What is the time complexity of Minimum Moves to Capture The Queen?

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

What data structures are used to solve Minimum Moves to Capture The Queen?

This problem uses Math, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Moves to Capture The Queen?

Insight 1: We need to check if the rook and bishop can directly capture the queen in one move, and if not, return 2 since the constraints don't allow for more than 2 moves being necessary. Insight 2: The key is determining if the path between the capturing piece (rook or bishop) and the queen is blocked by the other white piece. We must check for obstruction.

Which companies ask LeetCode 3001. Minimum Moves to Capture The Queen in interviews?

This problem is frequently asked at Wipro.

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Minimum Moves to Capture The Queen - LeetCode 3001 Solution

Minimum Moves to Capture The Queen - Complete Solution Guide

Minimum Moves to Capture The Queen is LeetCode problem 3001, 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

There is a 1-indexed 8 x 8 chessboard containing 3 pieces. You are given 6 integers a , b , c , d , e , and f where: (a, b) denotes the position of the white rook. (c, d) denotes the position of the white bishop. (e, f) denotes the position of the black queen. Given that you can only move the white pieces, return the minimum number of moves required to capture the black queen . Note that: Rooks can move any number of squares either vertically or horizontally, but cannot jump over other pieces. B

Detailed Explanation

The problem asks us to determine the minimum number of moves required for white pieces (a rook and a bishop) to capture the black queen on an 8x8 chessboard. We are given the positions of the rook, bishop, and queen. The rook moves horizontally or vertically, and the bishop moves diagonally. Pieces cannot jump over each other. We need to return 1 if either piece can capture the queen in one move, and 2 otherwise, considering the obstacle caused by other pieces.

Solution Approach

The solution checks if the rook or bishop can capture the queen in one move. To do this, it first checks if the rook or bishop is on the same row/column or diagonal as the queen, respectively. If so, it then checks if the other white piece is blocking the path between the capturing piece and the queen. If no piece is blocking, it returns 1. If neither the rook nor the bishop can capture the queen in one move, it returns 2.

Step-by-Step Algorithm

Step 1: Check if the rook can capture the queen in one move. This involves checking if they share the same row or column (a == e or b == f).
Step 2: If the rook and queen are on the same row or column, check if the bishop is obstructing the path between them. The bishop is obstructing if it is on the same row/column as the rook and queen, and its coordinate along that row/column falls between the rook's and queen's coordinates.
Step 3: If the rook can capture the queen without obstruction, return 1.
Step 4: Check if the bishop can capture the queen in one move. This involves checking if they are on the same diagonal (abs(c - e) == abs(d - f)).
Step 5: If the bishop and queen are on the same diagonal, check if the rook is obstructing the path between them. The rook is obstructing if it is on the same diagonal as the bishop and queen, and its coordinate falls between the bishop's and queen's coordinates along the diagonal.
Step 6: If the bishop can capture the queen without obstruction, return 1.
Step 7: If neither the rook nor the bishop can capture the queen in one move without obstruction, return 2.

Key Insights

Insight 1: We need to check if the rook and bishop can directly capture the queen in one move, and if not, return 2 since the constraints don't allow for more than 2 moves being necessary.
Insight 2: The key is determining if the path between the capturing piece (rook or bishop) and the queen is blocked by the other white piece. We must check for obstruction.
Insight 3: The problem has a fixed board size and only considers capturing within at most two moves. Thus, brute force checking whether a direct attack is possible, with obstacle consideration, is a viable approach.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, Enumeration.

Companies

Asked at: Wipro.

Minimum Moves to Capture The Queen | LeetCode 3001

Medium

Math Enumeration

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

Solution Code

Available in:

class Solution {
    public int minMovesToCaptureTheQueen(int a, int b, int c, int d, int e, int f) {
        // Check if the rook can capture in one move
        if (a == e) {  // Same column
            if (!(c == a && Math.min(b, f) < d && d < Math.max(b, f))) {
                return 1;
            }
        }
        if (b == f) {  // Same row
            if (!(d == b && Math.min(a, e) < c && c < Math.max(a, e))) {
                return 1;
            }
        }

        // Check if the bishop can capture in one move
        if (Math.abs(c - e) == Math.abs(d - f)) {  // Same diagonal
            if (!(Math.abs(c - a) == Math.abs(d - b) && Math.min(c, e) < a && a < Math.max(c, e))) {
                return 1;
            }
        }

        return 2;
    }
}

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Problem Info

Description

Problem 3001: Minimum Moves to Capture The Queen - Detailed explanation and solution approaches.

Topics

Math Enumeration

Companies

Wipro

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3001

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: Check if the rook can capture the queen in one move. This involves checking if they share the same row or column (a == e or b == f).
2
Step 2: If the rook and queen are on the same row or column, check if the bishop is obstructing the path between them. The bishop is obstructing if it is on the same row/column as the rook and queen, and its coordinate along that row/column falls between the rook's and queen's coordinates.
3
Step 3: If the rook can capture the queen without obstruction, return 1.
4
Step 4: Check if the bishop can capture the queen in one move. This involves checking if they are on the same diagonal (abs(c - e) == abs(d - f)).
5
Step 5: If the bishop and queen are on the same diagonal, check if the rook is obstructing the path between them. The rook is obstructing if it is on the same diagonal as the bishop and queen, and its coordinate falls between the bishop's and queen's coordinates along the diagonal.
6
Step 6: If the bishop can capture the queen without obstruction, return 1.
7
Step 7: If neither the rook nor the bishop can capture the queen in one move without obstruction, return 2.

💡

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 solution performs a fixed number of comparisons and arithmetic operations. The number of operations doesn't depend on the input size, hence O(1) time complexity. The solution uses a constant amount of extra space, regardless of the input, resulting in O(1) space complexity.

📖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 Helper Functions for Obstacle Detection

You could separate the obstacle detection logic into dedicated helper functions for rook and bishop moves. This makes the main function slightly more readable but doesn't change the overall complexity.

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

Trade-offs: This improves readability but doesn't offer any performance benefits. It might be preferred for better code organization, especially in larger codebases.

✨

Key Insights

1Insight 1: We need to check if the rook and bishop can directly capture the queen in one move, and if not, return 2 since the constraints don't allow for more than 2 moves being necessary.
2Insight 2: The key is determining if the path between the capturing piece (rook or bishop) and the queen is blocked by the other white piece. We must check for obstruction.
3Insight 3: The problem has a fixed board size and only considers capturing within at most two moves. Thus, brute force checking whether a direct attack is possible, with obstacle consideration, is a viable approach.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: The rook or bishop is positioned such that the other piece is in between it and the queen. This case is handled correctly by checking if the other piece's coordinates lie between the capturing piece's and the queen's.
Edge case 2: The rook or bishop can directly capture the queen without any obstructions. This is the primary case that the algorithm is designed to handle.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to check for obstructions. If you only check if the rook or bishop is on the same row/column/diagonal as the queen, without checking for the other piece blocking, you will get the wrong answer.

✗Common mistake 2: Incorrectly implementing the obstruction check, particularly the comparison of coordinates to determine if one coordinate lies between two others. Careful attention to inequalities is important.

✅

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: Simulating piece movements and determining if a piece can attack another piece without being blocked.

🏦Real-world scenario 2: Pathfinding algorithms: While this problem's solution is not directly a general pathfinding algorithm, the concept of checking for obstructions is relevant in more complex pathfinding scenarios, especially in games or robotics where agents need to navigate around obstacles.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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 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

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

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

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 "Minimum Moves to Capture The Queen"?

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

🤖 Need help with this problem?