What is the time complexity of Zuma Game?

The optimal solution has a time complexity of O(m^2 * n * 3^n).

What is the space complexity of Zuma Game?

The space complexity is O(n + 3^n).

What data structures are used to solve Zuma Game?

This problem uses String, Dynamic Programming, Stack, Breadth-First Search, Memoization. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Zuma Game?

Insight 1: The order of insertions matters significantly. We need to explore different insertion possibilities to find the optimal solution. Insight 2: The 'cleanup' operation, removing consecutive balls of the same color (>=3), is crucial and needs to be performed after each insertion.

Which companies ask LeetCode 488. Zuma Game in interviews?

This problem is frequently asked at Baidu.

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Zuma Game - LeetCode 488 Solution

Zuma Game - Complete Solution Guide

Zuma Game is LeetCode problem 488, 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

You are playing a variation of the game Zuma. In this variation of Zuma, there is a single row of colored balls on a board, where each ball can be colored red 'R' , yellow 'Y' , blue 'B' , green 'G' , or white 'W' . You also have several colored balls in your hand. Your goal is to clear all of the balls from the board. On each turn: Pick any ball from your hand and insert it in between two balls in the row or on either end of the row. If there is a group of three or more consecutive balls of the

Detailed Explanation

The Zuma Game problem presents a single row of colored balls on a board. The goal is to clear all balls by inserting balls from a hand into the board. After each insertion, if there are three or more consecutive balls of the same color, they are removed. This process repeats until no more such groups exist. The task is to find the minimum number of balls needed to clear the board, or return -1 if it's impossible. The input consists of a string `board` representing the initial balls and a string `hand` representing the available balls. The colors are represented by 'R', 'Y', 'B', 'G', and 'W'.

Solution Approach

The provided solutions use a Breadth-First Search (BFS) algorithm to find the minimum number of balls to insert. The core idea is to treat each possible board configuration and hand configuration as a state. The BFS explores all possible states reachable by inserting a ball from the hand into the board and then cleaning up the board. The algorithm maintains a queue of states to explore and a set of visited states to prevent cycles. The algorithm keeps track of the number of steps taken to reach each state. The search terminates when an empty board is reached, returning the number of steps. If the queue is exhausted without finding an empty board, it returns -1.

Step-by-Step Algorithm

Step 1: **Initialize:** Create a queue to store states and a set to track visited states. The initial state consists of the initial `board`, the initial `hand` (represented as counts of each color), and the initial step count (0).
Step 2: **Cleanup:** Define a `cleanup` function that repeatedly removes consecutive groups of three or more balls of the same color from the board until no such groups remain.
Step 3: **BFS Loop:** While the queue is not empty, dequeue a state (current `board`, current `hand`, number of `steps`).
Step 4: **Check for Completion:** If the current `board` is empty, return the current `steps` (success).
Step 5: **Check for Empty Hand:** If the hand is empty, and the board is not empty, this state is not a solution, continue to the next state in the queue
Step 6: **Iterate Hand:** Iterate through each color of balls available in the current `hand`.
Step 7: **Iterate Board:** For each available ball, iterate through all possible insertion positions on the current `board`.
Step 8: **Insert and Cleanup:** Insert the ball at the current position, creating a new `board`. Then, apply the `cleanup` function to the new `board` to remove any consecutive groups of three or more balls.
Step 9: **Create New State:** Create a new state consisting of the cleaned `board`, the updated `hand` (with one less of the inserted color), and the incremented `steps`.
Step 10: **Check Visited:** If this new state hasn't been visited, add it to the `visited` set and enqueue it for further exploration.
Step 11: **No Solution:** If the queue becomes empty without finding an empty `board`, return -1 (failure).
Step 12: **Representing the hand:** The algorithm uses a `Counter` (Python), `HashMap`(Java), and `map`(C++) to represent the number of each ball that is in the hand. Representing the hand in this way makes it easy to check which balls can be placed.

Key Insights

Insight 1: The order of insertions matters significantly. We need to explore different insertion possibilities to find the optimal solution.
Insight 2: The 'cleanup' operation, removing consecutive balls of the same color (>=3), is crucial and needs to be performed after each insertion.
Insight 3: We can use Breadth-First Search (BFS) to explore the possible states (board and hand) and find the shortest path (minimum number of balls inserted) to an empty board.
Insight 4: To prevent infinite loops, we must keep track of visited states to avoid re-exploring previously encountered board and hand configurations.

Complexity Analysis

Time Complexity: O(m^2 * n * 3^n)

Space Complexity: O(n + 3^n)

Topics

This problem involves: String, Dynamic Programming, Stack, Breadth-First Search, Memoization.

Companies

Asked at: Baidu.

