Easy C++ solution using heap
\nclass Solution {\npublic:\n    // comparison logic to build a min-heap of strings\n    struct comparator {\n        bool operator()(string &s1,string &s2) {\n
```\nclass Solution {\npublic:\n    // comparison logic to build a min-heap of strings\n    struct comparator {\n        bool operator()(string &s1,string &s2) {\n            // if length of first string is greater that means first string denotes greater number hence it should come next to lesser string. Hence swapping should be done\n            if(s1.length()>s2.length()) {\n                return true;\n            }\n            \n            // if length of first string is lesser that means first string denotes smaller number hence ordering is fine. No need to swap. \n            else if(s1.length()<s2.length()) {\n                return false;\n            }\n            \n            // if lengths are equal then we have to iterate them\n            else {\n                for(int i=0;i<s1.length();i++) {\n                    // if at a particular index digits vary then decide which one is greater or lesser and return \n                    if(s1[i]-\'0\'>s2[i]-\'0\') {\n                        return true;\n                    }\n                    else if(s1[i]-\'0\'<s2[i]-\'0\') {\n                        return false;\n                    }\n                }\n                return false;\n            }\n        }\n    };\n    string kthLargestNumber(vector<string>& nums, int k) {\n        // declare a min-heap of strings\n        priority_queue<string,vector<string>,comparator>q;\n        \n        // iterate the vector \n        for(auto it:nums) {\n            q.push(it);\n            \n            // whenever the size becomes greater than k remove the smallest string available in the heap.\n            if(q.size()>k) {\n                q.pop();\n            }\n        }\n        \n        // k\'th smallest string will be the topmost string of the heap since it\'s a min-heap\n        return q.top();\n    }\n};\n```\n***Please upvote if you have got any help from my code. Thank you.***
['C', 'Heap (Priority Queue)']