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Formatted question description: https://leetcode.ca/all/1947.html

1947. Maximum Compatibility Score Sum

Level

Medium

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 i-th 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 j-th 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.

Solution

Since both m and n do not exceed 8, each row in students and mentors can be considered as a binary representation and can be converted into an integer that does not exceed 255. After converting into integers, calculate the permutations of the m rows. For each permutation, calculate the compatibility score sum using bitwise operation, and return the maximum compatibility score sum.

  • class Solution {
    public int maxCompatibilitySum(int[][] students, int[][] mentors) {
        int m = students.length, n = students[0].length;
        int[] studentsNums = new int[m];
        int[] mentorsNums = new int[m];
        for (int i = 0; i < m; i++) {
            studentsNums[i] = convertToNum(students[i]);
            mentorsNums[i] = convertToNum(mentors[i]);
        }
        int[] nums = new int[m];
        for (int i = 0; i < m; i++)
            nums[i] = i;
        int maxSum = 0;
        List<List<Integer>> permutations = permute(nums);
        for (List<Integer> permutation : permutations) {
            int sum = 0;
            for (int i = 0; i < m; i++)
                sum += n - Integer.bitCount(studentsNums[i] ^ mentorsNums[permutation.get(i)]);
            maxSum = Math.max(maxSum, sum);
        }
        return maxSum;
    }

    public int convertToNum(int[] arr) {
        int num = 0;
        int length = arr.length;
        for (int i = 0; i < length; i++)
            num += arr[i] << i;
        return num;
    }

    public List<List<Integer>> permute(int[] nums) {
        int length = nums.length;
        List<List<Integer>> permutations = new ArrayList<List<Integer>>();
        List<Integer> permutation = new ArrayList<Integer>();
        for (int i = 0; i < length; i++)
            permutation.add(nums[i]);
        backtrack(length, permutations, 0, permutation);
        return permutations;
    }

    public void backtrack(int length, List<List<Integer>> permutations, int start, List<Integer> permutation) {
        if (start == length)
            permutations.add(new ArrayList<Integer>(permutation));
        else {
            for (int i = start; i < length; i++) {
                Collections.swap(permutation, start, i);
                backtrack(length, permutations, start + 1, permutation);
                Collections.swap(permutation, start, i);
            }
        }
    }
}

############

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:
    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

############

# 1947. Maximum Compatibility Score Sum
# https://leetcode.com/problems/maximum-compatibility-score-sum/

class Solution:
    def maxCompatibilitySum(self, students: List[List[int]], mentors: List[List[int]]) -> int:
        n = len(students)
        
        def dfs(i, used, score):
            if i == n: return score
            
            res = float(-inf)
            
            for j, mentor in enumerate(mentors):
                if j in used: continue
                    
                ss = sum(int(a == b) for a, b in zip(students[i], mentor))
                used.add(j)
                score += ss
                
                res = max(res, dfs(i + 1, used, score))

                used.remove(j)
                score -= ss
            
            return res
        
        return dfs(0, set(), 0)


  • 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;
    }
};

  • 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
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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