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Weekly Contest 251 Q3
Bit Manipulation
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Dynamic Programming
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1947. Maximum Compatibility Score Sum

中文文档

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

Solution 1

Python3

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

Java

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;
            }
        }
    }
}

C++

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;
    }
};

Go

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
}