Easy 0 ms 100% (Fully Explained)(Java, C++, Python, JS, C, Python3)
Java Solution:\nRuntime: 0 ms, faster than 100.00% of Java online submissions for Poor Pigs.\n\nclass Solution {\n    public int poorPigs(int buckets, int minut
# **Java Solution:**\nRuntime: 0 ms, faster than 100.00% of Java online submissions for Poor Pigs.\n```\nclass Solution {\n    public int poorPigs(int buckets, int minutesToDie, int minutesToTest) {\n        // Calculate the max time for a pig to test buckets...\n        // Note that, max time will not be (minutesToTest / minutesToDie)...\n        // Thinking about all pigs drinking all buckets at last, but no one died immediately, so the poison bucket is the last bucket...\n        int max_time = minutesToTest / minutesToDie + 1;\n        // Initialize the required minimum number of pigs...\n        int req_pigs = 0;\n        // To find the minimum number of pigs, find the minimum req_pigs such that Math.pow(max_time, req_pigs) >= buckets...\n        while (Math.pow(max_time, req_pigs) < buckets)\n            // Increment until it will be greater or equals to bucket...\n            ++req_pigs;\n        // Return the required minimum number of pigs...\n        return req_pigs;\n    }\n}\n```\n\n# **C++ Solution:**\n```\nclass Solution {\npublic:\n    int poorPigs(int buckets, int minutesToDie, int minutesToTest) {\n        // Calculate the max time for a pig to test buckets...\n        // Note that, max time will not be (minutesToTest / minutesToDie)...\n        // Thinking about all pigs drinking all buckets at last, but no one died immediately, so the poison bucket is the last bucket...\n        int max_time = minutesToTest / minutesToDie + 1;\n        // To find the minimum number of pigs, return ceil(log(buckets) / log(max_time))...\n        return ceil(log(buckets) / log(max_time));\n    }\n};\n```\n\n# **Python Solution:**\nRuntime: 21 ms, faster than 79.11% of Python online submissions for Poor Pigs.\n```\nclass Solution(object):\n    def poorPigs(self, buckets, minutesToDie, minutesToTest):\n        # Calculate the max time for a pig to test buckets...\n        # Note that, max time will not be (minutesToTest / minutesToDie)...\n        # Thinking about all pigs drinking all buckets at last, but no one died immediately, so the poison bucket is the last bucket...\n        max_time = minutesToTest / minutesToDie + 1\n        # Initialize the required minimum number of pigs...\n        req_pigs = 0\n        # To find the minimum number of pigs, find the minimum req_pigs such that Math.pow(max_time, req_pigs) >= buckets...\n        while (max_time) ** req_pigs < buckets:\n            # Increment until it will be greater or equals to bucket...\n            req_pigs += 1\n        # Return the required minimum number of pigs...\n        return req_pigs\n```\n        \n# **JavaScript Solution:**\n```\nvar poorPigs = function(buckets, minutesToDie, minutesToTest) {\n    // Calculate the max time for a pig to test buckets...\n    // Note that, max time will not be (minutesToTest / minutesToDie)...\n    // Thinking about all pigs drinking all buckets at last, but no one died immediately, so the poison bucket is the last bucket...\n    let max_time = minutesToTest / minutesToDie + 1;\n    // Initialize the required minimum number of pigs...\n    let req_pigs = 0;\n    // To find the minimum number of pigs, find the minimum req_pigs such that Math.pow(max_time, req_pigs) >= buckets...\n    while (Math.pow(max_time, req_pigs) < buckets)\n        // Increment until it will be greater or equals to bucket...\n        ++req_pigs;\n    // Return the required minimum number of pigs...\n    return req_pigs;\n};\n```\n\n# **C Language:**\nRuntime: 0 ms, faster than 100.00% of C online submissions for Poor Pigs.\n```\nint poorPigs(int buckets, int minutesToDie, int minutesToTest){\n    // Base condition...\n    if (buckets --== 1){\n            return 0;\n    }\n    // Calculate the max time for a pig to test buckets...\n        // Note that, max time will not be (minutesToTest / minutesToDie)...\n        // Thinking about all pigs drinking all buckets at last, but no one died immediately, so the poison bucket is the last bucket...\n    int max_time = minutesToTest / minutesToDie + 1;\n    // Initialize the required minimum number of pigs...\n    int req_pigs = 0;\n    // Process to find the minimum number of pigs...\n    while (buckets > 0){\n        buckets /= max_time;\n        req_pigs++;\n    }\n    // Return the required minimum number of pigs...\n    return req_pigs;\n}\n```\n\n# **Python3 Solution:**\nRuntime: 38 ms, faster than 76.64% of Python3 online submissions for Poor Pigs.\n```\nclass Solution:\n    def poorPigs(self, buckets: int, minutesToDie: int, minutesToTest: int) -> int:\n        # Calculate the max time for a pig to test buckets...\n        # Note that, max time will not be (minutesToTest / minutesToDie)...\n        # Thinking about all pigs drinking all buckets at last, but no one died immediately, so the poison bucket is the last bucket...\n        max_time = minutesToTest / minutesToDie + 1\n        # Initialize the required minimum number of pigs...\n        req_pigs = 0\n        # To find the minimum number of pigs, find the minimum req_pigs such that Math.pow(max_time, req_pigs) >= buckets...\n        while (max_time) ** req_pigs < buckets:\n            # Increment until it will be greater or equals to bucket...\n            req_pigs += 1\n        # Return the required minimum number of pigs...\n        return req_pigs\n```\n**I am working hard for you guys...\nPlease upvote if you find any help with this code...**
['C', 'Python', 'Java', 'Python3', 'JavaScript']