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419. Battleships in a Board

DifficultyMedium

Description

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).

 

Example 1:

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

Example 2:

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

 

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 board?

Solutions

Solution 1: Direct Iteration

Thinking

Ships are horizontal or vertical and never touch. A flood fill can mark a whole ship, but it writes the board or needs extra flags. The follow-up asks for one pass and constant extra memory.

Each ship has a unique top-left \(\texttt{X}\): the cell above and the cell to the left are not \(\texttt{X}\). Count those corners.

Because ships do not touch, that corner is unique, so there is neither under-count nor double-count.

We can iterate through the matrix, find the top-left corner of each battleship, i.e., the position where the current position is X and both the top and left are not X, and increment the answer by one.

After the iteration ends, return the answer.

The time complexity is \(O(m \times n)\), where \(m\) and \(n\) are the number of rows and columns of the matrix, respectively. The space complexity is \(O(1)\).

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class Solution:
    def countBattleships(self, board: List[List[str]]) -> int:
        m, n = len(board), len(board[0])
        ans = 0
        for i in range(m):
            for j in range(n):
                if board[i][j] == '.':
                    continue
                if i > 0 and board[i - 1][j] == 'X':
                    continue
                if j > 0 and board[i][j - 1] == 'X':
                    continue
                ans += 1
        return ans

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class Solution {
    public int countBattleships(char[][] board) {
        int m = board.length, n = board[0].length;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == '.') {
                    continue;
                }
                if (i > 0 && board[i - 1][j] == 'X') {
                    continue;
                }
                if (j > 0 && board[i][j - 1] == 'X') {
                    continue;
                }
                ++ans;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int countBattleships(vector<vector<char>>& board) {
        int m = board.size(), n = board[0].size();
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == '.') {
                    continue;
                }
                if (i > 0 && board[i - 1][j] == 'X') {
                    continue;
                }
                if (j > 0 && board[i][j - 1] == 'X') {
                    continue;
                }
                ++ans;
            }
        }
        return ans;
    }
};

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func countBattleships(board [][]byte) (ans int) {
    for i, row := range board {
        for j, c := range row {
            if c == '.' {
                continue
            }
            if i > 0 && board[i-1][j] == 'X' {
                continue
            }
            if j > 0 && board[i][j-1] == 'X' {
                continue
            }
            ans++
        }
    }
    return
}

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function countBattleships(board: string[][]): number {
    const m = board.length;
    const n = board[0].length;
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (board[i][j] === '.') {
                continue;
            }
            if (i && board[i - 1][j] === 'X') {
                continue;
            }
            if (j && board[i][j - 1] === 'X') {
                continue;
            }
            ++ans;
        }
    }
    return ans;
}

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