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Zuma Game

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ArraysStringsRecursionDynamic Programming

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


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What are the possible values for the balls' colors in the input string and hand? Are they represented by single characters, numbers, or something else?
  2. What is the maximum length of the board string, and the maximum number of balls in the hand string?
  3. If there are multiple possible sequences of inserting balls that achieve the minimum number of balls to insert, is any valid sequence acceptable, or is there a specific ordering/tie-breaker to consider?
  4. If the board can be cleared completely, what should I return? If it's impossible to clear the board, what should I return?
  5. Can the input strings 'board' and 'hand' be null or empty? If so, what is the expected behavior?

Brute Force Solution

Approach

The core idea is to try every possible placement of balls from your hand into the board. We keep trying until we eliminate all the balls on the board or we run out of balls in our hand. It's like trying every combination until something works.

Here's how the algorithm would work step-by-step:

  1. Start with the original board state and the balls in your hand.
  2. Consider every possible position on the board where you can insert a ball from your hand.
  3. For each possible insertion, create a new board state by adding the ball and then eliminating any sequences of three or more same-colored balls.
  4. If the new board state is empty, you have found a solution. Count the number of balls used.
  5. If the new board state is not empty, and you have balls left in your hand, repeat steps 2-4 with the new board state and remaining balls.
  6. Keep track of the minimum number of balls needed to clear the board across all the possible insertion sequences.
  7. If you run out of balls in your hand and the board is not empty, that insertion sequence doesn't work. Try another sequence.
  8. Return the minimum number of balls used to clear the board, or indicate that it's impossible if no sequence clears the board.

Code Implementation

def zuma_game_brute_force(board, hand):
    minimum_balls_needed = float('inf')

    def clear_board(current_board, current_hand, balls_used):
        nonlocal minimum_balls_needed

        # If the board is empty, update the minimum
        if not current_board:
            minimum_balls_needed = min(minimum_balls_needed, balls_used)
            return

        if not current_hand:
            return

        for ball_index, ball_color in enumerate(current_hand):
            for insert_index in range(len(current_board) + 1):
                # Simulate inserting a ball and reacting
                new_board = current_board[:insert_index] + ball_color + current_board[insert_index:]

                new_board = react(new_board)

                remaining_hand = current_hand[:ball_index] + current_hand[ball_index+1:]

                clear_board(new_board, remaining_hand, balls_used + 1)

    def react(board):
        while True:
            reduced = False
            start = 0
            while start < len(board):
                end = start
                while end < len(board) and board[start] == board[end]:
                    end += 1

                if end - start >= 3:
                    board = board[:start] + board[end:]
                    reduced = True
                    break
                start = end
            if not reduced:
                break
        return board

    clear_board(board, hand, 0)

    # Impossible to clear the board
    if minimum_balls_needed == float('inf'):
        return -1
    return minimum_balls_needed

Big(O) Analysis

Time Complexity
O((m+n)! / (n!))The provided solution uses a backtracking approach, trying every possible placement of balls from the hand into the board. 'm' represents the number of balls in the hand, and 'n' represents the initial length of the board. In the worst case, every possible combination of placing 'm' balls into 'n+m-1' possible slots before elimination has to be explored. This involves considering all permutations which is roughly (m+n)! divided by n! to account for the identical board positions and m! to account for the identical hand ball positions leading to ((m+n)!/(m!n!))*m!, simplifying to (m+n)! / (n!). Therefore the time complexity grows factorially with the sum of the board length and number of hand balls.
Space Complexity
O(B^H)The primary driver of space complexity is the recursive nature of the solution, where we explore every possible insertion sequence. In the worst case, we might explore almost every combination of placements. Let B be the maximum length of the board and H be the number of balls in hand. The recursion depth could reach H, and at each level, we store a copy of the board (of length at most B) and the remaining balls. Since we're effectively building a call stack branching out for each possible insertion, the space complexity becomes exponential relative to balls in hand in relation to the length of the board, approximating to O(B^H), where B is the board's length and H is hand's length, as each board state copies the majority of the board.

Optimal Solution

Approach

The best way to solve the Zuma game efficiently involves repeatedly eliminating adjacent balls of the same color. The key is to quickly identify and group balls to trigger chain reactions, aiming to clear the board with the fewest moves possible.

