1136HardDynamic ProgrammingBit ManipulationBreadth-First SearchGraph TheoryBitmask

Calling at Every Hub

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Trains the technique from

LeetCode 847Shortest Path Visiting All Nodes

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 network has hubs numbered 0 through n - 1, and links[i] lists the hubs joined directly to hub i. Every link works both ways, no hub is joined to itself, and every hub can be reached from every other one.

A courier may set off from whichever hub it likes, travels one link at a time, and may pass through any hub as often as it pleases.

Return the fewest links the courier has to cross to have called at every hub.

Examples

Example 1

Input: links = [[1], [0, 2], [1, 3], [2]]
Output: 3

The hubs form a single chain, so setting off at one end and walking to the other calls at all four hubs across three links.

Example 2

Input: links = [[1, 2, 3, 4], [0], [0], [0], [0]]
Output: 6

Four hubs hang off one middle hub. Setting off from a hanging hub, the courier has to come back through the middle between each pair of them, which is six links.

Example 3

Input: links = [[]]
Output: 0

A single hub is already called at, so nothing has to be crossed.

Constraints

1 <= links.length <= 12
0 <= links[i].length < links.length
no hub is listed among its own links
if hub b is listed for hub a then hub a is listed for hub b
every hub can be reached from every other one

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 shortest_path_length(links: list[list[int]]) -> int:

Java

public int shortestPathLength(int[][] links)

The intended approach

Time O(2^n * n^2) over the positions and their links.Space O(2^n * n) for the positions already reached.

Where the courier can usefully go next depends on two things: which hub it stands on and which hubs it has already called at. Since there are at most twelve hubs, the set of hubs called at fits in a twelve-bit pattern, so the whole position is a hub together with a pattern, at most twelve times four thousand and ninety-six of them. Every link costs the same, so walking outwards a layer at a time settles the fewest links to reach each position, and the first position reached whose pattern covers every hub gives the answer. Because the courier may start anywhere, every hub is seeded as a starting position at no cost, all in the first layer. Revisiting a hub is allowed and matters: the same hub with a different set of hubs called at is a different position, which is why the pattern belongs in the position and not merely a visited mark on the hub.

Pythonentry point: shortest_path_length

def shortest_path_length(links):
    from collections import deque
    n = len(links)
    full = (1 << n) - 1
    seen = set()
    queue = deque()
    for i in range(n):
        seen.add((i, 1 << i))
        queue.append((i, 1 << i, 0))
    while queue:
        node, mask, steps = queue.popleft()
        if mask == full:
            return steps
        for nb in links[node]:
            grown = mask | (1 << nb)
            if (nb, grown) not in seen:
                seen.add((nb, grown))
                queue.append((nb, grown, steps + 1))
    return -1

The obvious version

Work out the fewest links between every pair of hubs first, by letting each hub in turn serve as a stepping stone. Then treat the walk as an order in which the hubs are first called at, and build up the cheapest order over growing sets of hubs, adding the pair distance for each hub brought in. Any walk calling at every hub costs at least the pair distances between successive first calls, and stitching those shortest routes together reaches that cost, so the two agree.

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_shortest_path_length(links):
    n = len(links)
    inf = 1 << 30
    gap = [[inf] * n for _ in range(n)]
    for i in range(n):
        gap[i][i] = 0
        for j in links[i]:
            gap[i][j] = 1
    for mid in range(n):
        for i in range(n):
            for j in range(n):
                if gap[i][mid] + gap[mid][j] < gap[i][j]:
                    gap[i][j] = gap[i][mid] + gap[mid][j]
    full = (1 << n) - 1
    best = [[inf] * n for _ in range(1 << n)]
    for i in range(n):
        best[1 << i][i] = 0
    for mask in range(1 << n):
        for i in range(n):
            here = best[mask][i]
            if here >= inf:
                continue
            for j in range(n):
                if mask >> j & 1:
                    continue
                grown = mask | (1 << j)
                step = here + gap[i][j]
                if step < best[grown][j]:
                    best[grown][j] = step
    answer = inf
    for i in range(n):
        if best[full][i] < answer:
            answer = best[full][i]
    return answer

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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