# 1947. Maximum Compatibility Score Sum

## Description

There is a survey that consists of n questions where each question's answer is either 0 (no) or 1 (yes).

The survey was given to m students numbered from 0 to m - 1 and m mentors numbered from 0 to m - 1. The answers of the students are represented by a 2D integer array students where students[i] is an integer array that contains the answers of the ith student (0-indexed). The answers of the mentors are represented by a 2D integer array mentors where mentors[j] is an integer array that contains the answers of the jth mentor (0-indexed).

Each student will be assigned to one mentor, and each mentor will have one student assigned to them. The compatibility score of a student-mentor pair is the number of answers that are the same for both the student and the mentor.

• For example, if the student's answers were [1, 0, 1] and the mentor's answers were [0, 0, 1], then their compatibility score is 2 because only the second and the third answers are the same.

You are tasked with finding the optimal student-mentor pairings to maximize the sum of the compatibility scores.

Given students and mentors, return the maximum compatibility score sum that can be achieved.

Example 1:

Input: students = [[1,1,0],[1,0,1],[0,0,1]], mentors = [[1,0,0],[0,0,1],[1,1,0]]
Output: 8
Explanation: We assign students to mentors in the following way:
- student 0 to mentor 2 with a compatibility score of 3.
- student 1 to mentor 0 with a compatibility score of 2.
- student 2 to mentor 1 with a compatibility score of 3.
The compatibility score sum is 3 + 2 + 3 = 8.


Example 2:

Input: students = [[0,0],[0,0],[0,0]], mentors = [[1,1],[1,1],[1,1]]
Output: 0
Explanation: The compatibility score of any student-mentor pair is 0.


Constraints:

• m == students.length == mentors.length
• n == students[i].length == mentors[j].length
• 1 <= m, n <= 8
• students[i][k] is either 0 or 1.
• mentors[j][k] is either 0 or 1.

## Solutions

• class Solution {
private int[][] g;
private boolean[] vis;
private int m;
private int ans;

public int maxCompatibilitySum(int[][] students, int[][] mentors) {
m = students.length;
g = new int[m][m];
vis = new boolean[m];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < m; ++j) {
for (int k = 0; k < students[i].length; ++k) {
g[i][j] += students[i][k] == mentors[j][k] ? 1 : 0;
}
}
}
dfs(0, 0);
return ans;
}

private void dfs(int i, int t) {
if (i == m) {
ans = Math.max(ans, t);
return;
}
for (int j = 0; j < m; ++j) {
if (!vis[j]) {
vis[j] = true;
dfs(i + 1, t + g[i][j]);
vis[j] = false;
}
}
}
}

• class Solution {
public:
int maxCompatibilitySum(vector<vector<int>>& students, vector<vector<int>>& mentors) {
int m = students.size();
int n = students[0].size();
int g[m][m];
memset(g, 0, sizeof g);
bool vis[m];
memset(vis, 0, sizeof vis);
for (int i = 0; i < m; ++i) {
for (int j = 0; j < m; ++j) {
for (int k = 0; k < n; ++k) {
g[i][j] += students[i][k] == mentors[j][k];
}
}
}
int ans = 0;
function<void(int, int)> dfs = [&](int i, int t) {
if (i == m) {
ans = max(ans, t);
return;
}
for (int j = 0; j < m; ++j) {
if (!vis[j]) {
vis[j] = true;
dfs(i + 1, t + g[i][j]);
vis[j] = false;
}
}
};
dfs(0, 0);
return ans;
}
};

• class Solution:
def maxCompatibilitySum(
self, students: List[List[int]], mentors: List[List[int]]
) -> int:
def dfs(i, t):
if i == m:
nonlocal ans
ans = max(ans, t)
return
for j in range(m):
if not vis[j]:
vis[j] = True
dfs(i + 1, t + g[i][j])
vis[j] = False

m = len(students)
g = [[0] * m for _ in range(m)]
for i in range(m):
for j in range(m):
g[i][j] = sum(a == b for a, b in zip(students[i], mentors[j]))
vis = [False] * m
ans = 0
dfs(0, 0)
return ans


• func maxCompatibilitySum(students [][]int, mentors [][]int) (ans int) {
m, n := len(students), len(students[0])
g := make([][]int, m)
vis := make([]bool, m)
for i := range g {
g[i] = make([]int, m)
for j := range g {
for k := 0; k < n; k++ {
if students[i][k] == mentors[j][k] {
g[i][j]++
}
}
}
}
var dfs func(int, int)
dfs = func(i, t int) {
if i == m {
ans = max(ans, t)
return
}
for j := 0; j < m; j++ {
if !vis[j] {
vis[j] = true
dfs(i+1, t+g[i][j])
vis[j] = false
}
}
}
dfs(0, 0)
return
}