Here's how the algorithm would work step-by-step:

  1. Look at all possible places where you can insert a new ball into the existing sequence.
  2. For each possible insertion point, simulate the result of adding that ball.
  3. After inserting a ball, check if there are any consecutive balls of the same color with a length of three or more.
  4. If such a sequence exists, remove those balls and repeat the process of checking and removing adjacent sequences until no more can be removed.
  5. Calculate the length of the remaining sequence of balls and keep track of which initial insertion led to the shortest sequence. A completely empty sequence is better than any other.
  6. Repeat this process for every possible place to insert the ball. After doing all possible insertions, choose to insert the ball where the remaining sequence is the shortest.
  7. Continue this process until the initial sequence of balls is empty or no more moves can be made.

Code Implementation

def zuma_game(board_string, hand_string):
    # This function simulates inserting the hand ball at index. 
    def insert_and_eliminate(current_board, index, hand_ball):
        new_board = list(current_board)
        new_board.insert(index, hand_ball)
        new_board = "".join(new_board)

        while True:
            original_length = len(new_board)
            new_board = remove_adjacent(new_board)
            if len(new_board) == original_length:
                break
        return new_board

    def remove_adjacent(board):
        i = 0
        while i < len(board):
            j = i
            while j < len(board) and board[i] == board[j]:
                j += 1
            if j - i >= 3:
                board = board[:i] + board[j:]
                i = 0 # restart search from the beginning
                continue
            else:
                i += 1
        return board

    shortest_board = float('inf')
    best_move = -1

    for i in range(len(board_string) + 1):
        # Trying all possible insertion points for the ball.
        resulting_board = insert_and_eliminate(board_string, i, hand_string)
        board_length = len(resulting_board)

        if board_length < shortest_board:
            shortest_board = board_length
            best_move = i

    if shortest_board == float('inf'):
        return -1

    resulting_board = insert_and_eliminate(board_string, best_move, hand_string)

    if len(resulting_board) == 0:
      return 0
    else:
      return len(resulting_board)

Big(O) Analysis

Time Complexity
O(n^3)The algorithm iterates through all possible insertion points in the initial sequence, which takes O(n) time. For each insertion, it simulates the insertion and then repeatedly removes adjacent balls of the same color of length three or more. The removal process, in the worst case, might need to traverse the entire sequence multiple times if the sequence is constructed to trigger multiple removals after each insertion. Each traversal and potential removal takes O(n) time, and this removal process could repeat up to n times (in a contrived worst case scenario where each removal only eliminates a small number of balls and creates more chains). Therefore, each insertion can cost up to O(n^2). Since we perform this insertion and removal process for each of the n possible insertion positions, the overall time complexity becomes O(n * n^2) = O(n^3).
Space Complexity
O(N)The space complexity is primarily determined by the simulation process where, for each possible insertion, a copy of the current ball sequence, which is of size N in the worst case, needs to be created and potentially modified. The recursive calls to handle chain reactions after inserting a ball could lead to a recursion depth proportional to the length of the sequence (N). Each recursive call requires stack space to store the state of the function, so the auxiliary space used could be proportional to N in the worst-case scenario, where N is the length of the initial sequence of balls. Therefore, the overall auxiliary space complexity is O(N).

Edge Cases

Null or empty hand string
How to Handle:
Return 0 immediately if the hand is null or empty as no insertion is possible.
Null or empty board string
How to Handle:
Return 0 if the board is null or empty, because the empty hand cannot be used.
Hand contains all identical balls and board has a single ball
How to Handle:
Iteratively add balls from the hand until a matching sequence eliminates the single ball or the hand is empty.
Board contains only one type of ball in a very long sequence
How to Handle:
Check if the hand can completely eliminate it by inserting enough balls to form a sequence of 3 or more.
Hand contains a very large number of balls
How to Handle:
Consider the constraints on the hand size for possible optimizations or limitations on memory usage.
Board contains multiple identical sequences with varying lengths separated by different colors
How to Handle:
The solution should correctly handle collapsing the board from left to right after each insertion and removal.
The hand cannot eliminate the board, regardless of the order the balls are inserted
How to Handle:
Return -1 when all possible combinations have been tried, indicating no solution exists.
Integer overflow potential with recursive depth calculations
How to Handle:
Ensure that memoization and pruning techniques are employed to reduce unnecessary recursive calls to avoid exceeding the maximum recursion depth or integer limits.