0184HardArrayHash TableBreadth-First Search

Fewest Loops to the Dock

Tracked in this browser only

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 hospital campus is served by a fixed set of shuttle loops. Loop i calls at the docks listed in loops[i], cycling through them again and again without ever stopping.

A courier standing at a dock may board any loop that calls at that dock, stay aboard as long as they like, and step off at any other dock the loop calls at. Moving from one loop to another is only possible at a dock the two loops have in common, and the courier is not allowed to walk between docks.

The courier begins at dock start and needs to reach dock finish. Return the fewest loops the courier must board. When the courier is already standing at finish, the answer is 0. When no chain of loops joins the two docks, return -1.

Within a single loop every dock number is listed once, but two loops may well share docks. Dock numbers are only labels, so a number that appears in no loop simply has no shuttle service.

Examples

Example 1

Input: loops = [[4, 8, 12], [12, 20, 5]], start = 4, finish = 5
Output: 2

Board the first loop at dock 4 and ride to dock 12, which both loops call at, then board the second loop there and ride to dock 5.

Example 2

Input: loops = [[3, 9], [11, 14]], start = 3, finish = 14
Output: -1

The two loops share no dock, so a courier who boards at dock 3 can never reach dock 14.

Example 3

Input: loops = [[7, 2, 5]], start = 5, finish = 5
Output: 0

The courier is already where they need to be, so no shuttle is boarded at all.

Constraints

1 <= loops.length <= 500
1 <= loops[i].length <= 10^5
Every dock number inside one loop is distinct
The total number of dock entries across all loops is at most 10^5
0 <= loops[i][j] < 10^6
0 <= start, finish < 10^6

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 fewest_loops(loops: list[list[int]], start: int, finish: int) -> int:

Java

public int fewestLoops(int[][] loops, int start, int finish)

The intended approach

Time O(S), where S is the total number of dock entries across all loopsSpace O(S)

Treat each loop, not each dock, as a node. Index every dock to the loops that call there, seed a breadth-first search with the loops serving `start` at a cost of one boarding, and expand a loop by scanning its docks and enqueuing any loop not yet boarded at one more boarding. Because layers grow by one boarding at a time, the first time `finish` shows up the count is minimal.

Pythonentry point: fewest_loops

def fewest_loops(loops, start, finish):
    # Search over loops, not docks: each BFS layer is one extra boarding.
    # Time O(S) where S = total docks listed, since every dock and every loop
    # is expanded at most once. Space O(S) for the dock index plus the queue.
    if start == finish:
        return 0
    from collections import deque
    dock_loops = {}
    for idx, docks in enumerate(loops):
        for dock in docks:
            if dock in dock_loops:
                dock_loops[dock].append(idx)
            else:
                dock_loops[dock] = [idx]
    if start not in dock_loops or finish not in dock_loops:
        return -1
    seen_loops = set()
    seen_docks = {start}
    queue = deque()
    for idx in dock_loops[start]:
        seen_loops.add(idx)
        queue.append((idx, 1))
    while queue:
        idx, boarded = queue.popleft()
        for dock in loops[idx]:
            if dock == finish:
                return boarded
            if dock in seen_docks:
                continue
            seen_docks.add(dock)
            for nxt in dock_loops[dock]:
                if nxt not in seen_loops:
                    seen_loops.add(nxt)
                    queue.append((nxt, boarded + 1))
    return -1

The obvious version

Grow the set of docks the courier can stand at, one boarding at a time: each round consumes every not-yet-used loop that touches the current set and merges all of its docks in afterwards, so nothing added this round is credited to this round's expansion. Stop when the finish appears or when a round adds no loop.

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_fewest_loops(loops, start, finish):
    if start == finish:
        return 0
    reachable = {start}
    consumed = [False] * len(loops)
    rides = 0
    while True:
        rides += 1
        added = set()
        grew = False
        for i, docks in enumerate(loops):
            if consumed[i]:
                continue
            touches = False
            for dock in docks:
                if dock in reachable:
                    touches = True
                    break
            if touches:
                consumed[i] = True
                grew = True
                for dock in docks:
                    added.add(dock)
        reachable |= added
        if finish in reachable:
            return rides
        if not grew:
            return -1

How this problem was checked

Reference solution executed against 12 test cases — 12 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in