✅"2" Lines of Code with "2" Step Explanation❤️💯🔥
Approach\nStep 1: Initialization and Loop\n- Two integer variables a and b are initialized to 0. These variables will be used to keep track of the number of con
# Approach\nStep 1: Initialization and Loop\n- Two integer variables `a` and `b` are initialized to 0. These variables will be used to keep track of the number of consecutive triplets \'AAA\' and \'BBB\' in the input string `s`, respectively.\n- The code then enters a `for` loop that iterates over the characters of the string `s` starting from the second character (index 1) up to the second-to-last character (index `s.length() - 2`). It stops at the second-to-last character because it\'s checking triplets of characters (i-1, i, i+1), so the loop stops one character before the end of the string.\n\nStep 2: Counting Triplets and Determining the Winner\n- Within the loop, there is an `if` statement that checks the following condition:\n  - If the current character at index `i` is equal to both the previous character at index `i-1` and the next character at index `i+1`, i.e., it\'s part of a triplet like \'XXX\' (where \'X\' can be \'A\' or \'B\'), then it checks the character itself to see if it\'s \'A\' or \'B\'.\n  - If the current character is \'A\', it increments the count `a` by 1. This checks for consecutive triplets of \'AAA\'.\n  - If the current character is \'B\', it increments the count `b` by 1. This checks for consecutive triplets of \'BBB\'.\n- After the loop, it compares the counts `a` and `b`.\n  - If `a` is greater than `b`, it means that player A has more consecutive \'AAA\' triplets and returns `true`, indicating that player A is the winner.\n  - Otherwise, it returns `false`, indicating that player B is the winner or it\'s a tie if `b` is greater than or equal to `a`.\n# Complexity\n- Time complexity:O(n)\n- Space complexity:O(1)\n# Code\n```\nclass Solution {\n    public boolean winnerOfGame(String s) {\n        int a=0;\n        int b=0;\n        for(int i=1;i<s.length()-1;i++){\n            if(s.charAt(i)==s.charAt(i-1) && s.charAt(i)==s.charAt(i+1)){\n                if(s.charAt(i)==\'A\'){\n                    a++;\n                }\n                else{\n                    b++;\n                }\n            }\n        }\n        if(a>b){\n            return true;\n        }\n        return false;\n    }\n}\n\n```
['Java']