What is the time complexity of Number of Valid Move Combinations On Chessboard?

The optimal solution has a time complexity of O(8^n * n^2).

What data structures are used to solve Number of Valid Move Combinations On Chessboard?

This problem uses Array, String, Backtracking, Simulation. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Valid Move Combinations On Chessboard?

Insight 1: The number of pieces is small (up to 4), making a brute-force approach with backtracking feasible. We can try all possible destination squares for each piece. Insight 2: The key is to efficiently determine if a move combination is valid, which means no two pieces should occupy the same square at any point in time.

Which companies ask LeetCode 2056. Number of Valid Move Combinations On Chessboard in interviews?

This problem is frequently asked at PayPal.

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Number of Valid Move Combinations On Chessboard - LeetCode 2056 Solution

Number of Valid Move Combinations On Chessboard - Complete Solution Guide

Number of Valid Move Combinations On Chessboard is LeetCode problem 2056, a Hard 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 an 8 x 8 chessboard containing n pieces (rooks, queens, or bishops). You are given a string array pieces of length n , where pieces[i] describes the type (rook, queen, or bishop) of the i th piece. In addition, you are given a 2D integer array positions also of length n , where positions[i] = [r i , c i ] indicates that the i th piece is currently at the 1-based coordinate (r i , c i ) on the chessboard. When making a move for a piece, you choose a destination square that the piece will

Detailed Explanation

The problem asks us to find the number of valid move combinations for a set of chess pieces (rooks, queens, or bishops) on an 8x8 chessboard. Each piece moves simultaneously towards its destination square at a rate of one square per second. A move combination is invalid if two or more pieces occupy the same square at any given time. The task is to count how many possible valid destination assignments exist for all pieces.

Solution Approach

The solution uses a backtracking algorithm to explore all possible move combinations for the given chess pieces. For each piece, we generate all possible destination squares (including the current square). The backtracking function recursively explores these possibilities. Before considering a move combination as valid, the `collides` function is used to determine if any two pieces in the current combination occupy the same square at any point during their movement. If no collisions occur, the move combination is considered valid, and the counter is incremented.

Step-by-Step Algorithm

Step 1: Define the possible directions for each piece type (rook, queen, bishop).
Step 2: Generate all possible moves for each piece. For each piece, iterate through possible directions and steps to find all valid destination squares within the 8x8 board. Include the option of not moving (staying at the original position). Store these moves as a list of (direction, steps) pairs.
Step 3: Define a `collides` function that takes two pieces' starting positions and moves as input and determines whether they will collide at any point in time during their movement. The collision is detected by checking each time step if the positions coincide.
Step 4: Implement the backtracking function. The function takes the current piece index and the current move combination as input. If all pieces have been assigned a move, increment the valid combination count. Otherwise, for the current piece, try each possible move generated in Step 2. Before adding a move to the combination, check for collisions with previously placed pieces using the `collides` function. If no collisions exist, add the move to the combination and recursively call the backtracking function for the next piece. After the recursive call returns, remove the move to explore other possibilities.
Step 5: Start the backtracking with the first piece and an empty move combination.

Key Insights

Insight 1: The number of pieces is small (up to 4), making a brute-force approach with backtracking feasible. We can try all possible destination squares for each piece.
Insight 2: The key is to efficiently determine if a move combination is valid, which means no two pieces should occupy the same square at any point in time.
Insight 3: Representing the state of the board at each time step is unnecessary. We can directly check if the paths of two pieces intersect at any point.
Insight 4: The problem constraints guarantee that no two pieces start at the same square, simplifying collision detection.

Complexity Analysis

Time Complexity: O(8^n * n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Backtracking, Simulation.

Companies

Asked at: PayPal.

