Easy to understand solution
like you have 15 minutes timeToDie & 60 minutes TimesToTest\nnow if you have 1 pig then you can test 5 buckets\nSuch that for the 0th round you will give bucket
*like you have 15 minutes timeToDie & 60 minutes TimesToTest\nnow if you have 1 pig then you can test 5 buckets\nSuch that for the 0th round you will give bucket 1 if pig dies it means you found else you will give 2 bucket if dies you found and so on then if you gave first 4 buckets to pig and pig is still alive then it means poison is in last buckets that is why we are taking T+1.\n\nso this solution will work in nd matrix like we are having 2d marix we will check from row & column where both will intersect thats our answer.\n\n\n**UPVote if it helps**\n```\nclass Solution {\npublic:\n    int poorPigs(int n, int d, int t) {\n        int cnt = t/d;\n        cnt++;\n        int ans = 0;\n        int x = 1;\n        while(x < n){\n             x *= cnt;\n            ans++;\n        }\n        return ans;\n    }\n};\n```
['C++']