Battleships in a Board Python Solution

Tested Python solution for LeetCode 419 with 34 pytest cases. Generate a practice environment with lcpy.

LeetCode 419, Medium. Topics: Array, Depth-First Search, Matrix. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 34 parametrized pytest cases, and a playground notebook:

lcpy gen -n 419   # by problem number
lcpy gen -s battleships_in_a_board   # by problem name

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Problem

Given an m x n matrix board where each cell is a battleship 'X' or empty '.', return the number of the battleships on board.

Battleships can only be placed horizontally or vertically on board. In other words, they can only be made of the shape 1 x k (1 row, k columns) or k x 1 (k rows, 1 column), where k can be of any size. At least one horizontal or vertical cell separates between two battleships (i.e., there are no adjacent battleships).

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Examples

Input: board = [["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]]
Output: 2

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

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Constraints

• m == board.length
• n == board[i].length
• 1 <= m, n <= 200
• board[i][j] is either ’.’ or ‘X’.

Follow up: Could you do it in one-pass, using only O(1) extra memory and without modifying the values of board?

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(m * n)
    # Space: O(1)
    def count_battleships(self, board: list[list[str]]) -> int:
        count = 0
        for r, row in enumerate(board):
            for c, cell in enumerate(row):
                if cell != "X":
                    continue
                if r > 0 and board[r - 1][c] == "X":
                    continue
                if c > 0 and row[c - 1] == "X":
                    continue
                count += 1
        return count

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Complexity

Time	Space
O(m * n)	O(1)

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