Simple O(N) approach, with detailed explanation
Intuition\nHere are the initial steps that come to mind:\n\nUnderstanding the Game Rules: To devise a solution, it\'s crucial to understand the rules of the gam
# Intuition\nHere are the initial steps that come to mind:\n\nUnderstanding the Game Rules: To devise a solution, it\'s crucial to understand the rules of the game thoroughly. Each player aims to avoid making the sum of the values of removed stones divisible by 3. This implies that players need to be strategic in their choices to force their opponent into making a losing move.\n\nAnalyzing Stone Values: We need to consider the values of the stones and how they affect the game. Stones with values divisible by 3 are crucial, as removing them could change the game\'s dynamics significantly.\n\nIdentifying Winning Conditions: Understanding the conditions under which Alice can win is essential. It might involve identifying patterns in the sum of stone values or exploiting specific configurations of stones to force Bob into making a losing move.\n\n# Approach\nCount Remainders: First, we count the number of stones with remainders 0, 1, and 2 when divided by 3. Let\'s denote these counts as R0, R1, R2\n\n**Analysis of Winning Conditions:**\n\nIf R0 is even:\nAlice cannot win because she will always make the sum divisible by 3 with her moves.\nIf R0 is odd:\nAlice can potentially win if she can create a sequence of moves that results in a non-divisible sum. This depends on the relative counts of R1 and R2.\nIf |R1-R2| \u2265 3Alice can win because she can alternate removing stones from remainders 1 and 2 to keep the sum non-divisible by 3.\nIf |R1-R2| <3, Alice cannot win as Bob can always counter Alice\'s moves to make the sum divisible by 3.\n\n**Optimal Strategy:**\n\nIf R0 is odd and |R1-R2|\u22653, Alice should aim to create an alternating sequence of moves, removing stones from remainders 1 and 2.\nIf R0 is odd and |R1-R2|<3, Alice cannot win regardless of her moves.\n\n**Corner Cases Handling:**\n\nIf there are no stones left (R0=R1=R2=0), Bob wins automatically.\nIf R0 is even, Bob wins automatically because Alice cannot avoid making the sum divisible by 3.\n\n# Complexity\n- Time complexity: O(N)\n\n\n- Space complexity: O(1)\n\n\n# Code\n```\nclass Solution {\n    public boolean stoneGameIX(int[] stones) {\n        int[] modCounts = new int[3]; // To store count of stones with remainder 0, 1, and 2 respectively when divided by 3.\n        for (int stone : stones) {\n            modCounts[stone % 3]++;\n        }\n        \n        // Check for corner cases where either Alice or Bob automatically wins.\n        if (modCounts[0] % 2 == 0) {\n            // If there are even number of stones with remainder 0, Alice cannot win since she\'ll make the sum divisible by 3.\n            return modCounts[1] != 0 && modCounts[2] != 0;\n        } else {\n            // If there are odd number of stones with remainder 0, Alice can win if she has stones with remainder 1 and remainder 2 respectively.\n            return Math.abs(modCounts[1] - modCounts[2]) >= 3;\n        }\n    }\n}\n```
['Math', 'Java']