#488

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Zuma Game

The Zuma Game involves clearing balls from the board using a limited hand, applying dynamic programming and state transitions.

string dynamic programming stack breadth first search memoization

Problem Statement

In the Zuma Game variation, you're tasked with clearing a row of colored balls on a board, where each ball can be one of five colors: red 'R', yellow 'Y', blue 'B', green 'G', or white 'W'. You are given a certain number of colored balls in your hand. The goal is to remove all the balls from the board by inserting balls from your hand. The board is cleared by forming groups of three or more consecutive balls of the same color, which are then removed. You can only insert balls from your hand to form these groups.

In each turn, you can insert a ball from your hand into the board at any position. After insertion, groups of three or more consecutive balls of the same color are automatically cleared. Your task is to determine the minimum number of balls needed to be inserted from your hand to clear the board. If it is impossible to clear the board, return -1.

Examples

Example 1

Input: board = "WRRBBW", hand = "RB"

Output: -1

It is impossible to clear all the balls. The best you can do is:

Insert 'R' so the board becomes WRRRBBW. WRRRBBW -> WBBW.
Insert 'B' so the board becomes WBBBW. WBBBW -> WW. There are still balls remaining on the board, and you are out of balls to insert.

Example 2

Input: board = "WWRRBBWW", hand = "WRBRW"

Output: 2

To make the board empty:

Insert 'R' so the board becomes WWRRRBBWW. WWRRRBBWW -> WWBBWW.
Insert 'B' so the board becomes WWBBBWW. WWBBBWW -> WWWW -> empty. 2 balls from your hand were needed to clear the board.

Example 3

Input: board = "G", hand = "GGGGG"

Output: 2

To make the board empty:

Insert 'G' so the board becomes GG.
Insert 'G' so the board becomes GGG. GGG -> empty. 2 balls from your hand were needed to clear the board.

Constraints

1 <= board.length <= 16
1 <= hand.length <= 5
board and hand consist of the characters 'R', 'Y', 'B', 'G', and 'W'.
The initial row of balls on the board will not have any groups of three or more consecutive balls of the same color.

Solution Approach

State Transition Dynamic Programming

The key to solving this problem is using dynamic programming with state transitions. Each state represents a configuration of the board with a particular subset of balls inserted from the hand. Transitioning between these states allows you to efficiently compute the minimum number of insertions needed to clear the board.

Simulate Ball Insertion and Removal

Simulate inserting a ball from the hand at different positions on the board. After each insertion, check if any groups of three or more consecutive balls are formed and automatically removed. Repeat this until the board is empty or there are no more valid moves.

Memoization and Pruning

Memoization helps store intermediate states to avoid redundant calculations. Prune invalid paths where inserting a ball cannot lead to a solution, reducing the number of recursive calls and improving performance.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity of this problem depend on the approach used. A brute force solution may require exponential time due to the recursive nature of the problem and the state transitions. However, by using dynamic programming with memoization, the problem can be solved more efficiently. The space complexity will involve storing intermediate states and subproblems.

What Interviewers Usually Probe

Tests candidate's ability to apply dynamic programming concepts to a problem with state transitions.
Assesses problem-solving skills related to handling recursive problems and pruning invalid paths.
Evaluates efficiency in optimizing solutions using memoization to reduce redundant calculations.

Common Pitfalls or Variants

Common pitfalls

Forgetting to prune invalid paths, which can lead to excessive recursion and inefficient solutions.
Not handling state transitions properly, resulting in incorrect intermediate states.
Failing to consider edge cases where the board cannot be cleared, resulting in an incorrect output of -1.

Follow-up variants

Modify the problem to allow a different number of balls in hand, affecting the complexity of the solution.
Change the board's configuration to have more than one row of balls, introducing a new dimension to the problem.
Allow a larger set of colors for the balls, increasing the branching factor and the number of possible transitions.

FAQ

What is the key approach to solving the Zuma Game problem?

The main approach is dynamic programming with state transitions, which allows you to simulate ball insertion and removal, minimizing the number of insertions needed to clear the board.

Why is memoization important in solving Zuma Game?

Memoization helps store intermediate board states, reducing redundant calculations and improving the efficiency of the solution.

What should I consider when handling the state transitions in Zuma Game?

Ensure that each insertion of a ball results in valid group formations and automatic removal, while keeping track of the minimal insertions required to clear the board.

How does GhostInterview assist in solving Zuma Game?

GhostInterview provides a step-by-step guide to understanding the problem and applying dynamic programming, helping you practice the optimal approach and refine your solution.

Can Zuma Game be solved with a brute force approach?

While brute force is possible, it is inefficient for larger inputs. Dynamic programming with memoization provides a much more efficient solution by pruning unnecessary paths.

terminal

Solution

Solution 1

#### Python3

class Solution:
    def findMinStep(self, board: str, hand: str) -> int:
        def remove(s):
            while len(s):
                next = re.sub(r'B{3,}|G{3,}|R{3,}|W{3,}|Y{3,}', '', s)
                if len(next) == len(s):
                    break
                s = next
            return s

        visited = set()
        q = deque([(board, hand)])
        while q:
            state, balls = q.popleft()
            if not state:
                return len(hand) - len(balls)
            for ball in set(balls):
                b = balls.replace(ball, '', 1)
                for i in range(1, len(state) + 1):
                    s = state[:i] + ball + state[i:]
                    s = remove(s)
                    if s not in visited:
                        visited.add(s)
                        q.append((s, b))
        return -1

Continue Practicing

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