0424MediumArrayDepth-First SearchBreadth-First SearchMatrix

Test Pit on the Survey Grid

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 construction site is surveyed square by square before anyone digs. board is the survey sheet, one character per square:

'M' a buried shell that the survey has located.
'E' earth nobody has opened up yet.
'B' earth already opened up with no shell in any square touching it.
a digit '1' through '8', earth already opened up, the digit being how many of the squares touching it hold a shell.

Two squares touch when they share an edge or just a corner, so a square away from the border touches eight others.

The surveyor sinks one test pit at click = [click_r, click_c], and that square is always 'M' or 'E'. Work the sheet as follows:

If the pit lands on 'M', the shell is struck: write 'X' on that square and the survey stops there.
Otherwise the square is opened up. If any touching square holds a shell, write the digit counting them on the square and go no further from it.
If no touching square holds a shell, write 'B' on the square and then open up every touching square that is still 'E' by these same rules, continuing outward from each square that turns out to be 'B'.

Squares that already carry 'B' or a digit when the pit is sunk are left exactly as they are, are never opened up again, and the opening up never spreads outward from them. Modify board in place and return it.

Examples

Example 1

Input: board = [["E", "E", "M", "E"], ["E", "E", "E", "E"], ["M", "E", "E", "E"], ["E", "E", "E", "E"]], click = [0, 0]
Output: [["B", "1", "M", "E"], ["1", "2", "E", "E"], ["M", "E", "E", "E"], ["E", "E", "E", "E"]]

No shell touches the pit square, so it becomes `"B"` and the survey opens up the three squares touching it. Each of those has a shell somewhere among the squares it touches, so each takes a digit and the survey goes no further.

Example 2

Input: board = [["E", "E", "E"], ["E", "M", "E"], ["E", "E", "E"]], click = [1, 1]
Output: [["E", "E", "E"], ["E", "X", "E"], ["E", "E", "E"]]

The pit lands straight on a located shell, so that square is marked struck and nothing else on the sheet changes.

Example 3

Input: board = [["E", "E", "E"], ["E", "M", "E"], ["E", "E", "E"]], click = [0, 0]
Output: [["1", "E", "E"], ["E", "M", "E"], ["E", "E", "E"]]

The corner square touches the shell in the middle, so it takes the digit for one shell and the rest of the sheet is left alone.

Example 4

Input: board = [["M", "M", "M"], ["M", "E", "M"], ["M", "M", "M"]], click = [1, 1]
Output: [["M", "M", "M"], ["M", "8", "M"], ["M", "M", "M"]]

The middle square is surrounded on all eight sides by located shells, so it takes the digit for eight.

Example 5

Input: board = [["E", "E", "E", "E"], ["E", "E", "E", "E"], ["E", "E", "E", "E"]], click = [1, 2]
Output: [["B", "B", "B", "B"], ["B", "B", "B", "B"], ["B", "B", "B", "B"]]

The sheet holds no shells at all, so the pit square becomes `"B"` and the opening up carries on until every square on the sheet is `"B"`.

Constraints

m == board.length
n == board[i].length
1 <= m, n <= 50
board[i][j] is either 'M', 'E', 'B', or a digit from '1' to '8'.
click.length == 2
0 <= click_r < m
0 <= click_c < n
board[click_r][click_c] is either 'M' or 'E'.

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 update_board(board: list[list[str]], click: list[int]) -> list[list[str]]:

Java

public char[][] updateBoard(char[][] board, int[] click)

The intended approach

Time O(m*n) for an m by n sheet: each square is written at most once and each write inspects its eight neighbours.Space O(m*n) for the pending list in the worst case, where the whole sheet is clear earth; the sheet itself is edited in place.

A struck shell is settled in one line. Otherwise count the shells touching the pit; a non-zero count writes the digit and finishes. A zero count starts a spread held in an explicit pending list: pop a square that was written `'B'`, look at its eight touching squares, and for each one still `'E'` count its own touching shells, writing either a digit, which ends that branch, or `'B'`, which pushes it on to be spread from in turn. Writing the character before queueing means a square is never handled twice, and the explicit list keeps a wide sheet of clear earth from nesting thousands of calls deep.

Pythonentry point: update_board

def update_board(board, click):
    # Excavate outward from an explicit pending list, never nested calls, so a
    # 50 by 50 sheet of clear earth cannot exhaust the interpreter's frames.
    # Squares are stamped the moment they are queued, so none is queued twice.
    rows = len(board)
    cols = len(board[0])
    start_r, start_c = click[0], click[1]
    if board[start_r][start_c] == "M":
        board[start_r][start_c] = "X"
        return board

    steps = ((-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1))

    def shells(r, c):
        found = 0
        for dr, dc in steps:
            rr, cc = r + dr, c + dc
            if 0 <= rr < rows and 0 <= cc < cols and board[rr][cc] == "M":
                found += 1
        return found

    pending = [(start_r, start_c)]
    near = shells(start_r, start_c)
    if near > 0:
        board[start_r][start_c] = str(near)
        return board
    board[start_r][start_c] = "B"
    while pending:
        r, c = pending.pop()
        for dr, dc in steps:
            rr, cc = r + dr, c + dc
            if not (0 <= rr < rows and 0 <= cc < cols):
                continue
            if board[rr][cc] != "E":
                continue
            near = shells(rr, cc)
            if near > 0:
                board[rr][cc] = str(near)
            else:
                board[rr][cc] = "B"
                pending.append((rr, cc))
    return board

The obvious version

Note which squares already carried `'B'` before the pit, write the pit square, then sweep the whole sheet over and over. On each sweep, any square carrying `'B'` that was not one of those opens up every touching square that is still `'E'`. Stop when a full sweep changes nothing. O((m*n)^2).

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_update_board(board, click):
    rows = len(board)
    cols = len(board[0])
    steps = ((-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1))

    def shells(r, c):
        found = 0
        for dr, dc in steps:
            rr = r + dr
            cc = c + dc
            if 0 <= rr < rows and 0 <= cc < cols and board[rr][cc] == "M":
                found += 1
        return found

    already = set()
    for r in range(rows):
        for c in range(cols):
            if board[r][c] == "B":
                already.add((r, c))

    sr = click[0]
    sc = click[1]
    if board[sr][sc] == "M":
        board[sr][sc] = "X"
        return board
    hits = shells(sr, sc)
    board[sr][sc] = str(hits) if hits > 0 else "B"
    moved = True
    while moved:
        moved = False
        for r in range(rows):
            for c in range(cols):
                if board[r][c] != "B" or (r, c) in already:
                    continue
                for dr, dc in steps:
                    rr = r + dr
                    cc = c + dc
                    if not (0 <= rr < rows and 0 <= cc < cols):
                        continue
                    if board[rr][cc] != "E":
                        continue
                    near = shells(rr, cc)
                    board[rr][cc] = str(near) if near > 0 else "B"
                    moved = True
    return board

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
5 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