Battleships in a Board

LeetCode 419 | Difficulty: Medium

Medium

Problem 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?

Topics: Array, Depth-First Search, Matrix

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Approach

Matrix

Treat the matrix as a 2D grid. Common techniques: directional arrays (dx, dy) for movement, BFS/DFS for connected regions, in-place marking for visited cells, boundary traversal for spiral patterns.

When to use

Grid traversal, island problems, path finding, rotating/transforming matrices.

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Solutions

Solution 1: C# (Best: 92 ms)

Metric	Value
Runtime	92 ms
Memory	39 MB
Date	2022-02-14

Solution

public class Solution {
    public int CountBattleships(char[][] board) {
        int m = board.Length;
    int n = board[0].GetLength(0);
    bool[,] used = new bool[m,n];
    int ships = 0;
    
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            if (board[i][j] == 'X' && !used[i,j])
            {
                Dfs(board, i, j, m, n, used);
                ships++;
            }           
        }
    }
    return ships;
}

private void Dfs(char[][] board, int i, int j, int m, int n,bool[,] used)
{
    List<int[]> dirs = new List<int[]>() {
        new int[] {1, 0},
        new int[] {0, 1}
    };
    foreach (var dir in dirs)
    {
        int x = i + dir[0];
        int y = j + dir[1];
        if (x < 0 || x >= m || y < 0 || y >= n || used[x, y] || board[x][y] != 'X') continue;
        while (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'X')
        {
            used[x, y] = true;
            x = x + dir[0];
            y = y + dir[1];
        }
        break;
    }
}
}

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Complexity Analysis

Approach	Time	Space
Graph BFS/DFS	$O(V + E)$	$O(V)$

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Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.