Parallel Courses III

// Accepted solution for LeetCode #2050: Parallel Courses III
class Solution {
    public int minimumTime(int n, int[][] relations, int[] time) {
        List<Integer>[] g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        int[] indeg = new int[n];
        for (int[] e : relations) {
            int a = e[0] - 1, b = e[1] - 1;
            g[a].add(b);
            ++indeg[b];
        }
        Deque<Integer> q = new ArrayDeque<>();
        int[] f = new int[n];
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int v = indeg[i], t = time[i];
            if (v == 0) {
                q.offer(i);
                f[i] = t;
                ans = Math.max(ans, t);
            }
        }
        while (!q.isEmpty()) {
            int i = q.pollFirst();
            for (int j : g[i]) {
                f[j] = Math.max(f[j], f[i] + time[j]);
                ans = Math.max(ans, f[j]);
                if (--indeg[j] == 0) {
                    q.offer(j);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2050: Parallel Courses III
func minimumTime(n int, relations [][]int, time []int) int {
	g := make([][]int, n)
	indeg := make([]int, n)
	for _, e := range relations {
		a, b := e[0]-1, e[1]-1
		g[a] = append(g[a], b)
		indeg[b]++
	}
	f := make([]int, n)
	q := []int{}
	ans := 0
	for i, v := range indeg {
		if v == 0 {
			q = append(q, i)
			f[i] = time[i]
			ans = max(ans, time[i])
		}
	}
	for len(q) > 0 {
		i := q[0]
		q = q[1:]
		for _, j := range g[i] {
			indeg[j]--
			if indeg[j] == 0 {
				q = append(q, j)
			}
			f[j] = max(f[j], f[i]+time[j])
			ans = max(ans, f[j])
		}
	}
	return ans
}

# Accepted solution for LeetCode #2050: Parallel Courses III
class Solution:
    def minimumTime(self, n: int, relations: List[List[int]], time: List[int]) -> int:
        g = defaultdict(list)
        indeg = [0] * n
        for a, b in relations:
            g[a - 1].append(b - 1)
            indeg[b - 1] += 1
        q = deque()
        f = [0] * n
        ans = 0
        for i, (v, t) in enumerate(zip(indeg, time)):
            if v == 0:
                q.append(i)
                f[i] = t
                ans = max(ans, t)
        while q:
            i = q.popleft()
            for j in g[i]:
                f[j] = max(f[j], f[i] + time[j])
                ans = max(ans, f[j])
                indeg[j] -= 1
                if indeg[j] == 0:
                    q.append(j)
        return ans

// Accepted solution for LeetCode #2050: Parallel Courses III
/**
 * [2050] Parallel Courses III
 *
 * 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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/10/07/ex1.png" style="width: 392px; height: 232px;" />
 *
 * 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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/10/07/ex2.png" style="width: 500px; height: 365px;" />
 *
 * 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 * 10^4
 * 	0 <= relations.length <= min(n * (n - 1) / 2, 5 * 10^4)
 * 	relations[j].length == 2
 * 	1 <= prevCoursej, nextCoursej <= n
 * 	prevCoursej != nextCoursej
 * 	All the pairs [prevCoursej, nextCoursej] are unique.
 * 	time.length == n
 * 	1 <= time[i] <= 10^4
 * 	The given graph is a directed acyclic graph.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/parallel-courses-iii/
// discuss: https://leetcode.com/problems/parallel-courses-iii/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn minimum_time(n: i32, relations: Vec<Vec<i32>>, time: Vec<i32>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2050_example_1() {
        let n = 3;
        let relations = vec![vec![1, 3], vec![2, 3]];
        let time = vec![3, 2, 5];

        let result = 8;

        assert_eq!(Solution::minimum_time(n, relations, time), result);
    }

    #[test]
    #[ignore]
    fn test_2050_example_2() {
        let n = 5;
        let relations = vec![vec![1, 5], vec![2, 5], vec![3, 5], vec![3, 4], vec![4, 5]];
        let time = vec![1, 2, 3, 4, 5];

        let result = 12;

        assert_eq!(Solution::minimum_time(n, relations, time), result);
    }
}

// Accepted solution for LeetCode #2050: Parallel Courses III
function minimumTime(n: number, relations: number[][], time: number[]): number {
    const g: number[][] = Array(n)
        .fill(0)
        .map(() => []);
    const indeg: number[] = Array(n).fill(0);
    for (const [a, b] of relations) {
        g[a - 1].push(b - 1);
        ++indeg[b - 1];
    }
    const q: number[] = [];
    const f: number[] = Array(n).fill(0);
    let ans: number = 0;
    for (let i = 0; i < n; ++i) {
        if (indeg[i] === 0) {
            q.push(i);
            f[i] = time[i];
            ans = Math.max(ans, f[i]);
        }
    }
    while (q.length > 0) {
        const i = q.shift()!;
        for (const j of g[i]) {
            f[j] = Math.max(f[j], f[i] + time[j]);
            ans = Math.max(ans, f[j]);
            if (--indeg[j] === 0) {
                q.push(j);
            }
        }
    }
    return ans;
}