How do you solve Zuma Game on LeetCode?

Zuma Game (LeetCode #488) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m²) time complexity and O(n * m) space complexity. Key insight: Understanding how to remove consecutive elements is crucial.

What is the brute force solution for Zuma Game?

The brute force approach for Zuma Game is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible way to insert the balls from the hand into the board. It checks all combinations of ball placements and simulates the game until either the board is cleared or no more moves are possible. The optimal Optimal Solution approach improves this to O(n * m²).

What algorithm or data structure is used to solve Zuma Game?

Zuma Game uses the following concepts: String, Dynamic Programming, Stack, Breadth-First Search, Memoization. The recommended patterns to study are: Depth-First Search, Memoization, Backtracking.

What are the interview tips for Zuma Game?

Explain your thought process clearly while coding. Discuss the time and space complexity of your solution. Be prepared to optimize your solution if asked.

What are common mistakes when solving Zuma Game?

Not handling edge cases where the board is already empty. Failing to optimize the recursive calls with memoization.

#488

Zuma Game

Hard

StringDynamic ProgrammingStackBreadth-First SearchMemoizationDepth-First SearchMemoizationBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m²)
Space	O(1)	O(n * m)

💡

Intuition

Time O(n * m²)Space O(n * m)

The optimal solution uses a depth-first search (DFS) combined with memoization to efficiently explore the possible placements of balls while avoiding redundant calculations. It minimizes the number of balls needed by leveraging the hand's available balls strategically.

⚙️

Algorithm

3 steps

1Step 1: Create a function to remove consecutive balls from the board.
2Step 2: Use a recursive DFS function to explore all possible placements of balls from the hand.
3Step 3: Use memoization to store results of previously computed states to avoid recalculating.

solution.py36 lines

1def findMinStep(board, hand):
2    from collections import Counter
3    hand_count = Counter(hand)
4    memo = {}
5    def remove_consecutive(b):
6        i = 0
7        while i < len(b):
8            j = i
9            while j < len(b) and b[j] == b[i]:
10                j += 1
11            if j - i >= 3:
12                return remove_consecutive(b[:i] + b[j:])
13            i = j
14        return b
15    def dfs(b, hand_count):
16        if (b, tuple(hand_count.items())) in memo:
17            return memo[(b, tuple(hand_count.items()))]
18        b = remove_consecutive(b)
19        if not b:
20            return 0
21        if not hand_count:
22            return float('inf')
23        min_steps = float('inf')
24        for i in range(len(b) + 1):
25            for color in hand_count:
26                if hand_count[color] > 0:
27                    hand_count[color] -= 1
28                    new_board = b[:i] + color + b[i:]
29                    steps = dfs(new_board, hand_count)
30                    if steps != float('inf'):
31                        min_steps = min(min_steps, steps + 1)
32                    hand_count[color] += 1
33        memo[(b, tuple(hand_count.items()))] = min_steps
34        return min_steps
35    result = dfs(board, hand_count)
36    return result if result != float('inf') else -1

ℹ

Complexity note: The time complexity is O(n * m²) where n is the length of the board and m is the number of distinct colors in hand. This is due to the recursive nature of the DFS and the memoization storing results for each unique state.

1Understanding how to remove consecutive elements is crucial.
2Memoization can significantly reduce redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.