Greedy Approach
Intuition:\nThe goal is to maximize the binary string by transforming pairs of adjacent \'0\'s into \'1\'s and maintaining the position of a single \'0\'. This 
**Intuition:**\nThe goal is to maximize the binary string by transforming pairs of adjacent \'0\'s into \'1\'s and maintaining the position of a single \'0\'. This can be achieved by moving all \'0\'s to the rightmost part of the string and then converting all possible \'0\'s to \'1\'s while keeping one \'0\' intact in the optimal position.\n\n**Approach:**\n1. Traverse the string and identify positions of \'0\'s.\n2. Move \'0\'s towards the rightmost part of the string by flipping adjacent pairs to maximize the number of \'1\'s.\n3. Ensure a single \'0\' remains at the position just after the first series of \'1\'s if any \'0\'s were found initially.\n4. Return the transformed binary string.\n\n**Time Complexity:** $$O(n)$$, where \\(n\\) is the length of the string, as we make a single pass through the string.\n\n**Space Complexity:** $$O(1)$$, as we use only a few additional variables.\n\n**Code with Comments:**\n\n```cpp\nclass Solution {\npublic:\n    string maximumBinaryString(string binary) {\n        int n = binary.size(); // Get the size of the binary string\n        int lastZero = -1; // Initialize the last position of \'0\'\n        \n        if (binary[0] == \'0\') {\n            lastZero = 0; // Set lastZero if the first character is \'0\'\n        }\n\n        for (int i = 1; i < n; ++i) {\n            if (binary[i] == binary[i - 1]) {\n                binary[i - 1] = \'1\'; // Convert the previous \'0\' to \'1\'\n            } else {\n                if (binary[i] == \'0\') {\n                    // Check and update the position of lastZero\n                    if (lastZero != -1) {\n                        binary[lastZero] = \'1\'; // Convert lastZero position to \'1\'\n                        binary[lastZero + 1] = \'0\'; // Move \'0\' one position to the right\n                        binary[i] = \'1\'; // Convert current position to \'1\'\n                        lastZero++; // Update the lastZero position\n                    }\n                }\n            }\n\n            if (binary[i] == \'0\') {\n                lastZero = i; // Update lastZero to current position\n            }\n        }\n\n        return binary; // Return the transformed binary string\n    }\n};\n```\n\n**Explanation:**\n- **Initialization:** Start by checking if the first character is \'0\'. If so, initialize `lastZero` to 0.\n- **Traversal and Transformation:** Iterate through the string from the second character. \n  - If two consecutive characters are the same, set the previous character to \'1\'.\n  - If a \'0\' is found, use `lastZero` to move the \'0\' one position to the right and convert the positions involved to \'1\'s, updating `lastZero`.\n- **Updating `lastZero`:** Continuously update `lastZero` to keep track of the rightmost \'0\'.\n- **Final Return:** After processing, return the modified binary string.\n\nThis approach ensures that the binary string is maximized by moving all \'0\'s to the rightmost part and converting all possible positions to \'1\'s while keeping the necessary \'0\' in the optimal position.
['String', 'Greedy', 'C++']