[C++] Concise solution O(n) [Explained]
result string will contain either no zeros or single zero.\n result will contain no zeros if and only if the binary string is all 1s\n Otherwise, the task is to
* `result` string will contain either no zeros or single zero.\n* `result` will contain no zeros if and only if the `binary` string is all 1s\n* Otherwise, the task is to push the left-most 0 to the right as much as possible\n* Every zero appearing after the left-most zero can shift it right by 1 as a result of some *2nd Operations* and then *1st Operation*\n\n**Idea:** Count the number of zeros and find the index that contains the 0\n\n```\nstring maximumBinaryString(string binary) {\n        int n = binary.size();\n        string result = string(n, \'1\');\n        int zero_count = count(binary.begin(), binary.end(), \'0\');\n        if (zero_count == 0) {\n            return result;\n        }\n        int first_zero = binary.find(\'0\');\n        result[first_zero+zero_count-1] = \'0\';\n        return result;\n}\n```
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