Number of Valid Move Combinations On Chessboard | LeetCode 2056

Hard

Array String Backtracking Simulation

Time: O(8^n * n^2)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public int countCombinations(String[] pieces, int[][] positions) {
        int n = pieces.length;
        
        Map<String, int[][]> DIRECTIONS = new HashMap<>();
        DIRECTIONS.put("rook", new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}});
        DIRECTIONS.put("bishop", new int[][]{{1, 1}, {1, -1}, {-1, 1}, {-1, -1}});
        DIRECTIONS.put("queen", new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}});

        List<List<Pair>> all_moves = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            all_moves.add(generate_moves(pieces[i], positions[i], DIRECTIONS));
        }
        
        List<Pair> pos_tuples = new ArrayList<>();
        for (int[] p : positions) {
            pos_tuples.add(new Pair(p[0], p[1]));
        }
        
        return backtrack(0, new ArrayList<>(), n, all_moves, pos_tuples);
    }
    
    private List<Pair> generate_moves(String piece_type, int[] position, Map<String, int[][]> DIRECTIONS) {
        int r = position[0];
        int c = position[1];
        List<Pair> moves = new ArrayList<>();
        moves.add(new Pair(new Pair(0, 0), 0));
        
        for (int[] direction : DIRECTIONS.get(piece_type)) {
            int dr = direction[0];
            int dc = direction[1];
            
            for (int steps = 1; steps <= 7; steps++) {
                int nr = r + steps * dr;
                int nc = c + steps * dc;
                if (1 <= nr && nr <= 8 && 1 <= nc && nc <= 8) {
                    moves.add(new Pair(new Pair(dr, dc), steps));
                } else {
                    break;
                }
            }
        }
        return moves;
    }
    
    private boolean collides(int k, Pair move_k, int j, Pair move_j, List<Pair> pos_tuples) {
        int r_k = pos_tuples.get(k).r;
        int c_k = pos_tuples.get(k).c;
        int r_j = pos_tuples.get(j).r;
        int c_j = pos_tuples.get(j).c;
        
        int dr_k = move_k.dir.r;
        int dc_k = move_k.dir.c;
        int steps_k = move_k.steps;
        
        int dr_j = move_j.dir.r;
        int dc_j = move_j.dir.c;
        int steps_j = move_j.steps;
        
        int max_s = Math.max(steps_k, steps_j);
        if (max_s == 0) {
            return false;
        }
        
        for (int t = 1; t <= max_s; t++) {
            int pos_k_t_r = r_k + Math.min(t, steps_k) * dr_k;
            int pos_k_t_c = c_k + Math.min(t, steps_k) * dc_k;
            int pos_j_t_r = r_j + Math.min(t, steps_j) * dr_j;
            int pos_j_t_c = c_j + Math.min(t, steps_j) * dc_j;
            
            if (pos_k_t_r == pos_j_t_r && pos_k_t_c == pos_j_t_c) {
                return true;
            }
        }
        return false;
    }

    private int backtrack(int k, List<Pair> current_combo, int n, List<List<Pair>> all_moves, List<Pair> pos_tuples) {
        if (k == n) {
            return 1;
        }
        
        int count = 0;
        for (Pair move_k : all_moves.get(k)) {
            boolean is_valid_with_prev = true;
            for (int j = 0; j < k; j++) {
                if (collides(k, move_k, j, current_combo.get(j), pos_tuples)) {
                    is_valid_with_prev = false;
                    break;
                }
            }
            
            if (is_valid_with_prev) {
                current_combo.add(move_k);
                count += backtrack(k + 1, current_combo, n, all_moves, pos_tuples);
                current_combo.remove(current_combo.size() - 1);
            }
        }
        return count;
    }
    
    static class Pair {
        int r;
        int c;
        int steps;
        Pair dir;

        public Pair(int r, int c) {
            this.r = r;
            this.c = c;
        }

        public Pair(Pair dir, int steps) {
            this.dir = dir;
            this.steps = steps;
        }

        @Override
        public String toString() {
            return "(" + r + ", " + c + ")";
        }
    }
}

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Problem Info

Description

Problem 2056: Number of Valid Move Combinations On Chessboard - Detailed explanation and solution approaches.

