0539HardArrayBreadth-First SearchMatrix

Siting the Campus Bike Rack

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Trains the technique from

LeetCode 317Shortest Distance from All Buildings

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 campus plan arrives as grid. Each cell holds one of three codes:

0 is paved walkway, which people may stand on and walk across.
1 is a lecture hall.
2 is a fenced-off works area, which nobody may enter.

A bike rack has to go on one walkway cell. From a walkway cell a person may step to the cell directly above, below, left or right. A route from the chosen cell to a lecture hall is a chain of such steps that begins on the chosen cell, ends on that hall, and whose every other cell is walkway; the length of the route is the number of steps it contains. The reach of a walkway cell is the sum, over all lecture halls, of the length of the shortest route from that cell to that hall.

Return the smallest reach over all walkway cells from which every lecture hall has at least one route. If no walkway cell can route to every hall, return -1.

Examples

Example 1

Input: grid = [[0, 0, 1, 0], [0, 2, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1]]
Output: 6

Put the rack on the walkway cell at row 2 column 2. Its shortest route to the hall at row 0 column 2 is two steps, to the hall at row 2 column 0 is two steps, and to the hall at row 3 column 3 is two steps, so its reach is 6, and no walkway cell has a smaller reach.

Example 2

Input: grid = [[0, 1, 0, 0], [2, 0, 2, 0], [0, 0, 0, 1]]
Output: 4

The cell at row 0 column 2 is one step from the hall at row 0 column 1 and three steps from the hall at row 2 column 3, a reach of 4, which is the smallest reach on this plan.

Example 3

Input: grid = [[1, 0, 0], [2, 1, 2], [1, 0, 0]]
Output: -1

The works areas at row 1 column 0 and row 1 column 2 leave the two walkway cells of row 0 and the two walkway cells of row 2 with no chain of walkway steps between them, so no single cell has a route to all three halls and the answer is -1.

Constraints

rows == grid.length
cols == grid[i].length
1 <= rows, cols <= 50
grid[i][j] is 0, 1, or 2.
grid holds at least one lecture hall.

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_distance(grid: list[list[int]]) -> int:

Java

public int shortestDistance(int[][] grid)

The intended approach

Time O(h * rows * cols) where h is the number of lecture halls, since each hall's sweep touches every cell at most once.Space O(rows * cols) for the running totals, the counters and one sweep's visited flags.

Run one breadth-first sweep outward from each lecture hall instead of from each candidate cell, because there are usually far fewer halls than walkway cells and a single sweep already gives that hall's shortest route to every walkway cell at once. Each sweep only steps onto walkway cells, adds its step count into a running total kept per walkway cell, and increments a per-cell counter of how many halls have arrived. After all sweeps, scan the walkway cells whose counter equals the hall count and report the smallest running total; if the scan finds none, report `-1`.

Pythonentry point: shortest_distance

def shortest_distance(grid):
    # One breadth-first sweep per lecture hall. Each sweep adds its step count
    # into a running total for every walkway cell it can reach and bumps that
    # cell's reach counter, so a cell is a candidate only when its counter
    # equals the number of halls.
    from collections import deque
    rows = len(grid)
    cols = len(grid[0])
    total = [[0] * cols for _ in range(rows)]
    reached = [[0] * cols for _ in range(rows)]
    halls = 0
    for hr in range(rows):
        for hc in range(cols):
            if grid[hr][hc] != 1:
                continue
            halls += 1
            seen = [[False] * cols for _ in range(rows)]
            seen[hr][hc] = True
            queue = deque([(hr, hc, 0)])
            while queue:
                r, c, d = queue.popleft()
                for nr, nc in ((r + 1, c), (r - 1, c), (r, c + 1), (r, c - 1)):
                    if 0 <= nr < rows and 0 <= nc < cols and not seen[nr][nc]:
                        if grid[nr][nc] == 0:
                            seen[nr][nc] = True
                            total[nr][nc] += d + 1
                            reached[nr][nc] += 1
                            queue.append((nr, nc, d + 1))
    best = -1
    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == 0 and reached[r][c] == halls:
                if best < 0 or total[r][c] < best:
                    best = total[r][c]
    return best

The obvious version

Start a separate breadth-first sweep from every walkway cell in turn. The sweep spreads across walkway cells and records the step count the first time it touches each lecture hall, without walking onward through halls or works areas. A cell qualifies when the recorded set covers all halls, and its reach is the sum of those recorded step counts. Keep the smallest reach found, or -1 if no cell qualifies.

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_distance(grid):
    from collections import deque
    rows = len(grid)
    cols = len(grid[0])
    hall_count = 0
    for row in grid:
        for v in row:
            if v == 1:
                hall_count += 1
    best = -1
    for sr in range(rows):
        for sc in range(cols):
            if grid[sr][sc] != 0:
                continue
            found = {}
            seen = [[False] * cols for _ in range(rows)]
            seen[sr][sc] = True
            queue = deque([(sr, sc, 0)])
            while queue:
                r, c, d = queue.popleft()
                for nr, nc in ((r + 1, c), (r - 1, c), (r, c + 1), (r, c - 1)):
                    if 0 <= nr < rows and 0 <= nc < cols and not seen[nr][nc]:
                        if grid[nr][nc] == 1:
                            seen[nr][nc] = True
                            found[(nr, nc)] = d + 1
                        elif grid[nr][nc] == 0:
                            seen[nr][nc] = True
                            queue.append((nr, nc, d + 1))
            if len(found) == hall_count:
                walk = 0
                for v in found.values():
                    walk += v
                if best < 0 or walk < best:
                    best = walk
    return best

How this problem was checked

Reference solution executed against 11 test cases — 11 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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