0466HardArrayBit ManipulationBreadth-First SearchMatrix

Wrench Run Through the Utility Vault

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Trains the technique from

LeetCode 864Shortest Path to Get All Keys

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 splice crew has a floor plan of an underground utility vault, handed over as vault: a list of equal-length strings, one per row, read top to bottom and left to right.

Each character says what is in that cell:

"@" is the access hatch the crew climbs in through. Exactly one cell holds it.
"." is bare floor.
"#" is solid rock.
a lowercase letter is a coded wrench lying on the floor.
an uppercase letter is a bolted gate that only the wrench with the same letter can turn.

From the cell it stands on the crew may step to the cell directly above, below, left or right. Each such step counts as one move. It may never step outside the plan and never into rock. Stepping onto a wrench picks it up at no cost, and a wrench once picked up is carried for the rest of the run. Stepping onto a gate is only allowed while carrying its matching wrench, and costs nothing beyond the step itself; the gate stays open afterwards. Bare floor and the hatch cell are always free to step on.

Return the fewest moves needed until the crew is carrying every wrench on the plan. Return -1 when no sequence of moves collects them all.

Examples

Example 1

Input: vault = ["@..a#", "##.#.", "b.A.."]
Output: 8

One run of eight moves works: three steps east onto wrench `"a"`, one step back west, two steps south through gate `"A"`, then two steps west onto wrench `"b"`.

Example 2

Input: vault = ["@#a", "###", "A#."]
Output: -1

Rock sits east and south of the hatch, so the crew cannot take a single step and wrench `"a"` is never reached.

Example 3

Input: vault = ["@.a.#.", ".#.#.b", "..A..#", "c#..B.", "...#.C"]
Output: 14

Three wrenches lie on this plan and a run of fourteen moves ends with all three in hand, having opened gates `"A"` and `"B"` along the way.

Constraints

rows == len(vault)
cols == len(vault[i])
1 <= rows, cols <= 30
vault[i][j] is an English letter, '.', '#', or '@'.
There is exactly one '@' cell.
The number of wrenches k satisfies 1 <= k <= 6, and they are labelled with the first k lowercase letters.
Each wrench letter appears in exactly one cell, and its uppercase gate appears in exactly one cell.

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_moves_for_wrenches(vault: list[str]) -> int:

Java

public int fewestMovesForWrenches(String[] vault)

The intended approach

Time O(rows * cols * 2^k), with k the number of wrenchesSpace O(rows * cols * 2^k) for the queue and the set of visited triples

A position on the plan is not enough to describe the crew: two visits to the same cell differ when different wrenches are in hand. Pack the carried wrenches into a bitmask, one bit per letter, and treat `(row, col, mask)` as the thing being explored. Sweep those triples outward from the hatch with a queue so every triple is first reached by the fewest moves, refusing a gate cell unless its bit is set and setting a bit on entering a wrench cell. The first time the mask covers every letter, report the move count.

Pythonentry point: fewest_moves_for_wrenches

from collections import deque


def fewest_moves_for_wrenches(vault):
    # BFS over (row, col, wrenches carried); at most rows*cols*2^k states.
    # time O(rows * cols * 2^k), space O(rows * cols * 2^k)
    rows, cols = len(vault), len(vault[0])
    every = 0
    sr = sc = 0
    for r in range(rows):
        line = vault[r]
        for c in range(cols):
            ch = line[c]
            if ch == "@":
                sr, sc = r, c
            elif "a" <= ch <= "f":
                every |= 1 << (ord(ch) - 97)
    seen = {(sr, sc, 0)}
    queue = deque([(sr, sc, 0, 0)])
    while queue:
        r, c, carried, moves = queue.popleft()
        for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
            nr, nc = r + dr, c + dc
            if nr < 0 or nr >= rows or nc < 0 or nc >= cols:
                continue
            ch = vault[nr][nc]
            if ch == "#":
                continue
            if "A" <= ch <= "F" and not (carried >> (ord(ch) - 65)) & 1:
                continue
            nxt = carried
            if "a" <= ch <= "f":
                nxt |= 1 << (ord(ch) - 97)
            if nxt == every:
                return moves + 1
            state = (nr, nc, nxt)
            if state in seen:
                continue
            seen.add(state)
            queue.append((nr, nc, nxt, moves + 1))
    return -1

The obvious version

Keep a dictionary of the best move count found so far for each reachable (row, col, carried) triple and relax it repeatedly: sweep every recorded triple, push its four neighbours, and lower any count that improves. Repeat whole sweeps until one sweep changes nothing, then take the smallest count over triples holding every wrench.

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_moves_for_wrenches(vault):
    rows, cols = len(vault), len(vault[0])
    letters = set()
    sr = sc = 0
    for r in range(rows):
        for c in range(cols):
            ch = vault[r][c]
            if ch == "@":
                sr, sc = r, c
            elif ch.islower():
                letters.add(ch)
    every = (1 << len(letters)) - 1
    best = {(sr, sc, 0): 0}
    moved = True
    while moved:
        moved = False
        for (r, c, carried), d in list(best.items()):
            for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
                nr, nc = r + dr, c + dc
                if nr < 0 or nr >= rows or nc < 0 or nc >= cols:
                    continue
                ch = vault[nr][nc]
                if ch == "#":
                    continue
                if ch.isupper() and ch != "@":
                    if not (carried >> (ord(ch.lower()) - 97)) & 1:
                        continue
                nxt = carried
                if ch.islower():
                    nxt = carried | (1 << (ord(ch) - 97))
                state = (nr, nc, nxt)
                if state not in best or best[state] > d + 1:
                    best[state] = d + 1
                    moved = True
    answer = -1
    for (r, c, carried), d in best.items():
        if carried == every and (answer < 0 or d < answer):
            answer = d
    return answer

How this problem was checked

Reference solution executed against 7 test cases — 7 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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