NOOB CODE : Easy to Understand
Approach\n\n1. It initializes two variables al and bo to 0 to keep track of the number of consecutive colors for player \'A\' and player \'B\', respectively.\n\
# Approach\n\n1. It initializes two variables `al` and `bo` to 0 to keep track of the number of consecutive colors for player \'A\' and player \'B\', respectively.\n\n2. It then iterates through the string `colors` from the second character (index 1) to the second-to-last character (index `len(colors) - 2`).\n\n3. Inside the loop, it checks if the current character is \'A\' and if it\'s the same as the previous and next characters. If all these conditions are met, it increments the `al` counter. Similarly, it checks for \'B\' and increments the `bo` counter.\n\n4. Finally, it compares the `al` and `bo` counters and returns `True` if `al` is greater than `bo`, indicating that player \'A\' is the winner, or `False` otherwise.\n\n# Complexity\n- Time complexity: ***O(n)***\n\n- Space complexity: ***O(1)***\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def winnerOfGame(self, colors: str) -> bool:\n        al=0\n        bo=0\n        for i in range(1,len(colors)-1):\n            if (colors[i]==\'A\' and colors[i]==colors[i-1] and colors[i]==colors[i+1]):\n                al+=1\n            elif (colors[i]==\'B\' and colors[i]==colors[i-1] and colors[i]==colors[i+1]):\n                bo+=1\n        if(al>bo):\n            return(True)\n        else:\n            return(False)\n```
['Python3']