Why I am not getting TLE? passed in 208ms/6.2MB
Below is my Code which I think should get TLE.\n\n\nint mn,mx;\nint f,s,n;\nclass Solution {\npublic: \n    void dfs(int deadmask,int i,int j,int round){\n     
Below is my Code which I think should get TLE.\n\n```\nint mn,mx;\nint f,s,n;\nclass Solution {\npublic: \n    void dfs(int deadmask,int i,int j,int round){\n        \n        while(i<n && deadmask &(1<<i)) i++;\n        while(j>=0 && deadmask &(1<<j)) j--;\n        // next round\n        if(i>=j) dfs(deadmask,1,n,round+1);\n        // both immortal , call will defintely reach here.\n        else if(i==f && j==s){ mn=min(mn,round),mx=max(mx,round); return;} \n        else {\n            // atleast 1 mortal\n            if(i!=f && i!=s) dfs(deadmask|(1<<i),i+1,j-1,round);\n            if(j!=f && j!=s) dfs(deadmask|(1<<j),i+1,j-1,round);\n        }\n        \n    }\npublic:\n    vector<int> earliestAndLatest(int n1, int firstPlayer, int secondPlayer) {\n        mn=INT_MAX,mx=INT_MIN;\n        f=firstPlayer,s=secondPlayer,n=n1;\n        dfs(0,1,n,1);\n        return {mn,mx};\n    }\n};\n```\n\nLet\'s caculate number of operations.\nFor first round we have `2*2*2*2*2.... till(28/2) times`. therefore 2^14 operations for Round 1.\nNow, for second we have `2^14*(2*2*2.... till 7 times)`. Therefore 2^(14+7) operations.\nNow, for third round `2^21*(2*2*2*2)`. Therefore 2^25 operations.\nNow, for 4th round `2^25*(2*2)`. So, 2^27 operations.\nNow, for 5th round 2^27 operations. Since it is deciding match.\n\nSo, clearly sum of all operations is greater than 2^28 (which is 2.6*10^8) operations. So, it should get TLE, right?\n\nInstead it is passing in 200 ms.\nAlso, by seeing 208 ms we can say that in 1 sec `10^9(2.6*10^8*(1000/208))` operations can be performed. Is it true?
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