# Maximize count of indices with same element by pairing rows from given Matrices

• Difficulty Level : Medium
• Last Updated : 09 Jun, 2022

Given two 2D binary arrays, a[][] and b[][] both of size M*N, the task is to pair each row in the array a[][] with any row in the array b[][] such that the total score can be maximized and the score for each pair is calculated as the total indexes at which values of both rows are identical.

Note: Each row of the array b[][] can only be paired with a single row of vector a[][].

Examples:

Input: a[][] = {{1, 1, 0}, {1, 0, 1}, {0, 0, 1}}, b[][] = {{1, 0, 0}, {0, 0, 1}, {1, 1, 0}}
Output: 8
Explanation:
Consider the pairing of rows in the following order, to maximize the total score obtained:

• Row 0 of a[][] paired with row 2 of b[][] has the score of 3.
• Row 1 of a[][] paired with row 0 of b[][] with score of 2.
• Row 2 of a[][] paired with row 1 of b[][] with score of 3.

Therefore, the sum of scores obtained is 3 + 2 + 3 = 8.

Input: a[][] = {{0, 0}, {0, 0}, {0, 0}}, b[][] = {{1, 1}, {1, 1}, {1, 1}}
Output: 0

Naive Approach: The simplest approach to solve the given problem is to generate all possible permutations of the rows of the arrays a[][]  and for each permutation of the array a[][], find the sum of scores of each corresponding pair, and if it’s greater than the current answer, update the answer to the value of the current sum of scores. After checking for all the pairs, print the maximum score obtained.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the maximum score``void` `maxScoreSum(vector >& a,``                 ``vector >& b)``{``    ``// Stores the maximum sum of scores``    ``int` `maxSum = 0;` `    ``vector<``int``> pos;``    ``for` `(``int` `i = 0; i < a.size(); i++) {``        ``pos.push_back(i);``    ``}` `    ``// For each permutation of pos vector``    ``// calculate the score``    ``do` `{``        ``int` `curSum = 0;``        ``for` `(``int` `i = 0; i < a.size(); i++) {` `            ``for` `(``int` `j = 0;``                 ``j < a[pos[i]].size(); j++) {` `                ``// If values at current indexes``                ``// are same then increment the``                ``// current score``                ``curSum += (a[pos[i]][j] == b[i][j]);``                ``maxSum = max(maxSum, curSum);``            ``}``        ``}``    ``} ``while` `(next_permutation(pos.begin(), pos.end()));` `    ``// Print the maximum score``    ``cout << maxSum;``}` `// Driver Code``int` `main()``{``    ``int` `N = 3, M = 3;``    ``vector > a``        ``= { { 1, 1, 0 }, { 1, 0, 1 }, { 0, 0, 1 } };``    ``vector > b``        ``= { { 1, 0, 0 }, { 0, 0, 1 }, { 1, 1, 0 } };` `    ``maxScoreSum(a, b);` `    ``return` `0;``}`

## Python3

 `# Python Program to implement``# the above approach``def` `next_permutation(array):``    ``i ``=` `len``(array) ``-` `1``    ``while` `(i > ``0` `and` `array[i ``-` `1``] >``=` `array[i]):``        ``i ``-``=` `1` `    ``if` `(i <``=` `0``):``        ``return` `False` `    ``j ``=` `len``(array) ``-` `1` `    ``while` `(array[j] <``=` `array[i ``-` `1``]):``        ``j ``-``=` `1` `    ``temp ``=` `array[i ``-` `1``]``    ``array[i ``-` `1``] ``=` `array[j]``    ``array[j] ``=` `temp` `    ``j ``=` `len``(array) ``-` `1` `    ``while` `(i < j):``        ``temp ``=` `array[i];``        ``array[i] ``=` `array[j];``        ``array[j] ``=` `temp;``        ``i ``+``=` `1``        ``j ``-``=` `1` `    ``return` `array` `# Function to find the maximum score``def` `maxScoreSum(a, b):``  ` `    ``# Stores the maximum sum of scores``    ``maxSum ``=` `0` `    ``pos ``=` `[]``    ``for` `i ``in` `range``(``len``(a)):``        ``pos.append(i)` `    ``# For each permutation of pos vector``    ``# calculate the score``    ``while``(``True``):``        ``curSum ``=` `0``        ``for` `i ``in` `range``(``len``(a)):` `            ``for` `j ``in` `range``(``len``(a[pos[i]])):` `                ``# If values at current indexes``                ``# are same then increment the``                ``# current score``                ``curSum ``+``=` `(a[pos[i]][j] ``=``=` `b[i][j])``                ``maxSum ``=` `max``(maxSum, curSum)` `        ``if``(next_permutation(pos) ``=``=` `False``):``            ``break` `    ``# Print the maximum score``    ``print``(maxSum)` `# Driver Code``N, M ``=` `3``, ``3``a ``=` `[[``1``, ``1``, ``0``], [``1``, ``0``, ``1``], [``0``, ``0``, ``1``]]``b ``=` `[[``1``, ``0``, ``0``], [``0``, ``0``, ``1``], [``1``, ``1``, ``0``]]` `maxScoreSum(a, b)` `# This code is contributed by shinjanpatra`

