0788MediumArrayGraph TheoryHeap (Priority Queue)Shortest PathDijkstra's Algorithm

Most Reliable Parcel Route

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Trains the technique from

LeetCode 1514Path with Maximum Probability

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 company runs n depots, numbered 0 to n - 1, joined by two-way trunk links.

The links are given as two parallel arrays. links[i] = [u, v] is a link that a parcel may cross in either direction, and success[i] is the probability that a parcel crossing that link arrives undamaged. No pair of depots is joined by more than one link, and no link joins a depot to itself.

A route is a sequence of links leading from one depot to the next. A parcel survives the route only if it survives every link on it, and the links fail independently, so the route's success probability is the product of the success values of the links it uses.

Return the largest success probability of any route from depot origin to depot target. If no route joins them at all, return 0.0.

Report the probability as a decimal number; an answer within 1e-5 of the correct value is accepted.

Examples

Example 1

Input: n = 4, links = [[0,1],[1,3],[0,2],[2,3]], success = [0.5,0.75,0.875,0.25], origin = 0, target = 3
Output: 0.375

The route 0 to 1 to 3 uses the links with chances 0.5 and 0.75, so a parcel survives it with probability 0.5 * 0.75 = 0.375. The other route, 0 to 2 to 3, multiplies out to 0.875 * 0.25 = 0.21875.

Example 2

Input: n = 5, links = [[0,1],[1,2],[3,4]], success = [0.5,0.5,1.0], origin = 0, target = 4
Output: 0.0

Depots 0, 1 and 2 sit on one side and depots 3 and 4 on the other, with no link crossing between the two sides, so no route reaches depot 4 from depot 0.

Example 3

Input: n = 2, links = [[0,1]], success = [0.625], origin = 1, target = 0
Output: 0.625

The single link is written as [0,1] but may be crossed in either direction, so the one-link route from depot 1 to depot 0 carries its chance of 0.625.

Constraints

2 <= n <= 10^4
0 <= links.length <= 2 * 10^4
links.length == success.length
links[i].length == 2
0 <= links[i][j] <= n - 1
links[i][0] != links[i][1]
0 <= success[i] <= 1
0 <= origin <= n - 1
0 <= target <= n - 1
origin != target
At most one link joins any given pair of depots.

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 best_delivery_odds(n: int, links: list[list[int]], success: list[float], origin: int, target: int) -> float:

Java

public double bestDeliveryOdds(int n, int[][] links, double[] success, int origin, int target)

The intended approach

Time O((n + m) log n) for n depots and m links: each link relaxation may push one heap entry.Space O(n + m) for the adjacency lists, the per-depot odds and the heap.

Build adjacency lists holding both directions of every link, then settle depots from a max-heap keyed on the best odds found so far, starting with the origin at 1.0. Popping the best pending depot and relaxing its links records, for each neighbour, the product of the popped odds and the link's chance whenever that beats the neighbour's current figure. Because a chance is never above 1 the recorded odds only ever fall along a route, so the first pop of the target is its answer, and a target never popped keeps its initial 0.0.

Pythonentry point: best_delivery_odds

import heapq


def best_delivery_odds(n, links, success, origin, target):
    # Every hop multiplies the survival odds by a factor of at most 1, so a route's
    # odds never rise as it is extended. That lets the depots be settled in order of
    # decreasing odds from a max-heap: the first time a depot comes off the heap its
    # recorded odds cannot be beaten later.
    # Time O((V + E) log V), space O(V + E) for the adjacency lists and the heap.
    adjacent = [[] for _ in range(n)]
    for i in range(len(links)):
        u, v = links[i]
        chance = success[i]
        adjacent[u].append((v, chance))
        adjacent[v].append((u, chance))

    odds = [0.0] * n
    odds[origin] = 1.0
    heap = [(-1.0, origin)]
    while heap:
        negative, depot = heapq.heappop(heap)
        here = -negative
        if here < odds[depot]:
            continue
        if depot == target:
            return here
        for nxt, chance in adjacent[depot]:
            through = here * chance
            if through > odds[nxt]:
                odds[nxt] = through
                heapq.heappush(heap, (-through, nxt))
    return odds[target]

The obvious version

Walk every route that never revisits a depot, carried on an explicit stack of (depot, running product, visited set) frames, and keep the largest product that reaches the target. Nothing reaching the target leaves the answer at 0.0.

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_best_delivery_odds(n, links, success, origin, target):
    adjacent = [[] for _ in range(n)]
    for i in range(len(links)):
        u, v = links[i]
        adjacent[u].append((v, success[i]))
        adjacent[v].append((u, success[i]))

    best = 0.0
    frames = [(origin, 1.0, 1 << origin)]
    while frames:
        depot, running, seen = frames.pop()
        if depot == target:
            if running > best:
                best = running
            continue
        for nxt, chance in adjacent[depot]:
            if not (seen >> nxt) & 1:
                frames.append((nxt, running * chance, seen | (1 << nxt)))
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 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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