Leetcode #419: Battleships in a Board
In this guide, we solve Leetcode #419 Battleships in a Board in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Array, Matrix
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: board = [["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]]
Output: 2
Python Solution
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
Complexity
The time complexity is , where and are the number of rows and columns of the matrix, respectively. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.