last Updated : 19 Jul, 2022

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, , maximizing the left-most digit is more important than any other digit. Thus, any rows should be toggled such that the left most column should be either all 0 or all 1 (so that after toggling the left-most column [if necessary], the left column is all 1).
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++

 // C++ program to find the maximum score after// flipping a Binary Matrix atmost K times#include using namespace std; const int n = 3;const int m = 4; // Function to find maximum score of matrixint 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 programint 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

 // Java program to find the maximum score after// flipping a Binary Matrix atmost K timesimport java.util.*; class GFG{ static int n = 3;static int m = 4; // Function to find maximum score of matrixstatic int maxMatrixScore(int A[][], int K){    HashMap update =        new HashMap();     // 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 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 codepublic 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

 # Python3 program to find the maximum# score after flipping a Binary Matrix# atmost K times n = 3m = 4 # Function to find maximum score of matrixdef 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 Codeif __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#

 // C# program to find the maximum score after// flipping a Binary Matrix atmost K timesusing System;using System.Collections.Generic; class GFG{ static int n = 3;static int m = 4; // Function to find maximum score of matrixstatic 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 codepublic 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

## Javascript

 

Output:

36

Time Complexity: O(N*M)

Auxiliary Space: O(N)

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