Dry run of everyone's One liner code here !!!!
```\n / lets dry run this for a test case \n        \n        n=5 dieTime=15 , totaltim=60;   Ans=1\n        \n    pig will test bucket no 1 in 1st trial -->  p
```\n /* lets dry run this for a test case \n        \n        n=5 dieTime=15 , totaltim=60;   Ans=1\n        \n    pig will test bucket no 1 in 1st trial -->  pig will test bucket no 2 in 2nd trial \n    pig will test bucket no 3 in 3rd trial -->  pig will test bucket no 4 in 4th trial \nnow pig doesnt die and trials finished and we dont encounter a poison bucket ?? Really?\n\nif there are 5 buckets and pig identified 4 buckets as non poisnosus in 4 attempts so the 5th bucket is poisonous . So a pig can identify (totalTime/dieTime +1 ) buckets in total \n\n   1 pig can identify --> (totalTime/dieTime+1) buckets\n   p pigs can idenify --> (totalTime/dieTime+1)^p where p=no of pigs\n   \n   so (totalTime/dieTime+1)^p <= n   \n       \n      taking log on both sides\n      \n      p* log(trials) <= logn   where trials = (totaltime/dieTime+1)\n      p <= log(n)/log(trials)\n      p=ceil(log(n)/log(trials));\n      */\n    \n    int poorPigs(int n, int dieTime, int totalTime) {\n        \n       return ceil(log(n)/log(totalTime/dieTime+1));\n    }
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