Very simple Intuition solution || C++
Complexity\n- Time complexity: O(N^3)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(N)\n Add your space complexity here, e.g. O(n) \n\n# 
# Complexity\n- Time complexity: O(N^3)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    \n    void solve(vector<string> &ans, string &a, string &b){\n        // Here we have to validate and then add that to answer\n        vector<string> tem;\n        string t = "(";\n        // putting dot on\n        for(int i=0;i<a.length();i++){\n            t = "(";\n            if(i == 0) t += a;\n            else{\n                t += a.substr(0,i);\n                t += \'.\';\n                t += a.substr(i,a.length()-i);\n            }\n            // Now we got t\n            // check for 00. or 000. 01.\n            if(t[1] == \'0\'){\n                // if first char is zero we must see if we got any other char after that or not if we got that we break\n                if(t.length() > 2 && t[2] != \'.\') continue;\n            }\n            // check for .000 or .00\n            if(i != 0 && t[t.length()-1] == \'0\') continue;\n            tem.push_back(t);\n        }\n        t = "";\n        for(int i=0;i<b.length();i++){\n            if(i == 0) t = b;\n            else{\n                t = b.substr(0,i);\n                t += \'.\';\n                t += b.substr(i,b.length()-i);\n            }\n            // Now we got t\n            // check for 00. or 000. 01.\n            if(t[0] == \'0\'){\n                // if first char is zero we must see if we got any other char after that or not if we got that we break\n                if(t.length() > 1 && t[1] != \'.\') continue;\n            }\n            // check for .000 or .00\n            if(i != 0 && t[t.length()-1] == \'0\') continue;\n            for(auto s: tem){\n                string n = s;\n                n += ", ";\n                n += t;\n                n += \')\';\n                ans.push_back(n);\n            }\n        }\n    }\n    \n    vector<string> ambiguousCoordinates(string s) {\n        vector<string> ans;\n        string fi = "", se = "";\n        // we are making first coordinate and second then try to add point in all and validate \n        for(int j = 1;j<s.length()-2;j++){\n            fi += s[j];\n            se = s.substr(j+1,s.length()-1-(j+1));\n            // Now we have to pass this\n            solve(ans,fi,se);\n        }\n        return ans;\n    }\n};\n```
['String', 'C++']