1036MediumArrayBreadth-First SearchMatrix

The Nearest Way Out of the Vault

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Trains the technique from

LeetCode 1926Nearest Exit from Entrance in Maze

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 vault floor is given as the grid vault, where '.' marks an open cell and '+' marks a wall. A step moves to a cell sharing an edge with the current one, and only open cells may be stood on.

You start on the open cell at row start[0] and column start[1]. An exit is any open cell on the outer edge of the grid other than the cell you start on.

Return the fewest steps needed to reach an exit, or -1 when no exit can be reached.

Examples

Example 1

Input: vault = [[".", ".", "."], [".", "+", "."], [".", ".", "."]], start = [1, 0]
Output: 1

The starting cell sits on the left edge, but it cannot count as its own exit. The cell straight above it is open and also on the edge, so one step is enough.

Example 2

Input: vault = [["+", "+", "+"], ["+", ".", "+"], ["+", "+", "+"]], start = [1, 1]
Output: -1

The only open cell is walled in on all four sides, so no exit can be reached.

Example 3

Input: vault = [["+", "+", "+", "+", "+"], ["+", ".", ".", ".", "."], ["+", ".", "+", "+", "+"], ["+", ".", "+", "+", "+"], ["+", ".", "+", "+", "+"]], start = [1, 1]
Output: 3

Two exits are reachable: the open cell at the right end of the second row and the one at the bottom of the second column. Both lie three steps away.

Constraints

1 <= vault.length <= 100
1 <= vault[i].length <= 100
Every cell of vault is a full stop or a plus sign.
start.length == 2
0 <= start[0] <= vault.length - 1
The cell you start on is open.

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 nearest_exit(vault: list[list[str]], start: list[int]) -> int:

Java

public int nearestExit(char[][] vault, int[] start)

The intended approach

Time O(m * n) over the grid's cells.Space O(m * n) for the queue and the marks.

Steps all cost the same, so walking outward level by level from the starting cell reaches every cell by the fewest steps possible, and the first exit the walk meets is the nearest one. Put the starting cell into a queue at nothing and mark it. Then take cells off the front and step to their unmarked open neighbours. The moment a neighbour lands on the outer edge, report the step count that reached it. That test happens before the cell joins the queue, which is also why the starting cell never counts as its own exit even when it sits on the edge. Running the queue dry means no exit is reachable at all. Each cell is marked once and joins the queue at most once, so the walk costs the grid's size.

Pythonentry point: nearest_exit

def nearest_exit(vault, start):
    from collections import deque
    rows = len(vault)
    cols = len(vault[0])
    seen = [[False] * cols for _ in range(rows)]
    seen[start[0]][start[1]] = True
    queue = deque([(start[0], start[1], 0)])
    while queue:
        r, c, steps = queue.popleft()
        for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
            nr = r + dr
            nc = c + dc
            if nr < 0 or nr >= rows or nc < 0 or nc >= cols:
                continue
            if seen[nr][nc] or vault[nr][nc] == '+':
                continue
            if nr == 0 or nr == rows - 1 or nc == 0 or nc == cols - 1:
                return steps + 1
            seen[nr][nc] = True
            queue.append((nr, nc, steps + 1))
    return -1

The obvious version

Settle the step counts by relaxation instead. Start every cell at an impossibly large count except the starting cell at nothing, then sweep the whole grid over and over, lowering each open cell to one more than its cheapest open neighbour, until a full sweep changes nothing. Then take the smallest count among the open edge cells other than the start. Costs repeated sweeps of the grid rather than one outward walk.

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_nearest_exit(vault, start):
    rows = len(vault)
    cols = len(vault[0])
    big = rows * cols + 1
    far = [[big] * cols for _ in range(rows)]
    far[start[0]][start[1]] = 0
    changed = True
    while changed:
        changed = False
        for r in range(rows):
            for c in range(cols):
                if vault[r][c] == '+':
                    continue
                for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
                    nr = r + dr
                    nc = c + dc
                    if nr < 0 or nr >= rows or nc < 0 or nc >= cols:
                        continue
                    if vault[nr][nc] == '+':
                        continue
                    if far[nr][nc] + 1 < far[r][c]:
                        far[r][c] = far[nr][nc] + 1
                        changed = True
    best = -1
    for r in range(rows):
        for c in range(cols):
            if vault[r][c] == '+':
                continue
            if r == start[0] and c == start[1]:
                continue
            if r == 0 or r == rows - 1 or c == 0 or c == cols - 1:
                if far[r][c] < big and (best < 0 or far[r][c] < best):
                    best = far[r][c]
    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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