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:
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 <= 161 <= hand.length <= 5board and hand consist of the characters 'R', 'Y', 'B', 'G', and 'W'.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:
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:
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_neededThe 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:
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)| Case | How to Handle |
|---|---|
| Null or empty hand string | Return 0 immediately if the hand is null or empty as no insertion is possible. |
| Null or empty board string | 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 | 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 | 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 | 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 | 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 | Return -1 when all possible combinations have been tried, indicating no solution exists. |
| Integer overflow potential with recursive depth calculations | Ensure that memoization and pruning techniques are employed to reduce unnecessary recursive calls to avoid exceeding the maximum recursion depth or integer limits. |