Zuma Game | LeetCode 488

Hard

String Dynamic Programming Stack Breadth-First Search Memoization

Time: O(m^2 * n * 3^n)Space: O(n + 3^n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int findMinStep(String board, String hand) {
        
        // Cleanup function to remove consecutive balls of the same color with length >= 3
        Function<String, String> cleanup = (s) -> {
            while (true) {
                int startLen = s.length();
                int i = 0;
                while (i < s.length()) {
                    int j = i;
                    while (j < s.length() && s.charAt(j) == s.charAt(i)) {
                        j++;
                    }
                    if (j - i >= 3) {
                        s = s.substring(0, i) + s.substring(j);
                        break;
                    }
                    i = j;
                }
                if (s.length() == startLen) {
                    break;
                }
            }
            return s;
        };

        // Count the occurrences of each ball color in the hand
        Map<Character, Integer> handCounts = new HashMap<>();
        for (char c : hand.toCharArray()) {
            handCounts.put(c, handCounts.getOrDefault(c, 0) + 1);
        }
        
        // Convert hand counts to a tuple (sorted list of pairs)
        List<Map.Entry<Character, Integer>> sortedHandList = new ArrayList<>(handCounts.entrySet());
        Collections.sort(sortedHandList, Comparator.comparing(Map.Entry::getKey));
        Tuple initialHandTuple = new Tuple(sortedHandList);

        // Initialize queue and visited set
        Queue<State> q = new LinkedList<>();
        Set<State> visited = new HashSet<>();
        q.offer(new State(board, initialHandTuple, 0));
        visited.add(new State(board, initialHandTuple, 0));

        // BFS
        while (!q.isEmpty()) {
            State currState = q.poll();
            String currBoard = currState.board;
            Tuple currHandTuple = currState.hand;
            int steps = currState.steps;

            // If the board is empty, return the number of steps
            if (currBoard.isEmpty()) {
                return steps;
            }

            // If the hand is empty, continue to the next state
            if (currHandTuple.isEmpty()) {
                continue;
            }

            // Convert the hand tuple back to a hand counts map
            Map<Character, Integer> currHandCounts = new HashMap<>();
            for (Map.Entry<Character, Integer> entry : currHandTuple.toList()) {
                currHandCounts.put(entry.getKey(), entry.getValue());
            }

            // Iterate through each ball color in the hand
            for (char ballColor : currHandCounts.keySet()) {
                // Create a copy of the hand counts map
                Map<Character, Integer> newHandCounts = new HashMap<>(currHandCounts);
                newHandCounts.put(ballColor, newHandCounts.get(ballColor) - 1);
                if (newHandCounts.get(ballColor) == 0) {
                    newHandCounts.remove(ballColor);
                }

                // Convert the updated hand counts to a tuple
                List<Map.Entry<Character, Integer>> newSortedHandList = new ArrayList<>(newHandCounts.entrySet());
                Collections.sort(newSortedHandList, Comparator.comparing(Map.Entry::getKey));
                Tuple newHandTuple = new Tuple(newSortedHandList);

                // Iterate through each possible insertion position on the board
                for (int i = 0; i <= currBoard.length(); i++) {
                    // Insert the ball into the board
                    String newBoard = currBoard.substring(0, i) + ballColor + currBoard.substring(i);

                    // Clean up the board
                    String cleanedBoard = cleanup.apply(newBoard);

                    // Create a new state
                    State newState = new State(cleanedBoard, newHandTuple, 0);

                    // If the state has not been visited, add it to the queue and visited set
                    if (!visited.contains(newState)) {
                        visited.add(new State(cleanedBoard, newHandTuple, 0));
                        q.offer(new State(cleanedBoard, newHandTuple, steps + 1));
                    }
                }
            }
        }

        // If no solution is found, return -1
        return -1;
    }
    
    // Helper class to represent a state in the BFS
    private static class State {
        String board;
        Tuple hand;
        int steps;

        public State(String board, Tuple hand, int steps) {
            this.board = board;
            this.hand = hand;
            this.steps = steps;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            State state = (State) o;
            return Objects.equals(board, state.board) && Objects.equals(hand, state.hand);
        }

        @Override
        public int hashCode() {
            return Objects.hash(board, hand);
        }
    }
    
    // Helper class to represent a tuple of hand counts
    private static class Tuple {
        List<Map.Entry<Character, Integer>> data;

        public Tuple(List<Map.Entry<Character, Integer>> data) {
            this.data = data;
        }
        
        public List<Map.Entry<Character, Integer>> toList() {
            return data;
        }

        public boolean isEmpty() {
            return data.isEmpty();
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Tuple tuple = (Tuple) o;
            return Objects.equals(data, tuple.data);
        }

        @Override
        public int hashCode() {
            return Objects.hash(data);
        }
    }
    
    // Define a functional interface for the cleanup function
    private interface Function<T, R> {
        R apply(T t);
    }
}

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Problem Info

Description

Problem 488: Zuma Game - Detailed explanation and solution approaches.

