What is the time complexity of Battleships in a Board?

The optimal solution has a time complexity of O(m*n).

What data structures are used to solve Battleships in a Board?

This problem uses Array, Depth-First Search, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Battleships in a Board?

Insight 1: A battleship's starting point is the top-left cell if it is placed vertically or horizontally. This is because no two battleships are adjacent. Insight 2: We can identify the 'head' of a battleship by checking if the cell is an 'X' and if the cell above and to the left are not 'X's. This prevents overcounting segments of the same battleship.

Which companies ask LeetCode 419. Battleships in a Board in interviews?

This problem is frequently asked at Microstrategy, Tinkoff.

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Battleships in a Board - LeetCode 419 Solution

Battleships in a Board - Complete Solution Guide

Battleships in a Board is LeetCode problem 419, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 . 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"],[".","

Detailed Explanation

The problem asks us to count the number of battleships present in a given 2D board. The board contains 'X' representing a battleship and '.' representing an empty cell. Battleships are placed either horizontally or vertically, and two battleships are always separated by at least one horizontal or vertical cell. The goal is to efficiently count the number of distinct battleships without overcounting connected parts of the same ship.

Solution Approach

The provided solution iterates through each cell of the board. If a cell contains a battleship ('X'), it checks if it is the starting point (head) of the battleship. A cell is considered the starting point if the cell above it and the cell to its left are not part of another battleship (i.e., they are not 'X'). If it's a starting point, the battleship count is incremented.

Step-by-Step Algorithm

Step 1: Initialize a count variable to 0. This variable will store the number of battleships.
Step 2: Iterate through each cell in the board, row by row.
Step 3: For each cell, check if the cell contains 'X'.
Step 4: If the cell contains 'X', check if it's the starting point of a battleship. To do this, check if the cell above and the cell to the left are also 'X'. If either of them is 'X', then this cell is a part of a battleship that has already been counted, so we skip counting it again.
Step 5: If the current cell is 'X' and it's the starting point of a battleship (i.e., no 'X' above or to the left), increment the count.
Step 6: After iterating through the entire board, return the final count of battleships.

Key Insights

Insight 1: A battleship's starting point is the top-left cell if it is placed vertically or horizontally. This is because no two battleships are adjacent.
Insight 2: We can identify the 'head' of a battleship by checking if the cell is an 'X' and if the cell above and to the left are not 'X's. This prevents overcounting segments of the same battleship.
Insight 3: The problem can be solved in one pass and O(1) space, without modifying the input board, as hinted by the follow-up question.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(1)

Topics

This problem involves: Array, Depth-First Search, Matrix.

Companies

Asked at: Microstrategy, Tinkoff.

Battleships in a Board | LeetCode 419

Medium

Array Depth-First Search Matrix

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

Solution Code

Available in:

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

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Problem Info

Description

Problem 419: Battleships in a Board - Detailed explanation and solution approaches.

Topics

Array Depth-First Search Matrix

Companies

MicrostrategyTinkoff

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:419

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize a count variable to 0. This variable will store the number of battleships.
2
Step 2: Iterate through each cell in the board, row by row.
3
Step 3: For each cell, check if the cell contains 'X'.
4
Step 4: If the cell contains 'X', check if it's the starting point of a battleship. To do this, check if the cell above and the cell to the left are also 'X'. If either of them is 'X', then this cell is a part of a battleship that has already been counted, so we skip counting it again.
5
Step 5: If the current cell is 'X' and it's the starting point of a battleship (i.e., no 'X' above or to the left), increment the count.
6
Step 6: After iterating through the entire board, return the final count of battleships.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm visits each cell in the matrix once, performing a constant number of operations per cell. Therefore, the time complexity is directly proportional to the number of cells in the matrix (m*n). The space complexity is constant because we are not allocating any extra memory dependent on the size of the input. The variables used (count, i, j) take constant space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Depth-First Search (DFS)

We could use DFS to visit and mark all parts of a battleship once we find its 'head'. This would involve iterating through the board and when an 'X' is found, performing a DFS to 'sink' the battleship by marking all connected 'X's as visited. Each time DFS is called, it implies a new battleship.

Time: O(m*n), in the worst case, every cell is 'X' and DFS visits each cell.Space: O(m*n) in the worst case due to the recursion stack of the DFS algorithm if the entire grid is a single battleship. Note, the problem statement encourages an O(1) space solution.

Trade-offs: While DFS also has O(m*n) time complexity, it requires O(m*n) space in the worst case for the recursion stack, whereas the provided solution achieves O(1) space. The provided solution is much more efficient in terms of space. The DFS approach would also modify the board, which might not be desirable in some situations.

✨

Key Insights

1Insight 1: A battleship's starting point is the top-left cell if it is placed vertically or horizontally. This is because no two battleships are adjacent.
2Insight 2: We can identify the 'head' of a battleship by checking if the cell is an 'X' and if the cell above and to the left are not 'X's. This prevents overcounting segments of the same battleship.
3Insight 3: The problem can be solved in one pass and O(1) space, without modifying the input board, as hinted by the follow-up question.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Empty board. The code handles this by checking if the number of rows is 0. If it is, it returns 0, correctly indicating that there are no battleships.
Edge case 2: A single 'X' on the board. The code correctly identifies this as one battleship because there are no adjacent 'X's.
Edge case 3: A horizontal or vertical line of 'X's representing a single battleship. The code correctly counts this as one battleship because only the leftmost or topmost 'X' will be counted as the start of the battleship.
Edge case 4: Board filled entirely with '.'. The code correctly counts this as zero battleships.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Overcounting parts of the same battleship. Forgetting to check the cells above and to the left before incrementing the count will lead to counting each 'X' in a battleship as a separate battleship.

✗Common mistake 2: Not handling the edge case of an empty board or a board with only '.' characters.

✗Common mistake 3: Incorrectly implementing the bounds check when accessing board[i-1][j] or board[i][j-1], potentially leading to array out-of-bounds exceptions.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Image processing. Identifying and counting distinct objects (like connected components of pixels) in a binary image can use a similar approach. For example, counting cars in a parking lot from an aerial image.

🏦Real-world scenario 2: Network analysis. Identifying connected components in a network graph, where each 'X' represents a node and adjacent 'X's represent connections. This could be used to analyze social networks or infrastructure networks.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Battleships in a Board"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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