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 .
and decrements K by one.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 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<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 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 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 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
<script>// Javascript program to find the maximum score after// flipping a Binary Matrix atmost K timesvar n = 3;var m = 4;// Function to find maximum score of matrixfunction 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 programvar A = [ [ 0, 0, 1, 1 ], [ 1, 0, 1, 0 ], [ 1, 1, 0, 0 ] ];var K = 2;// function call to print required answerdocument.write( maxMatrixScore(A, K));// This code is contributed by noob2000.</script> |
Output:
36
Time Complexity: O(N*M)
Auxiliary Space: O(N)
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