Topics

String Dynamic Programming Stack Breadth-First Search Memoization

Companies

Baidu

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:488

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

12 steps

1
Step 1: **Initialize:** Create a queue to store states and a set to track visited states. The initial state consists of the initial `board`, the initial `hand` (represented as counts of each color), and the initial step count (0).
2
Step 2: **Cleanup:** Define a `cleanup` function that repeatedly removes consecutive groups of three or more balls of the same color from the board until no such groups remain.
3
Step 3: **BFS Loop:** While the queue is not empty, dequeue a state (current `board`, current `hand`, number of `steps`).
4
Step 4: **Check for Completion:** If the current `board` is empty, return the current `steps` (success).
5
Step 5: **Check for Empty Hand:** If the hand is empty, and the board is not empty, this state is not a solution, continue to the next state in the queue
6
Step 6: **Iterate Hand:** Iterate through each color of balls available in the current `hand`.
7
Step 7: **Iterate Board:** For each available ball, iterate through all possible insertion positions on the current `board`.
8
Step 8: **Insert and Cleanup:** Insert the ball at the current position, creating a new `board`. Then, apply the `cleanup` function to the new `board` to remove any consecutive groups of three or more balls.
9
Step 9: **Create New State:** Create a new state consisting of the cleaned `board`, the updated `hand` (with one less of the inserted color), and the incremented `steps`.
10
Step 10: **Check Visited:** If this new state hasn't been visited, add it to the `visited` set and enqueue it for further exploration.
11
Step 11: **No Solution:** If the queue becomes empty without finding an empty `board`, return -1 (failure).
12
Step 12: **Representing the hand:** The algorithm uses a `Counter` (Python), `HashMap`(Java), and `map`(C++) to represent the number of each ball that is in the hand. Representing the hand in this way makes it easy to check which balls can be placed.

💡

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^2 * n * 3^n)

💾

Space Complexity

O(n + 3^n)

📝

Why?

The time complexity is dominated by the BFS search. The number of possible board configurations after cleanup is hard to precisely determine, but since each color can potentially have multiple occurrences (up to the length of the hand + board), and the cleanup operation has to happen after each insertion, along with the nested loops for inserting and cleanup, the O(m^2 * n * 3^n) is a safe estimate. The space complexity is mainly from the queue and the visited set. Both of these potentially can hold 3^n board and hand combinations, where each board can be up to length of the original board plus the hand.

📖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) with Memoization

DFS can also be used, exploring the state space recursively. Memoization can be applied to store the results of previously computed states, avoiding redundant computations. The hand can be converted into sorted tuples to optimize state representation.

Time: O(m^2 * n * 3^n)Space: O(n + 3^n)

Trade-offs: DFS with memoization can be more efficient in some cases due to better caching or pruning of the search space. However, it is prone to stack overflow issues with very deep search trees. BFS guarantees the shortest path, but DFS with memoization can reduce memory overhead by exploring only relevant subtrees.

✨

Key Insights

1Insight 1: The order of insertions matters significantly. We need to explore different insertion possibilities to find the optimal solution.
2Insight 2: The 'cleanup' operation, removing consecutive balls of the same color (>=3), is crucial and needs to be performed after each insertion.
3Insight 3: We can use Breadth-First Search (BFS) to explore the possible states (board and hand) and find the shortest path (minimum number of balls inserted) to an empty board.
4Insight 4: To prevent infinite loops, we must keep track of visited states to avoid re-exploring previously encountered board and hand configurations.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

String

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

Edge Cases

Edge case 1: **Empty Board:** If the initial board is empty, the algorithm needs to check if the hand is sufficient to create and then clear three or more consecutive balls of the same color. This is handled correctly by the empty board check in the BFS loop.
Edge case 2: **Unsolvable Board:** If the balls in the hand are insufficient to clear the board, the BFS will explore all possible states and eventually exhaust the queue, returning -1. This indicates that no solution is possible.
Edge case 3: **Initial Board has groups of 3 or more**: According to the constraints of the problem, this will not happen, so it is not handled

⚠️

Common Mistakes to Avoid

✗Common mistake 1: **Not handling the cleanup operation correctly:** Failing to repeatedly remove consecutive balls of the same color after each insertion will lead to incorrect board states and an incorrect solution.

✗Common mistake 2: **Not tracking visited states:** Without a visited set, the BFS will enter infinite loops, exploring the same board and hand configurations repeatedly. Note, for more complex problems, hashing is not possible, as demonstrated in the C solution, and implementing a data structure that can compare boards becomes difficult.

✗Common mistake 3: **Incorrectly updating the hand:** Failing to correctly decrement the count of balls used from the hand or not creating copies of the hand at each step will lead to incorrect state exploration.

✗Common mistake 4: **Inefficient String manipulation:** Creating substrings repeatedly using naive methods can lead to performance issues. Instead, use efficient methods offered by the programming language.

✅

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 AI:** The Zuma Game problem is directly related to game AI development. The algorithm can be used to create an AI player that can optimally clear the board in a Zuma-like game.

🏦Real-world scenario 2: **Optimization problems:** The BFS approach with state tracking and pruning (cleanup) is applicable to a wider range of optimization problems where the goal is to find the shortest path to a desired state through a series of actions. This includes problems in logistics, planning, and resource management.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: October 7, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

❓

Frequently Asked Questions

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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 "Zuma Game"?

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

🤖 Need help with this problem?