0956MediumArrayDepth-First SearchMatrix

Counting Vessels on the Chart

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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 chart is given as board, each cell holding '.' for open water or 'X' for part of a vessel.

Every vessel lies in a single straight line, either along one row or down one column, and occupies a run of neighbouring cells. No two vessels touch, so between any two of them there is at least one cell of open water.

Return how many vessels are on the chart.

Examples

Example 1

Input: board = [["X", ".", ".", "X"], [".", ".", ".", "X"], [".", ".", ".", "X"], ["X", "X", ".", "."]]
Output: 3

There is a single cell at the top left, a run of three down the last column, and a run of two along the bottom row, which is three vessels.

Example 2

Input: board = [["X", "X", "X"]]
Output: 1

The three cells form one vessel lying along the row, counted once at its left-hand end.

Example 3

Input: board = [[".", ".", "."], [".", ".", "."], [".", ".", "."]]
Output: 0

The chart is all open water.

Constraints

1 <= board.length <= 200
1 <= board[i].length <= 200
Every row of board has the same length
board[i][j] is either '.' or 'X'

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 count_battleships(board: list[list[str]]) -> int:

Java

public int countBattleships(char[][] board)

The intended approach

Time O(m * n) for an m by n chart, one sweep.Space O(1).

Because no two vessels touch, every run of vessel cells belongs to exactly one vessel, and each vessel has exactly one cell that is its top-left end: the cell with no vessel cell directly above it and none directly to its left. So counting the vessels means counting those cells, which is one sweep with two neighbour checks per cell and nothing to remember. No marking, no searching outwards. Any one of the four corners would do just as well, so long as the choice is consistent: a vessel lying along a row has one leftmost cell and one rightmost, and one lying down a column has one topmost and one bottommost. What matters is picking one end per vessel and never two.

Pythonentry point: count_battleships

def count_battleships(board):
    m = len(board)
    n = len(board[0])
    total = 0
    for r in range(m):
        for c in range(n):
            if board[r][c] != "X":
                continue
            if r > 0 and board[r - 1][c] == "X":
                continue
            if c > 0 and board[r][c - 1] == "X":
                continue
            total += 1
    return total

The obvious version

Search outwards from each unvisited vessel cell, marking everything joined to it, and count how many such searches were needed. Keeps a visited mark for the whole chart rather than testing two neighbours.

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_count_battleships(board):
    m = len(board)
    n = len(board[0])
    seen = [[False] * n for _ in range(m)]
    total = 0
    for r in range(m):
        for c in range(n):
            if board[r][c] != "X" or seen[r][c]:
                continue
            total += 1
            stack = [(r, c)]
            seen[r][c] = True
            while stack:
                cr, cc = stack.pop()
                for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
                    nr = cr + dr
                    nc = cc + dc
                    if nr < 0 or nc < 0 or nr >= m or nc >= n:
                        continue
                    if board[nr][nc] != "X" or seen[nr][nc]:
                        continue
                    seen[nr][nc] = True
                    stack.append((nr, nc))
    return total

How this problem was checked

Reference solution executed against 10 test cases — 10 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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