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Maximum score after flipping a Binary Matrix atmost K times

  • Difficulty Level : Medium
  • Last Updated : 26 May, 2021

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

 

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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++




// 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




// 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




# 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#




// 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

Javascript




<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)
 




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