## C#

 `// C# program to implement above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `  ``public` `static` `List> perms = ``new` `List>();` `  ``public` `static` `void` `generate_permuatation(``int` `mask, ``int` `ind, ``int` `N, List<``int``> current){``    ``if``(ind == N){``      ``List<``int``> temp = ``new` `List<``int``>(current);``      ``perms.Add(temp);``      ``return``;``    ``}``    ``for``(``int` `i = 0 ; i < N ; i++){``      ``if``(((mask >> i) & 1) == 0){``        ``current.Add(i);``        ``generate_permuatation(mask | (1 << i), ind + 1, N, current);``        ``current.RemoveAt(current.Count-1);``      ``}``    ``}``  ``}` `  ``// Function to find the maximum score``  ``public` `static` `void` `maxScoreSum(List> a, List> b)``  ``{``    ``// Stores the maximum sum of scores``    ``int` `maxSum = 0;` `    ``generate_permuatation(0, 0, a.Count, ``new` `List<``int``>());` `    ``// For each permutation of {0, a.Count - 1} vector``    ``// calculate the score``    ``for``(``int` `ind = 0 ; ind < perms.Count ; ind++){``      ``List<``int``> pos = perms[ind];``      ``int` `curSum = 0;``      ``for` `(``int` `i = 0; i < a.Count ; i++) {` `        ``for` `(``int` `j = 0 ; j < a[pos[i]].Count ; j++) {` `          ``// If values at current indexes``          ``// are same then increment the``          ``// current score``          ``curSum += (a[pos[i]][j] == b[i][j] ? 1 : 0);``          ``maxSum = Math.Max(maxSum, curSum);``        ``}``      ``}``    ``}` `    ``// Print the maximum score``    ``Console.Write(maxSum);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args){` `    ``// int N = 3;``    ``// int M = 3;``    ``List> a = ``new` `List>{``      ``new` `List<``int``>{ 1, 1, 0 },``      ``new` `List<``int``>{ 1, 0, 1 },``      ``new` `List<``int``>{ 0, 0, 1 }``    ``};``    ``List> b = ``new` `List>{``      ``new` `List<``int``>{ 1, 0, 0 },``      ``new` `List<``int``>{ 0, 0, 1 },``      ``new` `List<``int``>{ 1, 1, 0 }``    ``};` `    ``maxScoreSum(a, b);` `  ``}``}` `// This code is contributed by subhamgoyal2014.`

## Javascript

 ``

Output:

`8`

Time Complexity: O(N*M*M!), where M! are the number of permutations and N*M for calculating the score of each pair.
Auxiliary Space: O(M)

Efficient Approach: The above approach can also be optimized using the concept of Bitmasking, The idea is for each row in vector a[][], try all rows in vector b[][] that haven’t been chosen before. Use a bitmask to represent already chosen rows of vector b[][]. To avoid re-computing the same subproblem, memoize the result for each bitmask. Follow the steps below to solve the problem:

