Problem
You are given an integer n
, which indicates that there are n
courses labeled from 1
to n
. You are also given a 2D integer array relations
where relations[j] = [prevCoursej, nextCoursej]
denotes that course prevCoursej
has to be completed before course nextCoursej
(prerequisite relationship). Furthermore, you are given a 0-indexed integer array time
where time[i]
denotes how many months it takes to complete the (i+1)th
course.
You must find the minimum number of months needed to complete all the courses following these rules:
You may start taking a course at any time if the prerequisites are met.
Any number of courses can be taken at the same time.
Return **the *minimum* number of months needed to complete all the courses**.
Note: The test cases are generated such that it is possible to complete every course (i.e., the graph is a directed acyclic graph).
Example 1:
Input: n = 3, relations = [[1,3],[2,3]], time = [3,2,5]
Output: 8
Explanation: The figure above represents the given graph and the time required to complete each course.
We start course 1 and course 2 simultaneously at month 0.
Course 1 takes 3 months and course 2 takes 2 months to complete respectively.
Thus, the earliest time we can start course 3 is at month 3, and the total time required is 3 + 5 = 8 months.
Example 2:
Input: n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], time = [1,2,3,4,5]
Output: 12
Explanation: The figure above represents the given graph and the time required to complete each course.
You can start courses 1, 2, and 3 at month 0.
You can complete them after 1, 2, and 3 months respectively.
Course 4 can be taken only after course 3 is completed, i.e., after 3 months. It is completed after 3 + 4 = 7 months.
Course 5 can be taken only after courses 1, 2, 3, and 4 have been completed, i.e., after max(1,2,3,7) = 7 months.
Thus, the minimum time needed to complete all the courses is 7 + 5 = 12 months.
Constraints:
1 <= n <= 5 * 104
0 <= relations.length <= min(n * (n - 1) / 2, 5 * 104)
relations[j].length == 2
1 <= prevCoursej, nextCoursej <= n
prevCoursej != nextCoursej
All the pairs
[prevCoursej, nextCoursej]
are unique.time.length == n
1 <= time[i] <= 104
The given graph is a directed acyclic graph.
Solution
/**
* @param {number} n
* @param {number[][]} relations
* @param {number[]} time
* @return {number}
*/
var minimumTime = function(n, relations, time) {
var graph = Array(n).fill(0).map(() => []);
for (var i = 0; i < relations.length; i++) {
var [a, b] = relations[i];
graph[a - 1].push(b - 1);
}
var max = 0;
var dp = Array(n);
for (var i = 0; i < n; i++) {
max = Math.max(max, dfs(i, graph, time, dp));
}
return max;
};
var dfs = function(i, graph, time, dp) {
if (dp[i] !== undefined) return dp[i];
var max = 0;
for (var j = 0; j < graph[i].length; j++) {
max = Math.max(max, dfs(graph[i][j], graph, time, dp));
}
dp[i] = max + time[i];
return dp[i];
};
Explain:
nope.
Complexity:
- Time complexity : O(n + m).
- Space complexity : O(n + m).