Clear Explanation (for lee215's solution)
Intuition\n Describe your first thoughts on how to solve this problem. \n\nIt should first be noted that only the $\text{mod }3$ of the stone values matter. So 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nIt should first be noted that only the $\\text{mod }3$ of the stone values matter. So it makes sense to just bucket them into \n- $n_0$ (counts of stone values $\\text{mod }3=0$)\n- $n_1$ (counts of stone values $\\text{mod }3=1$)\n- $n_2$ (counts of stone values $\\text{mod }3=2$)\n\nBelow, I will use the terminology "take 2" to indicate taking a stone s.t. the stone\'s value $\\text{mod }3=2$.\n\n\nNext, I will state the state transition graph for $X\\text{ mod }3$ where $X$ is the total sum of stones grabbed so far. $X\\text{ mod }3=0$  is the end of the game.\n\n![image.png](https://assets.leetcode.com/users/images/0afe2a44-46db-4783-a57d-ff0d057eaf43_1709085486.5147562.png)\n\nThere is a great property above, States $1$ and $2$ are basically symmetric. Lets first pretend $n_0=0$.\n\nWithout loss of generality, lets say $n_1 \\ge n_2$. We make the claim that Alice should take $2$ in her first move. Lets create a table for this below.\n\n\n| Curr Turn | Prev Turn            | State                        |\n|-----------|----------------------|------------------------------|\n| Bob       | Alice takes 2        | ($n_1$, $n_2-1$) @ $2$       |\n| Alice     | Bob takes 2          | ($n_1$, $n_2-2$) @ $1$       |\n| ...       | ...                  | ...                          |\n| Alice     | Bob takes (the last) 2 | ($n_1-(n_2-2)$, $0$) @ $1$   |\n| Bob       | Alice takes 1        | ($n_1-(n_2-2)-1$, $0$) @ $2$ |\n\nAll of the moves above are forced. At the end, we arrive at a state where Bob is forced to take $1$ (@ state $2$) to lose the game. **THUS, (when $n_0=0$) if you manage to put your opponent in the state with the lower frequency/count (yourself in the state of higher frequency), you win.** And indeed Alice should take $2$ (the lower frequency) on her first move.\n\nP.S. note that even if $n_1=n_2$, $n_2-1$ is strictly less than $n_1$ in the first row of the above table.\n\n\n---\n\nNow for dealing with $n_0$.\n\nAssuming $n_1 > 0$ & $n_2 > 0$, as stated in other answers, if $n_0 \\text{ mod } 2 = 0$ then it is equivalent to $n_0=0$. However, if $n_0 \\text{ mod } 2 = 1$, you should in general choose put your opponent in the state of the higher frequency since you can take a $0$ to flip the opponent to the state of the lower frequency and having $n_0 \\text{ mod } 2 = 0$.\n\nHowever, there are some edge cases for $n_0 \\text{ mod } 2 = 1$.\nI will demonstrate it below. Without loss of generality again assume $n_1 \\ge n_2$. Below Alice takes the higher frequency ($1$) instead as discussed.\n\n| Curr Turn | Prev Turn            | State                        |\n|-----------|----------------------|------------------------------|\n| Bob       | Alice takes 1        | ($n_1-1$, $n_2$) @ $1$       |\n| Alice     | Bob takes 1          | ($n_1-2$, $n_2$) @ $2$       |\n| Bob     | Alice takes 2          | ($n_1-2$, $n_2-1$) @ $1$       |\n| ...       | ...                  | ...\n\nNote if $n_1=n_2+1$, then $n_1-2=n_2-1<n_2$. Alice has been put in the lower frequency state in row 2 while still $n_0 \\text{ mod } 2 = 1$ **!!!**\n\nIf $n_1=n_2+2$, then then $n_1-2=n_2$. Alice will put Bob in the higher frequency state in row 3.\n\n\nThere are some additional things to think about like why Bob in general should not take $0$ in the above case (b.c. it will put Alice in state of higher frequency & $n_0 \\text{ mod } 2 = 0$). However, all the tools of analysis have been provided.\n\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nfrom collections import Counter\n\n# credit: lee215\nclass Solution:\n    def stoneGameIX(self, stones: List[int]) -> bool:\n        cnt = Counter(x % 3 for x in stones)\n\n        # when cnt[0] is even, \n        # the game is equivalent to having no cnt[0]=0\n\n        if cnt[0]%2 == 0:\n            # Alice will win if both cnt[1] & cnt[2] > 0, she will choose the one with lower frequency\n            # if cnt[1] or cnt[2] = 0 then whoever goes first loses or it\'s a tie (still Alice\'s loss)\n            # i.e. only cnt[1] > 0, 1 -> 2 -> 0; only cnt[2] > 0, 2 -> 1 -> 0;\n            return cnt[1] > 0 and cnt[2] > 0\n        \n        else:\n            # cnt[0] gives opportunity to flip to winning side\n            # choosing two 2s has same effect as choosing 1\n            # choosing two 1s has same effect as choosing 2\n            return abs(cnt[1]-cnt[2]) > 2\n```
['Python3']