[C++, Explanation, Easy to Understand with Comments]
First, we process the input string to remove the ( and ) characters at the beginning and end respectively. Now, we can place commas at n-1 positions, where n is
First, we process the input string to remove the `(` and `)` characters at the beginning and end respectively. Now, we can place commas at `n-1` positions, where `n` is the length of the processed string. For each of these cases, we divide the string into left and right subparts. For each subpart, we place a decimal point after each character, except in the following edge cases:\n1. If the last character of the subpart is `0`, we return, since there is no point in putting a decimal point with a zero at the end.\n2. If the subpart has a leading `0`, then we can place a decimal point only after the first character i.e. only after the first `0`\n\nHere is the code:\n```\nclass Solution {\npublic:\n    vector <string> putDecimal(string &s, int startIdx, int endIdx){\n        //if subpart\'s len == endIdx-startIdx+1 == 1, return itself\n        if(endIdx - startIdx + 1 == 1)\n            return {s.substr(startIdx, 1)};\n        vector <string> ret;\n        // if subpart does not contain a leading zero, we can add the whole subpart to the result without placing a decimal point  \n        if(s[startIdx] != \'0\')\n            ret.push_back(s.substr(startIdx, endIdx - startIdx + 1));\n        //edge case 1 mentioned above\n        if(s[endIdx] == \'0\')\n            return ret;\n        for(int i = startIdx ; i < endIdx ; i++){\n            //edge case 2 mentioned above\n            if(s[startIdx] == \'0\' && i > startIdx)\n                break;\n            string tmp = s.substr(startIdx, i - startIdx + 1) + \'.\' + s.substr(i + 1, endIdx - i);\n            ret.push_back(tmp);\n        }\n        return ret;\n    }\n    vector<string> ambiguousCoordinates(string s) {\n        int n = s.size();\n        //preprocess the input string\n        s = s.substr(1, n-2);\n        vector <string> ans;\n        n = s.size();\n        // trying to put commas at all the valid positions\n        for(int i = 0 ; i < n-1 ; i++){\n            // divide into left and right subparts\n            vector <string> part1 = putDecimal(s, 0, i), part2 = putDecimal(s, i+1, n-1);\n            //combine the results of the left and right subparts\n            for(string s1:part1){\n                for(string s2:part2)\n                    ans.push_back(\'(\' + s1 + ", " + s2 + \')\');\n            }\n        }\n        return ans;\n    }\n};\n```
['C++']