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Zuma Game

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 same color, remove the group of balls from the board.
    • If this removal causes more groups of three or more of the same color to form, then continue removing each group until there are none left.
  • If there are no more balls on the board, then you win the game.
  • Repeat this process until you either win or do not have any more balls in your hand.

Given a string board, representing the row of balls on the board, and a string hand, representing the balls in your hand, return the minimum number of balls you have to insert to clear all the balls from the board. If you cannot clear all the balls from the board using the balls in your hand, return -1.

 

Example 1:

Input: board = "WRRBBW", hand = "RB"
Output: -1
Explanation: 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
Explanation: 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
Explanation: 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 Explanation for Zuma Game

This problem asks for the minimum number of balls needed to clear a board in the Zuma game. The solution employs a Breadth-First Search (BFS) algorithm combined with memoization (though the provided Java solution uses a hash set for visited states, effectively achieving similar memoization). Let's break down the approach and the code.

Approach:

  1. State Representation: The core idea is to represent the game state efficiently. Both the Python and Java solutions use string manipulation for the board initially, then the Java solution converts to bit manipulation for efficiency. This allows for easy checking of visited states and fast transitions between states.

  2. BFS Algorithm: BFS is used to explore all possible game states systematically. The algorithm starts with the initial board and hand. In each iteration, it explores adding each ball from the hand at every possible position on the board.

  3. Ball Removal: After inserting a ball, the board is checked for consecutive groups of three or more balls of the same color. These groups are removed, potentially triggering cascading removals. This step is crucial and determines the efficiency of the game state exploration.

  4. Memoization/Visited States: The solutions keep track of visited game states to avoid redundant exploration. In the provided Java solution, a HashSet (visited) stores the bit representations of visited boards. This significantly reduces the search space and improves performance.

  5. Termination Condition: The BFS terminates when either the board is cleared (winning condition) or all possible moves have been explored without reaching a winning state.

Time Complexity Analysis:

The time complexity is difficult to express precisely due to the branching factor in the BFS. In the worst case, the number of possible game states is exponential with respect to the lengths of the board and hand. The memoization (or visited set in Java) helps significantly by pruning the search space. It prevents revisiting states, but it doesn't eliminate the inherent exponential nature of the problem.

However, given the constraints (board length ≤ 16, hand length ≤ 5), the search space, even though exponential, remains relatively manageable in practice due to the pruning by memoization. Thus, a precise, simple Big-O notation is hard to give, and it's more accurate to describe it as bounded exponential time.

Space Complexity Analysis:

The space complexity is dominated by the queue in BFS and the set of visited states (or the memoization table in a more explicit implementation). In the worst case, the queue could store a significant number of game states, while the visited set can, in the worst-case scenario, have an exponential number of entries. Hence, the space complexity is also considered bounded exponential in practice due to the input constraints.

Code Explanation (Python3):

The Python solution uses regular expressions (re.sub) for efficient ball removal. The BFS is implemented using a deque. The set visited stores strings representing visited board states.

Code Explanation (Java):

The Java solution is more intricate, employing bit manipulation for a compact and efficient representation of the board and hand. The Zuma class encapsulates the game state and provides methods for generating the next level of states and checking for groups. The bfs function implements the core BFS algorithm. The significant advantage of this Java solution lies in its improved performance due to the bit manipulation and careful use of data structures.

In summary: While the theoretical time and space complexity of this approach is exponential, the practical performance on the given input constraints is acceptable, due to the effective pruning done by the visited set (memoization). The use of bit manipulation in the Java solution further optimizes performance for this particular problem.