[C++] : O(nlog(n)) : Sorting
\n\tvector<int> getStrongest(vector<int>& a, int k) \n    {\n        int n = a.size();\n        sort(a.begin(),a.end());\n        int m = a[(n - 1) / 2];\n     
```\n\tvector<int> getStrongest(vector<int>& a, int k) \n    {\n        int n = a.size();\n        sort(a.begin(),a.end());\n        int m = a[(n - 1) / 2];\n        \n        vector<pair<int,int>> b;\n        for(auto x : a)\n            b.push_back({abs(x - m),x});\n        sort(b.begin(),b.end(),greater<pair<int,int>>());\n        \n        vector<int> ans;\n        for(int i = 0; i < k; i++)\n            ans.push_back(b[i].second);\n        return ans;\n    }\n```
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