ONLY MAP:EXPLAINED EASY TO UNDERSTAND
Intuition\nuse map to mark the friend that received the ball\n\n# Approach\nbe carefull when your next friends value come to zero then you make that value equal
# Intuition\nuse map to mark the friend that received the ball\n\n# Approach\nbe carefull when your next friends value come to zero then you make that value equal to n because 0 is not a friend and for finding the next value of friend simply int** **nf=(currf + (i+1)*k) % n;**** and make your current friend cf  equal to newfriend nf, \n\n1.Also store the number of friend at vector or any datastructure to keep track of marked one and lastly return the unmarked one\n\n# Complexity\n- Time complexity:\n appx O(N)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    vector<int> circularGameLosers(int n, int k) {\n        if(k>n) return {};\n        vector<int> ans;\n        vector<int> temp;\n        for(int i=0;i<n;i++)\n        {\n            temp.push_back(i+1);\n        }\n        map<int , int> mp;\n        mp[1]=1;\n        int currf=1;\n        for(int i=0;i<1000;i++)\n        {\n            \n            int nf=(currf + (i+1)*k) % n;\n            if (nf == 0) {\n            nf = n;\n            }\n            if(mp[nf]==1) break;\n            mp[nf]=1;\n            currf=nf; \n        }\n        for(auto i:temp)\n        {\n            if(mp[i]==0)\n            {\n                ans.push_back(i);\n            }\n        }\n        return ans;\n        \n    }\n};\n```
['C++']