Find how long a signal takes to reach every node of a weighted graph with Dijkstra's algorithm and heapq in O(E log V). Return -1 if a node is unreachable.
The problem
times is a list of directed edges [u, v, w] (travel time w ≥ 0) between nodes 1..n. A signal starts at node k. Return the time until all nodes have received it, or -1 if some node never does.
Examples
Example 1
Input
network_delay([[2, 1, 1], [2, 3, 1], [3, 4, 1]], 4, 2)
Expected output
2
Example 2
Input
network_delay([[1, 2, 1]], 2, 2)
Expected output
-1
+ 3 hidden tests on Submit — indirect path is shorter, zero-weight edge.
Edge cases to ask about
- Unreachable node
- Single node
- Zero-weight edges
Hints
0/3How an interviewer scores this
0/9Your code runs in real CPython inside your browser — nothing is sent anywhere. The first run downloads the interpreter (about 6 MB, once). Your code is saved on this device as you type.
Complexity Lab
What does this cost as n grows?
Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and measure your code against the optimal one at growing input sizes.
Pick both to reveal the answer.
Measure it
Runs the function on inputs of size 250 up to 16,000 and records the time and peak memory. Slow solutions stop early — a short curve is itself the answer.
From brute force to optimal
The progression an interviewer wants to hear, one step at a time.
| Approach | Time | Space | Idea |
|---|---|---|---|
| Bellman-Ford | O(V · E) | O(V) | Handles negative weights; not needed here. |
| bestDijkstra with a binary heap | O(E log V) | O(V + E) | Pop the closest unsettled node; its distance is final. |
Walkthrough of the optimal approach (try it yourself first)
Dijkstra: keep a min-heap of (distance, node). Pop the closest; if it's already settled, skip it (lazy deletion). Otherwise its distance is final — record it and push its neighbours with d + w. The answer is the largest settled distance, or -1 if not every node was reached.
Dijkstra needs non-negative weights; with negative edges use Bellman-Ford.
Complexity: O(E log V) time, O(V + E) space. Every edge may push one heap entry, and each push/pop costs O(log E) = O(log V).
Reveal the reference solution
import heapq def network_delay(times, n, k): graph = {i: [] for i in range(1, n + 1)} for u, v, w in times: graph[u].append((v, w)) dist = {} heap = [(0, k)] while heap: d, node = heapq.heappop(heap) if node in dist: continue dist[node] = d for nxt, w in graph[node]: if nxt not in dist: heapq.heappush(heap, (d + w, nxt)) return max(dist.values()) if len(dist) == n else -1
Follow-ups interviewers ask
- Return the path, not just the time.
- Cheapest flight with at most k stops (modified BFS/Bellman-Ford).
Frequently asked interview questions
Core interview concepts, complexities, and follow-ups scored by hiring teams.
What is the time complexity of Network Delay Time (Dijkstra) in Python?
The optimal solution runs in O(E log V) time and O(V + E) auxiliary space. Every edge may push one heap entry, and each push/pop costs O(log E) = O(log V).
What is the brute-force approach, and how do you optimise it?
Bellman-Ford: O(V · E) time, O(V) space. Handles negative weights; not needed here. Dijkstra with a binary heap: O(E log V) time, O(V + E) space. Pop the closest unsettled node; its distance is final.
What follow-up questions do interviewers ask about Network Delay Time (Dijkstra)?
Return the path, not just the time. Cheapest flight with at most k stops (modified BFS/Bellman-Ford).