How do you solve Stone Game IX on LeetCode?

Stone Game IX (LeetCode #2029) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the properties of modular arithmetic is crucial for solving this problem efficiently.

What is the brute force solution for Stone Game IX?

The brute force approach for Stone Game IX is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating all possible sequences of stone removals. By checking each possible combination, we can determine whether Alice can win or not. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Stone Game IX?

Stone Game IX uses the following concepts: Array, Math, Greedy, Counting, Game Theory. The recommended patterns to study are: Game Theory, Greedy Algorithms.

What are the interview tips for Stone Game IX?

Always clarify the rules of the game before diving into coding. Think about edge cases, such as all stones being the same or having only one stone. Practice explaining your thought process clearly as you work through the problem.

What are common mistakes when solving Stone Game IX?

Not considering the implications of the sum being divisible by 3 early in the game. Overcomplicating the problem by trying to simulate every possible move instead of recognizing patterns.

#2029

Stone Game IX

Medium

ArrayMathGreedyCountingGame TheoryGame TheoryGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages the properties of modular arithmetic and the game structure. By counting the stones based on their values modulo 3, we can determine the winner without simulating all possible moves.

⚙️

Algorithm

3 steps

1Step 1: Count the number of stones with values % 3 == 0, % 3 == 1, and % 3 == 2.
2Step 2: If there are no stones with value % 3 == 0, Alice can always win by removing stones strategically.
3Step 3: If there are stones with value % 3 == 0, Alice must avoid making the sum of removed stones divisible by 3.

solution.py5 lines

1def canAliceWin(stones):
2    count = [0, 0, 0]
3    for stone in stones:
4        count[stone % 3] += 1
5    return count[1] > count[2] or (count[1] == count[2] and count[0] == 0)

ℹ

Complexity note: This complexity is linear because we only need to traverse the array once to count the stones, leading to a very efficient solution.

1Understanding the properties of modular arithmetic is crucial for solving this problem efficiently.
2Optimal play involves recognizing when to force the opponent into a losing position.

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