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);
            }
        }
    }
}

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