Maximum Binary String After Change || O(n) Solution !!
\nclass Solution {\npublic:\n    string maximumBinaryString(string s) {\n        \n        int zero = 0, i = 0, len = s.length();\n        \n        while (s[i]
```\nclass Solution {\npublic:\n    string maximumBinaryString(string s) {\n        \n        int zero = 0, i = 0, len = s.length();\n        \n        while (s[i] == \'1\')\n        {\n            i++; \n            \n        } \n        for (int j = i; j < len; j++)\n        {\n            if(s[j]==\'0\')\n                zero++;\n        }\n        if (zero < 2) return s;\n        \n        while (--zero)\n        {\n            s[i++] = \'1\';\n        }\n        s[i++] = \'0\';\n        while (i < len)\n        {\n            s[i++] = \'1\';\n        }\n        \n        return s;\n    }\n};\n```
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