Easy || 1 liner || 0 ms || 100% || Explained || Java, C++, Python, JS, Python3
A pangram is a sentence where every letter of the English alphabet appears at least once...\nGiven a string sentence containing only lowercase English letters, 
A pangram is a sentence where every letter of the English alphabet appears at least once...\nGiven a string sentence containing only lowercase English letters, return true if sentence is a pangram, or false otherwise...\n\n# **Java Solution:**\n```\n// Time Complexity: O(n)\n// Space Complexity: O(1)\nclass Solution {\n    public boolean checkIfPangram(String sentence) {\n        // Create a hashSet containing only unique values...\n        Set<Character> hset = new HashSet<>();\n        // Add all elements to hset...\n        for (final char ch : sentence.toCharArray())\n            hset.add(ch);\n        // Check the size of the hset if it is 26 then the string is a pangram...\n        return hset.size() == 26;\n    }\n}\n```\n\n# **C++ Solution:**\n```\n/** Approach 1 **/\n// Time Complexity: O(n)\n// Space Complexity: O(1)\n// Runtime: 0 ms, faster than 100.00% of C++ online submissions for Check if the Sentence Is Pangram.\nclass Solution {\npublic:\n    bool checkIfPangram(string sentence) {\n        // Add all elements to set and check if there are 26 different letters...\n        return set<char>(begin(sentence), end(sentence)).size() == 26;\n    }\n};\n---------------------------------------------------------------------------------------------------------------------------------------------------------------\n/** Approach 2 **/\nclass Solution {\npublic:\n    bool checkIfPangram(string sentence) {\n        // Check if length of sentence is less than 26, then return false...\n        if (sentence.size() < 26) return false;\n        // Declare an Array with a size of 26 & initialize each index value as 0...\n        int arr[26] = {0};\n        // Iterate every character of the given string and store the occurrence of the particular character...\n        for(char ch : sentence) {\n            // If ch equal to space, continue...\n            if (ch == \' \') continue;\n            // Initialize a index...\n            // Suppose a character is \u2018a\u2018. then idx = (int)a-97 = 97 \u2013 97 = 0...\n            int idx = (int)ch - 97;\n            // arr[idx] = arr[idx] + 1 => arr[0] = arr[0] + 1 = 0 + 1 = 1...\n            arr[idx]++;\n        }\n        for (int i : arr) {\n            // Now check if any value of the arr is 0 then the string is not a Pangram...\n            if (i == 0)\n                return false;\n        }\n        // Otherwise the sentece is a Pangram...\n        return true;\n    }\n};\n```\n\n# **Python / Python3 Solution:**\n```\n# Time Complexity: O(n)\n# Space Complexity: O(1)\nclass Solution(object):\n    def checkIfPangram(self, sentence):\n        # Add all elements to set and check if there are 26 different letters...\n        return len(set(sentence)) == 26\n```\n            \n# **JavaScript Solution:**\n```\n// Time Complexity: O(n)\n// Space Complexity: O(1)\nvar checkIfPangram = function(sentence) {\n    return new Set(sentence.split("")).size === 26\n};\n```\n**I am working hard for you guys...\nPlease upvote if you found any help with this code...**
['C', 'Python', 'Java', 'Python3', 'JavaScript']