Longest Path in a Directed Acyclic Graph

You are given a directed acyclic graph with nodes 0 through n - 1 and directed edges [from, to]. Return the maximum number of edges in any directed path. An isolated node forms a path of length zero.

Function
longestDagPath(n: int, edges: int[][]) → int

Examples
Example 1
n = 5
edges = [[0,1],[0,2],[1,3],[2,3],[3,4]]
return = 3
A longest path such as 0 → 1 → 3 → 4 contains three edges.
Example 2
n = 4
edges = []
return = 0
Every node is isolated, so the longest path has zero edges.

Constraints
1 <= n <= 100000
0 <= edges.length <= 200000
The graph is acyclic and contains no duplicate edge.

Here's what it really reduces to: for every node, longest path ending there equals 1 + the max of its predecessors' values, or 0 if it has none. Compute in-degrees, push all zero in-degree nodes into a queue, and pop them in topological order. When you pop u, for each edge u to v, set dp[v] = max(dp[v], dp[u] + 1), then decrement v's in-degree and enqueue it at zero. Track the global max of dp. That's O(n + e). The common pitfall is recursing with DFS on 100000 nodes and blowing the stack, or skipping memoization and going exponential. Another trap is returning the node count instead of the edge count. Isolated nodes return 0, so initialize dp to zeros. If you freeze during the live OA, StealthCoder is the hedge that gives you the Kahn plus DP template fast.

Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.