LeetCode in Kotlin
1494. Parallel Courses II
Hard
You are given an integer n, which indicates that there are n courses labeled from 1 to n. You are also given an array relations where relations[i] = [prevCoursei, nextCoursei], representing a prerequisite relationship between course prevCoursei and course nextCoursei: course prevCoursei has to be taken before course nextCoursei. Also, you are given the integer k.
In one semester, you can take at most k courses as long as you have taken all the prerequisites in the previous semester for the courses you are taking.
Return the minimum number of semesters needed to take all courses. The testcases will be generated such that it is possible to take every course.
Example 1:
Input: n = 4, dependencies = [[2,1],[3,1],[1,4]], k = 2
Output: 3
Explanation: The figure above represents the given graph. In the first semester, you can take courses 2 and 3. In the second semester, you can take course 1. In the third semester, you can take course 4.
Example 2:
Input: n = 5, dependencies = [[2,1],[3,1],[4,1],[1,5]], k = 2
Output: 4
Explanation: The figure above represents the given graph. In the first semester, you can take courses 2 and 3 only since you cannot take more than two per semester. In the second semester, you can take course 4. In the third semester, you can take course 1. In the fourth semester, you can take course 5.
Example 3:
Input: n = 11, dependencies = [], k = 2
Output: 6
Constraints:
1 <= n <= 151 <= k <= n0 <= relations.length <= n * (n-1) / 2relations[i].length == 21 <= prevCoursei, nextCoursei <= nprevCoursei != nextCoursei- All the pairs
[prevCoursei, nextCoursei]are unique. - The given graph is a directed acyclic graph.
Solution
class Solution {
fun minNumberOfSemesters(n: Int, relations: Array<IntArray>, k: Int): Int {
val pres = IntArray(n)
for (r in relations) {
val prev = r[0] - 1
val next = r[1] - 1
pres[next] = pres[next] or (1 shl prev)
}
val dp = IntArray(1 shl n)
dp.fill(n)
dp[0] = 0
for (mask in dp.indices) {
var canTake = 0
for (i in 0 until n) {
// already taken
if (mask and (1 shl i) != 0) {
continue
}
// satisfy all pres
if (mask and pres[i] == pres[i]) {
canTake = canTake or (1 shl i)
}
}
// loop each sub-masks
var take = canTake
while (take > 0) {
if (Integer.bitCount(take) > k) {
take = take - 1 and canTake
continue
}
dp[take or mask] = Math.min(dp[take or mask], dp[mask] + 1)
take = take - 1 and canTake
}
}
return dp[(1 shl n) - 1]
}
}