Problem
Tasks 0..n−1 form a directed acyclic graph. Each edge [u, v, w] means task v can start w hours after task u. Find the length of the longest path (the critical path). Process nodes in topological order.
Implement longest_path(n, edges).
Input
n: The number of nodes. edges: Edges as [from, to, weight].
Output
Return the total weight of the longest path (0 if there are no edges).
Constraints
1 ≤ n ≤ 100,000; 0 ≤ edges ≤ 200,000; 1 ≤ weight ≤ 10,000; the graph has no cycles
Examples
Input:
n = 4
edges = [
[0, 1, 3],
[0, 2, 2],
[1, 3, 4],
[2, 3, 6]
]
Expected output:
expected = 8Input:
long long n = 4;
std::vector<std::vector<long long>> edges = {
{0, 1, 3},
{0, 2, 2},
{1, 3, 4},
{2, 3, 6}
};
Expected output:
long long expected = 8;Input:
long n = 4;
long[][] edges = new long[][] {
new long[] {0, 1, 3},
new long[] {0, 2, 2},
new long[] {1, 3, 4},
new long[] {2, 3, 6}
};
Expected output:
long expected = 8;Explanation
0→1→3 weighs 7, 0→2→3 weighs 8: the longest path is 8.
Input:
n = 3
edges = []
Expected output:
expected = 0Input:
long long n = 3;
std::vector<std::vector<long long>> edges = {};
Expected output:
long long expected = 0;Input:
long n = 3;
long[][] edges = new long[][] {};
Expected output:
long expected = 0;Explanation
Without edges every path has length 0.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Because the graph has no cycles, each node can be finalized after all its predecessors are processed.
Hint 2
Topologically order the nodes, then relax outgoing edges to build the best path length to each node.
Hint 3
If the graph has multiple components or no edges, the answer should still reflect the longest reachable path weight.
Expected complexity
O(n + m) time, O(n + m) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.