Java pure logic solution
Intuition\n Describe your first thoughts on how to solve this problem. \nGroup the stones into mod of 3. Consider the case which is a must win for Alice, since 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nGroup the stones into mod of 3. Consider the case which is a must win for Alice, since Alice goes first.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nGet number of stones that mod 3 equals to 0, 1, 2.\n\n0 is a very special case, we can talk about it in 2 different cases, odd 0 and even 0.\n\nIf 0 is even, it\'s very simple. Whenever Bob picks 0, Alice can pick 0 to counter, so it 0 doesn\'t affect Alice\'s strategy. If 1 is more or equal to 2, Alice can pick 2 first, then Bob has to pick 2, otherwise result in a 3. Alice then picks 1, Bob has to pick 2 again. Since there is not enough 2, Bob will run out first and lose. The same applies to if 2 is more or equal to 1. So as long as there is at least one 1 and 2 in the pool, Alice always wins. Otherwise, Bob wins.\n\nIf 0 is odd, that means someone can use 0 to escape their opponent\'s strategy once, but only once, because if one picks 0, the other can pick 0 to counter, resulting only one 0 is effective for Bob. In most cases, Bob would win because there is no effective strategy for Alice to start with. But if the difference between 1 and 2 is greater than 2, then Alice could have a strategy. Suppose we have 5 1s, 2 2s, 1 3, then Alice would pick 1 first, Bob has 2 options, pick 1 or 3. If Bob picks 3, Alice would pick 1, Bob has to pick 2, then Alice picks 1, resulting in a loop, but 1 is more than 2, so Bob would run out first. If Bob picks 1, Alice can pick 3, then Bob need to pick 2, Alice then picks 1, also resulting in a loop, Alice would win because there is more 1s. The same applies to 2 as well. Therefore, if the difference between 1 and 2 is greater than 2, Alice wins, otherwise Bob wins.\n\nBelow is the code (notice 0 is the variable "three")\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1)\n\n# Code\n```\nclass Solution {\n    public boolean stoneGameIX(int[] stones) {\n        if (stones.length == 1) {\n            return false;\n        }\n        int one = 0;\n        int two = 0;\n        int three = 0;\n        for (int stone : stones) {\n            if (stone % 3 == 1) {\n                one++;\n            } else if (stone % 3 == 2) {\n                two++;\n            } else {\n                three++;\n            }\n        }\n        if (three % 2 == 0) {\n            return one != 0 && two != 0;\n        } else {\n            return (one >= 3 && two == 0) || (two >= 3 && one == 0)\n                    || (Math.abs(one - two) >= 3);\n        }\n    }\n}\n```
['Java']