• Initialize the variables row as 0, mask as (2M – 1).
• Initialize the vector dp[] of size mask + 1 with values -1.
• If row is greater than equal to a.size() then return 0 and if dp[mask] is not equal to -1 then return dp[mask].
• Initialize the variable ans as 0 to store the answer.
• Iterate over the range [0, a.size()) using the variable i and perform the following tasks:
• If the bitwise AND of mask and 2i is true then initialize the variable newMask as mask^(1<<i) and curSum as 0.
• Iterate over the range [0, a[i].size()) using the variable j and if a[row][j] equals b[i][j] then increase the value of curSum by 1.
• Set the value of ans as the maximum of ans or curSum + maxScoreSum(a, b, row+1, newmask, dp) recursively.
• After performing the above steps, set the value of dp[mask] as ans and return the value of ans as the answer.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the maximum defined``// score``int` `maxScoreSum(vector >& a,``                ``vector >& b,``                ``int` `row, ``int` `mask,``                ``vector<``int``>& dp)``{``    ``// If all students are assigned``    ``if` `(row >= a.size()) {``        ``return` `0;``    ``}``    ``if` `(dp[mask] != -1) {``        ``return` `dp[mask];``    ``}` `    ``int` `ans = 0;` `    ``for` `(``int` `i = 0; i < a.size(); i++) {` `        ``// Check if row is not paired yet``        ``if` `(mask & (1 << i)) {``            ``int` `newMask = mask ^ (1 << i);``            ``int` `curSum = 0;` `            ``// Check for all indexes``            ``for` `(``int` `j = 0; j < a[i].size(); j++) {` `                ``// If values at current indexes``                ``// are same increase curSum``                ``if` `(a[row][j] == b[i][j]) {``                    ``curSum++;``                ``}``            ``}` `            ``// Further recursive call``            ``ans = max(``                ``ans, curSum``                         ``+ maxScoreSum(``                               ``a, b, row + 1,``                               ``newMask, dp));``        ``}``    ``}` `    ``// Store the ans for current``    ``// mask and return``    ``return` `dp[mask] = ans;``}` `// Utility function to find the maximum``// defined score``int` `maxScoreSumUtil(vector >& a,``                    ``vector >& b,``                    ``int` `N, ``int` `M)``{``    ``int` `row = 0;` `    ``// Create a mask with all set bits``    ``// 1 -> row is not paired yet``    ``// 0 -> row is already paired``    ``int` `mask = ``pow``(2, M) - 1;` `    ``// Initialise dp array with -1``    ``vector<``int``> dp(mask + 1, -1);` `    ``return` `maxScoreSum(a, b, row, mask, dp);``}` `// Driver Code``int` `main()``{``    ``int` `N = 3, M = 3;``    ``vector > a``        ``= { { 1, 1, 0 }, { 1, 0, 1 }, { 0, 0, 1 } };``    ``vector > b``        ``= { { 1, 0, 0 }, { 0, 0, 1 }, { 1, 1, 0 } };` `    ``cout << maxScoreSumUtil(a, b, N, M);` `    ``return` `0;``}`

