Maximum score after flipping a Binary Matrix atmost K times

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, 2^{n} > 2^{n-1}+2^{n-2}+2^{n-3}+...2^{0}, 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 1 \leq j \leq columns - 1 and decrements K by one.
  • Otherwise we update answer to ans = ans + one * pow( 2, columns – j – 1), for all 1 \leq j \leq columns - 1.

Below is the implementation of above approach:

C++

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// 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;
}

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Python3

# 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

Output:

36

Time Complexity: O(N*M)



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