Medium
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.
[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
.import java.util.Arrays;
public class Solution {
private int[][] dp;
private int m;
private int[][] memo;
public int maxCompatibilitySum(int[][] students, int[][] mentors) {
int n = students[0].length;
m = students.length;
dp = new int[m][m];
for (int i = 0; i < m; i++) {
for (int j = 0; j < m; j++) {
int tmp = 0;
for (int k = 0; k < n; k++) {
tmp += (students[i][k] == mentors[j][k] ? 1 : 0);
}
dp[i][j] = tmp;
}
}
memo = new int[m][(1 << m) + 1];
for (int[] x : memo) {
Arrays.fill(x, -1);
}
return dp(0, 0);
}
private int dp(int idx, int mask) {
if (idx == m) {
return 0;
}
if (memo[idx][mask] != -1) {
return memo[idx][mask];
}
int ans = 0;
for (int i = 0; i < m; i++) {
if ((mask & (1 << i)) == 0) {
ans = Math.max(ans, dp[idx][i] + dp(idx + 1, mask | (1 << i)));
}
}
memo[idx][mask] = ans;
return ans;
}
}