Topics

Array String Backtracking Simulation

Companies

PayPal

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2056

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

5 steps

1
Step 1: Define the possible directions for each piece type (rook, queen, bishop).
2
Step 2: Generate all possible moves for each piece. For each piece, iterate through possible directions and steps to find all valid destination squares within the 8x8 board. Include the option of not moving (staying at the original position). Store these moves as a list of (direction, steps) pairs.
3
Step 3: Define a `collides` function that takes two pieces' starting positions and moves as input and determines whether they will collide at any point in time during their movement. The collision is detected by checking each time step if the positions coincide.
4
Step 4: Implement the backtracking function. The function takes the current piece index and the current move combination as input. If all pieces have been assigned a move, increment the valid combination count. Otherwise, for the current piece, try each possible move generated in Step 2. Before adding a move to the combination, check for collisions with previously placed pieces using the `collides` function. If no collisions exist, add the move to the combination and recursively call the backtracking function for the next piece. After the recursive call returns, remove the move to explore other possibilities.
5
Step 5: Start the backtracking with the first piece and an empty move combination.

💡

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(8^n * n^2)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from the exponential nature of the backtracking algorithm (exploring all possible move combinations) multiplied by the cost of validating each combination (collision detection). The space complexity is determined by the depth of the recursion and the data structures used to store the current move combination, both of which are proportional to the number of pieces.

📖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

Dynamic Programming with Bitmasking

Represent the board as a bitmask, where each bit indicates whether a square is occupied. Use dynamic programming to store the number of valid combinations for subsets of pieces. This might improve performance if the number of states can be significantly reduced.

Time: Potentially O(n * 64 * 2^n) or worse, depending on collision detection efficiency. The 64 represents each square on the board, and we might need to iterate. The 2^n factor is due to considering all combinations of pieces, which gets worse as more pieces need to be considered.Space: O(64 * 2^n)

Trade-offs: This approach may be more efficient if collision detection can be optimized and the number of valid board states is significantly less than the total number of possible board states. However, the memory usage could be a concern due to the state representation.

✨

Key Insights

1Insight 1: The number of pieces is small (up to 4), making a brute-force approach with backtracking feasible. We can try all possible destination squares for each piece.
2Insight 2: The key is to efficiently determine if a move combination is valid, which means no two pieces should occupy the same square at any point in time.
3Insight 3: Representing the state of the board at each time step is unnecessary. We can directly check if the paths of two pieces intersect at any point.
4Insight 4: The problem constraints guarantee that no two pieces start at the same square, simplifying collision detection.

📚

Key Terms Explained

For Students

Array

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

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: A piece not moving at all. The solution includes the option for each piece to stay at its original position (move with 0 steps), handling this case.
Edge case 2: Pieces swapping positions. The `collides` function correctly detects collisions even when pieces are directly adjacent and move past each other in one second.
Edge case 3: A rook, bishop, or queen blocked by the edge of the board. The move generation logic ensures that only valid moves within the 8x8 board are considered.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the `collides` function, failing to account for simultaneous movements and potential swaps.

✗Common mistake 2: Not including the option for a piece to stay at its original position, leading to an incomplete set of possible combinations.

✗Common mistake 3: Inefficiently generating moves, leading to unnecessary computations and increased time complexity.

✗Common mistake 4: Using incorrect data structures or algorithms, leading to incorrect time complexity.

✅

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: Path planning for multiple robots in a constrained environment, where collision avoidance is critical.

🏦Real-world scenario 2: Resource allocation in a distributed system, where resources must be assigned to tasks without conflicts.

🎮Real-world scenario 3: Air traffic control, ensuring that no two aircrafts occupy the same three-dimensional space during flight

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: August 16, 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

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Number of Valid Move Combinations On Chessboard"?

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

🤖 Need help with this problem?