Most optimal solution with complete exaplanation
\n# Approach\nThe solution iterates through the input string colors. For each position i, it checks if the character at that position and its neighboring charac
\n# Approach\nThe solution iterates through the input string colors. For each position i, it checks if the character at that position and its neighboring characters (at positions i-1 and i+1) satisfy the conditions mentioned in the game rules. If the conditions are met, Alice or Bob can make a move, and their corresponding counter (alice or bob) is incremented.\n\nFinally, the function compares the counts of valid moves made by Alice and Bob. If Alice has made more valid moves, the function returns true, indicating that Alice wins; otherwise, it returns false, indicating that Bob wins.\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n```java []\nclass Solution {\n    public boolean winnerOfGame(String colors) {\n        int alice = 0, bob = 0;\n        for (int i = 1; i < colors.length() - 1; i++) {\n            char next = colors.charAt(i + 1);\n            char prev = colors.charAt(i - 1);\n            char curr = colors.charAt(i);\n            if (curr == \'A\' && next == \'A\' && prev == \'A\') {\n                alice++;\n            }\n            if (curr == \'B\' && next == \'B\' && prev == \'B\') {\n                bob++;\n            }\n        }\n        return alice > bob;\n    }\n}\n\n```\n```python3 []\nclass Solution:\n    def winnerOfGame(self, colors: str) -> bool:\n        alice, bob = 0, 0\n        for i in range(1, len(colors) - 1):\n            next_char = colors[i + 1]\n            prev_char = colors[i - 1]\n            curr_char = colors[i]\n            if curr_char == \'A\' and next_char == \'A\' and prev_char == \'A\':\n                alice += 1\n            if curr_char == \'B\' and next_char == \'B\' and prev_char == \'B\':\n                bob += 1\n        return alice > bob\n\n```\n```C++ []\nclass Solution {\npublic:\n    bool winnerOfGame(string colors) {\n        int alice = 0, bob = 0;\n        for(int i = 1; i< colors.size()-1; i++) {\n            char next = colors[i+1];\n            char prev = colors[i-1];\n            char curr = colors[i];\n            if(curr == \'A\' && next == \'A\' && prev == \'A\') alice++;\n            if(curr == \'B\' && next == \'B\' && prev == \'B\') bob++;\n        }\n        return alice>bob;\n    }\n};\n```\n
['Math', 'String', 'Greedy', 'Game Theory', 'Python', 'C++', 'Java', 'Python3']