0297MediumDepth-First SearchBreadth-First SearchGraph TheoryHeap (Priority Queue)Shortest PathDijkstra's Algorithm

Relay Broadcast Time

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A courier firm runs depots sorting depots numbered 1 through depots. Its one-way shuttle legs are listed in legs, where legs[i] = [u_i, v_i, w_i] means a shuttle leaves depot u_i for depot v_i and the crossing takes w_i minutes. A leg is usable in that direction only. No two legs share the same ordered pair of depots.

At minute 0 a dispatch note is posted at depot origin. A depot forwards a copy of the note along every leg leaving it the instant the note arrives, and copies travel at the same time without interfering. A depot counts as served the first moment any copy reaches it, and the origin is served at minute 0. Some legs take 0 minutes.

Return the minute at which the last depot is served. If at least one depot is never served, return -1.

Examples

Example 1

Input: legs = [[1,2,3],[2,3,4],[1,3,9]], depots = 3, origin = 1
Output: 7

Depot 1 holds the note at minute 0, depot 2 is served at minute 3, and a copy forwarded from depot 2 serves depot 3 at minute 7. The last depot is therefore served at minute 7.

Example 2

Input: legs = [[2,1,4]], depots = 2, origin = 1
Output: -1

The only leg runs from depot 2 to depot 1, and legs are one-way, so nothing ever leaves depot 1 and depot 2 stays unserved.

Constraints

1 <= origin <= depots <= 100
1 <= legs.length <= 6000
legs[i].length == 3
1 <= u_i, v_i <= depots
u_i != v_i
0 <= w_i <= 100
All the pairs (u_i, v_i) are unique. (i.e., no multiple legs.)

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def broadcast_time(legs: list[list[int]], depots: int, origin: int) -> int:

Java

public int broadcastTime(int[][] legs, int depots, int origin)

The intended approach

Time O(E log E)Space O(V + E)

Build the outgoing adjacency once, then run a heap-driven shortest-path sweep from the origin that always expands the depot with the smallest known arrival minute and relaxes its outgoing legs. Stale heap entries are skipped by comparing against the recorded arrival. Once the sweep ends, a depot still carrying no arrival forces -1; otherwise the answer is the largest recorded arrival.

Pythonentry point: broadcast_time

import heapq


def broadcast_time(legs, depots, origin):
    # Dijkstra outward from the origin: settle depots in order of arrival minute and
    # relax each outgoing leg once the tail is settled.  The answer is the largest
    # settled arrival minute, provided every depot got one.
    # O(E log E) time, O(V + E) space.
    adj = [[] for _ in range(depots + 1)]
    for u, v, w in legs:
        adj[u].append((v, w))
    unreached = float("inf")
    best = [unreached] * (depots + 1)
    best[origin] = 0
    heap = [(0, origin)]
    while heap:
        d, u = heapq.heappop(heap)
        if d > best[u]:
            continue
        for v, w in adj[u]:
            nd = d + w
            if nd < best[v]:
                best[v] = nd
                heapq.heappush(heap, (nd, v))
    arrivals = best[1:]
    for d in arrivals:
        if d == unreached:
            return -1
    return max(arrivals)

The obvious version

Relax every leg repeatedly, up to as many rounds as there are depots, stopping early when a full round changes nothing, then read the largest arrival.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_broadcast_time(legs, depots, origin):
    unreached = float("inf")
    dist = [unreached] * (depots + 1)
    dist[origin] = 0
    for _ in range(depots):
        changed = False
        for u, v, w in legs:
            if dist[u] + w < dist[v]:
                dist[v] = dist[u] + w
                changed = True
        if not changed:
            break
    arrivals = dist[1:]
    if any(d == unreached for d in arrivals):
        return -1
    return max(arrivals)

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
2 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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