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
.class Solution {
private lateinit var dp: Array<IntArray>
private var m = 0
private lateinit var memo: Array<IntArray>
fun maxCompatibilitySum(students: Array<IntArray>, mentors: Array<IntArray>): Int {
val n = students[0].size
m = students.size
dp = Array(m) { IntArray(m) }
for (i in 0 until m) {
for (j in 0 until m) {
var tmp = 0
for (k in 0 until n) {
tmp += if (students[i][k] == mentors[j][k]) 1 else 0
}
dp[i][j] = tmp
}
}
memo = Array(m) { IntArray((1 shl m) + 1) }
for (x in memo) {
x.fill(-1)
}
return dp(0, 0)
}
private fun dp(idx: Int, mask: Int): Int {
if (idx == m) {
return 0
}
if (memo[idx][mask] != -1) {
return memo[idx][mask]
}
var ans = 0
for (i in 0 until m) {
if (mask and (1 shl i) == 0) {
ans = Math.max(ans, dp[idx][i] + dp(idx + 1, mask or (1 shl i)))
}
}
memo[idx][mask] = ans
return ans
}
}