## Python3

 `# Python 3 program for the above approach` `# Function to find the maximum defined``# score``def` `maxScoreSum(a, b, row, mask, dp):` `    ``# If all students are assigned``    ``if` `(row >``=` `len``(a)):``        ``return` `0` `    ``if` `(dp[mask] !``=` `-``1``):``        ``return` `dp[mask]` `    ``ans ``=` `0` `    ``for` `i ``in` `range``(``len``(a)):` `        ``# Check if row is not paired yet``        ``if` `(mask & (``1` `<< i)):``            ``newMask ``=` `mask ^ (``1` `<< i)``            ``curSum ``=` `0` `            ``# Check for all indexes``            ``for` `j ``in` `range``(``len``(a[i])):` `                ``# If values at current indexes``                ``# are same increase curSum``                ``if` `(a[row][j] ``=``=` `b[i][j]):``                    ``curSum ``+``=` `1` `            ``# Further recursive call``            ``ans ``=` `max``(``                ``ans, curSum``                ``+` `maxScoreSum(``                    ``a, b, row ``+` `1``,``                    ``newMask, dp))` `    ``# Store the ans for current``    ``# mask and return``    ``dp[mask] ``=` `ans``    ``return` `dp[mask]` `# Utility function to find the maximum``# defined score``def` `maxScoreSumUtil(a,``                    ``b,``                    ``N, M):` `    ``row ``=` `0` `    ``# Create a mask with all set bits``    ``# 1 -> row is not paired yet``    ``# 0 -> row is already paired``    ``mask ``=` `pow``(``2``, M) ``-` `1` `    ``# Initialise dp array with -1``    ``dp ``=` `[``-``1``]``*``(mask ``+` `1``)` `    ``return` `maxScoreSum(a, b, row, mask, dp)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``N ``=` `3``    ``M ``=` `3``    ``a ``=` `[[``1``, ``1``, ``0``], [``1``, ``0``, ``1``], [``0``, ``0``, ``1``]]``    ``b ``=` `[[``1``, ``0``, ``0``], [``0``, ``0``, ``1``], [``1``, ``1``, ``0``]]` `    ``print``(maxScoreSumUtil(a, b, N, M))` `    ``# This code is contributed by ukasp.`

## C#

 `// C# program to implement above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ``// Function to find the maximum defined``    ``// score``    ``public` `static` `int` `maxScoreSum(List> a, List> b, ``int` `row, ``int` `mask, List<``int``> dp)``    ``{``        ``// If all students are assigned``        ``if` `(row >= a.Count){``            ``return` `0;``        ``}``        ``if` `(dp[mask] != -1){``            ``return` `dp[mask];``        ``}``    ` `        ``int` `ans = 0;`` ` `        ``for` `(``int` `i = 0 ; i < a.Count ; i++) {``    ` `            ``// Check if row is not paired yet``            ``if` `((mask & (1 << i)) > 0){``                ``int` `newMask = mask ^ (1 << i);``                ``int` `curSum = 0;``    ` `                ``// Check for all indexes``                ``for` `(``int` `j = 0; j < a[i].Count ; j++) {``    ` `                    ``// If values at current indexes``                    ``// are same increase curSum``                    ``if` `(a[row][j] == b[i][j]) {``                        ``curSum++;``                    ``}``                ``}``    ` `                ``// Further recursive call``                ``ans = Math.Max(ans, curSum + maxScoreSum(a, b, row + 1, newMask, dp));``            ``}``        ``}``    ` `        ``// Store the ans for current``        ``// mask and return``        ``return` `dp[mask] = ans;``    ``}``    ` `    ``// Utility function to find the maximum``    ``// defined score``    ``public` `static` `int` `maxScoreSumUtil(List> a, List> b, ``int` `N,``int` `M)``    ``{``        ``int` `row = 0;`` ` `        ``// Create a mask with all set bits``        ``// 1 -> row is not paired yet``        ``// 0 -> row is already paired``        ``int` `mask = (``int``)Math.Pow(2, M) - 1;``    ` `        ``// Initialise dp array with -1``        ``List<``int``> dp = ``new` `List<``int``>();``        ``for``(``int` `i = 0 ; i <= mask ; i++){``            ``dp.Add(-1);``        ``}``    ` `        ``return` `maxScoreSum(a, b, row, mask, dp);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(``string``[] args){``        ` `        ``int` `N = 3;``        ``int` `M = 3;``        ``List> a = ``new` `List>{``            ``new` `List<``int``>{ 1, 1, 0 },``            ``new` `List<``int``>{ 1, 0, 1 },``            ``new` `List<``int``>{ 0, 0, 1 }``        ``};``        ``List> b = ``new` `List>{``            ``new` `List<``int``>{ 1, 0, 0 },``            ``new` `List<``int``>{ 0, 0, 1 },``            ``new` `List<``int``>{ 1, 1, 0 }``        ``};` `        ``Console.Write(maxScoreSumUtil(a, b, N, M));` `    ``}``}` `// This code is contributed by subhamgoyal2014.`

## Javascript

 ``

Output:

`8`

Time Complexity: O(2M*M*N)
Auxiliary Space: O(2M)

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