Commented | Intuitive
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n    def stoneGameIX(self, stones: List[int]) -> bool:\n        # General idea: Game theory minimax\n        # state: (turn, count: tuple, runningSum)\n        # for winning we need to find a move that is losing for the other player.\n        # the greedy idea here is that picking a multiple of 3 will only shift the turn\n        # We can\'t start from picking a multiple of 3.\n        # If running sum is a multiple of 2, we can only pick multiple of 2s.\n        # If running sum is a multiple of 1, we can only pick multiple of 1s.\n        # Greedy, use 0s first then simulate the optimal play in linear time.\n        count = [0] * 3\n        for stone in stones:\n            count[stone % 3] += 1\n        count[0] %= 2\n        \n        def canWin(turn: int, count: List[int], currentSum: int) -> bool:\n            # base case \n            if sum(count) == 0:\n                return turn == 1\n            # Use 0 to flip the turn\n            if count[0] > 0:\n                count[0] -= 1\n                return not canWin(1 - turn, count, currentSum)\n            elif currentSum % 3 == 1:\n                if count[1] > 0:\n                    count[1] -= 1\n                    return not canWin(1 - turn, count, (currentSum + 1) % 3)\n                else:\n                    return False\n            elif currentSum % 3 == 2:\n                if count[2] > 0:\n                    count[2] -= 1\n                    return not canWin(1 - turn, count, (currentSum + 2) % 3)\n                else:\n                    return False \n            return False\n        \n        ans = False\n        # try starting with 1\n        if count[1] > 0:\n            temp = count.copy()\n            temp[1] -= 1\n            ans = not canWin(1, temp, 1)\n        # try starting with 2\n        if count[2] > 0:\n            temp = count.copy()\n            temp[2] -= 1\n            ans |= not canWin(1, temp, 2)\n        return ans\n        \n        # DP, In the winning sequence, zeros will only be used to pass the turn\n        count = [0] * 3\n        for stone in stones:\n            count[stone % 3] += 1\n        dp = {}\n        def canAliceWin(isAlicesTurn: bool, countZero: int, countOne: int, countTwo: int, currentSum: int) -> bool:\n            if countZero == countOne == countTwo == 0:\n                return False\n            \n            key = (isAlicesTurn, countZero, countOne, countTwo, currentSum)\n            if key in dp:\n                return dp[key]\n\n            if isAlicesTurn:\n                ans = False\n                if countZero > 0 and (currentSum) % 3 != 0:\n                    ans |= canAliceWin(False, countZero - 1, countOne, countTwo, (currentSum) % 3)\n                if countOne > 0 and (currentSum + 1) % 3 != 0:\n                    ans |= canAliceWin(False, countZero, countOne - 1, countTwo, (currentSum + 1) % 3)\n                if countTwo > 0 and (currentSum + 2) % 3 != 0:\n                    ans |= canAliceWin(False, countZero, countOne, countTwo - 1, (currentSum + 2) % 3)\n                dp[key] = ans\n                return ans\n            else:\n                ans = True\n                if countZero > 0 and (currentSum) % 3 != 0:\n                    ans &= canAliceWin(True, countZero - 1, countOne, countTwo, (currentSum) % 3)\n                if countOne > 0 and (currentSum + 1) % 3 != 0:\n                    ans &= canAliceWin(True, countZero, countOne - 1, countTwo, (currentSum + 1) % 3)\n                if countTwo > 0 and (currentSum + 2) % 3 != 0:\n                    ans &= canAliceWin(True, countZero, countOne, countTwo - 1, (currentSum + 2) % 3)\n                dp[key] = ans\n                return ans\n\n        return canAliceWin(True, count[0] % 2, count[1], count[2], 0)\n```
['Python3']