What is the time complexity of Available Captures for Rook?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Available Captures for Rook?

This problem uses Array, Matrix, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Available Captures for Rook?

Insight 1: The rook can only capture the first pawn encountered in each direction (up, down, left, right). Insight 2: The presence of a bishop blocks the rook's attack in that direction.

Which companies ask LeetCode 999. Available Captures for Rook in interviews?

This problem is frequently asked at Block.

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Available Captures for Rook - LeetCode 999 Solution

Available Captures for Rook - Complete Solution Guide

Available Captures for Rook is LeetCode problem 999, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an 8 x 8 matrix representing a chessboard. There is exactly one white rook represented by 'R' , some number of white bishops 'B' , and some number of black pawns 'p' . Empty squares are represented by '.' . A rook can move any number of squares horizontally or vertically (up, down, left, right) until it reaches another piece or the edge of the board. A rook is attacking a pawn if it can move to the pawn's square in one move. Note: A rook cannot move through other pieces, such as bi

Detailed Explanation

The problem asks us to find the number of black pawns ('p') a white rook ('R') can attack on an 8x8 chessboard. The rook can move horizontally or vertically any number of squares until it encounters another piece (bishop 'B' or pawn 'p') or the edge of the board. A rook attacks a pawn if it can move to the pawn's square in one move without being blocked by another piece. We need to count the number of pawns the rook is attacking.

Solution Approach

The solution involves first finding the position of the rook on the board. Then, from the rook's position, it iterates in four directions (up, down, left, and right) until it either hits the edge of the board or encounters a piece (either a bishop or a pawn). If it encounters a pawn, the capture count is incremented. If it encounters a bishop, the search stops in that direction. The process is repeated for all four directions.

Step-by-Step Algorithm

Step 1: Iterate through the board to find the row and column index of the rook ('R').
Step 2: Initialize a capture count variable to 0.
Step 3: Iterate upwards from the rook's position, checking each cell until the edge of the board or a piece is encountered. If a pawn ('p') is found, increment the capture count and stop the iteration in this direction. If a bishop ('B') is found, stop the iteration in this direction.
Step 4: Repeat step 3 for the downward direction.
Step 5: Repeat step 3 for the leftward direction.
Step 6: Repeat step 3 for the rightward direction.
Step 7: Return the final capture count.

Key Insights

Insight 1: The rook can only capture the first pawn encountered in each direction (up, down, left, right).
Insight 2: The presence of a bishop blocks the rook's attack in that direction.
Insight 3: We need to iterate through the board in all four directions from the rook's position until we hit the board edge or encounter another piece.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Array, Matrix, Simulation.

Companies

Asked at: Block.

Available Captures for Rook | LeetCode 999

Easy

Array Matrix Simulation

Time: O(1)Space: O(1)

Solution Code

Available in:

class Solution {
    public int numRookCaptures(char[][] board) {
        int rookRow = -1, rookCol = -1;
        for (int i = 0; i < 8; i++) {
            for (int j = 0; j < 8; j++) {
                if (board[i][j] == 'R') {
                    rookRow = i;
                    rookCol = j;
                    break;
                }
            }
        }

        int count = 0;

        // Check up
        for (int i = rookRow - 1; i >= 0; i--) {
            if (board[i][rookCol] == 'p') {
                count++;
                break;
            } else if (board[i][rookCol] == 'B') {
                break;
            }
        }

        // Check down
        for (int i = rookRow + 1; i < 8; i++) {
            if (board[i][rookCol] == 'p') {
                count++;
                break;
            } else if (board[i][rookCol] == 'B') {
                break;
            }
        }

        // Check left
        for (int j = rookCol - 1; j >= 0; j--) {
            if (board[rookRow][j] == 'p') {
                count++;
                break;
            } else if (board[rookRow][j] == 'B') {
                break;
            }
        }

        // Check right
        for (int j = rookCol + 1; j < 8; j++) {
            if (board[rookRow][j] == 'p') {
                count++;
                break;
            } else if (board[rookRow][j] == 'B') {
                break;
            }
        }

        return count;
    }
}

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Problem Info

Description

Problem 999: Available Captures for Rook - Detailed explanation and solution approaches.

Topics

Array Matrix Simulation

Companies

Block

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:999

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

7 steps

1
Step 1: Iterate through the board to find the row and column index of the rook ('R').
2
Step 2: Initialize a capture count variable to 0.
3
Step 3: Iterate upwards from the rook's position, checking each cell until the edge of the board or a piece is encountered. If a pawn ('p') is found, increment the capture count and stop the iteration in this direction. If a bishop ('B') is found, stop the iteration in this direction.
4
Step 4: Repeat step 3 for the downward direction.
5
Step 5: Repeat step 3 for the leftward direction.
6
Step 6: Repeat step 3 for the rightward direction.
7
Step 7: Return the final capture count.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The board size is fixed at 8x8, so the maximum number of cells we iterate through in each direction is 7. Therefore, the maximum number of operations is constant. We are only using a few integer variables, so space complexity is also constant.

📖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

Using Direction Vectors

Instead of writing separate loops for each direction, we could use direction vectors to simplify the code. Define vectors (0, 1), (0, -1), (1, 0), and (-1, 0) to represent the four directions. Then, iterate along each direction until a piece or boundary is hit.

Time: O(1)Space: O(1)

Trade-offs: This approach can make the code more concise and readable but does not change the time or space complexity.

✨

Key Insights

1Insight 1: The rook can only capture the first pawn encountered in each direction (up, down, left, right).
2Insight 2: The presence of a bishop blocks the rook's attack in that direction.
3Insight 3: We need to iterate through the board in all four directions from the rook's position until we hit the board edge or encounter another piece.

📚

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: The rook is at the edge of the board. The loops will naturally handle the edge cases since the loop condition checks for board boundaries.
Edge case 2: The rook is surrounded by bishops or empty cells. The loops correctly identify these scenarios and either increment the capture count or stop the iteration.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to stop the search after encountering a bishop. The rook cannot capture pawns through bishops.

✗Common mistake 2: Not stopping the search after capturing a pawn in a certain direction. The rook can only capture the first pawn it encounters in each direction.

✗Common mistake 3: Not checking for board boundaries. This could lead to out-of-bounds array accesses.

✅

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: Game development, specifically in board games like chess or checkers, where determining possible moves and captures is a core part of the gameplay.

🏦Real-world scenario 2: Pathfinding algorithms in constrained environments. The rook's movement restriction can be modeled as a simple pathfinding problem where certain cells are blocked.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: September 8, 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 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Available Captures for Rook"?

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

🤖 Need help with this problem?