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+---
+id: battleships-in-a-board
+title: Battleships in a Board
+sidebar_label: 0419-Battleships-in-a-Board
+tags:
+- Array
+- Depth-First Search
+- Matrix
+description: "Given an `m x n` board where each cell is a battleship 'X' or empty '.', count the number of battleships on the board."
+---
+
+## Problem
+
+Given an `m x n` board where each cell is a battleship `'X'` or empty `'.'`, return the number of battleships on the board.
+
+Battleships can only be placed horizontally or vertically on the board. In other words, they can only occupy a contiguous line of cells horizontally or vertically. Each battleship must be separated by at least one empty cell (either horizontally or vertically) from any other battleship.
+
+### Examples
+
+**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 `'X'` or `'.'`.
+
+### Follow-up
+
+Could you do it in one-pass, using only `O(1)` extra memory and without modifying the value of the board?
+
+---
+
+## Approach
+
+To count the number of battleships in one pass and without extra memory, we can traverse the board and count only the top-left cell of each battleship. A cell qualifies as the top-left cell if there are no battleship cells above it or to the left of it.
+
+### Steps:
+
+1. Initialize a counter to keep track of the number of battleships.
+2. Traverse each cell in the board.
+3. For each cell, check if it is a battleship (`'X'`):
+   - If it is, check if there is no battleship cell above it and no battleship cell to the left of it.
+   - If both conditions are met, increment the battleship counter.
+4. Return the counter after the traversal.
+
+### Solution
+
+#### Java Solution
+
+```java
+class Solution {
+    public int countBattleships(char[][] board) {
+        int count = 0;
+        for (int i = 0; i < board.length; i++) {
+            for (int j = 0; j < board[0].length; j++) {
+                if (board[i][j] == 'X') {
+                    if (i > 0 && board[i - 1][j] == 'X') continue;
+                    if (j > 0 && board[i][j - 1] == 'X') continue;
+                    count++;
+                }
+            }
+        }
+        return count;
+    }
+}
+```
+### C++ Solution
+
+```cpp
+class Solution {
+public:
+    int countBattleships(vector<vector<char>>& board) {
+        int count = 0;
+        for (int i = 0; i < board.size(); i++) {
+            for (int j = 0; j < board[0].size(); j++) {
+                if (board[i][j] == 'X') {
+                    if (i > 0 && board[i - 1][j] == 'X') continue;
+                    if (j > 0 && board[i][j - 1] == 'X') continue;
+                    count++;
+                }
+            }
+        }
+        return count;
+    }
+};
+```
+### Python Solution
+
+```python
+class Solution:
+    def countBattleships(self, board: List[List[str]]) -> int:
+        count = 0
+        for i in range(len(board)):
+            for j in range(len(board[0])):
+                if board[i][j] == 'X':
+                    if i > 0 and board[i - 1][j] == 'X':
+                        continue
+                    if j > 0 and board[i][j - 1] == 'X':
+                        continue
+                    count += 1
+        return count
+```
+### Complexity Analysis
+**Time Complexity:** O(m * n)
+>Reason: We traverse each cell of the board once.
+
+**Space Complexity:** O(1)
+>Reason: We do not use any extra space that scales with the input size.
+
+### References
+**LeetCode Problem:** Battleships in a Board
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