Java || using Frequency array || using Set || using HashMap
1. using Frequency Array\n\nclass Solution {\n    public boolean checkIfPangram(String sentence) {\n        int [] freq=new int[26];\n        for(int i=0;i<sent
**1. using Frequency Array**\n```\nclass Solution {\n    public boolean checkIfPangram(String sentence) {\n        int [] freq=new int[26];\n        for(int i=0;i<sentence.length();i++){\n            freq[sentence.charAt(i)-\'a\']++;\n        }\n        for(int i=0;i<26;i++){\n            if(freq[i]==0){\n                return false;\n            }\n        }\n        return true;\n    }\n}\n\n```\n\n**2. using HashSet**\n\n```\nclass Solution{\n    public boolean checkIfPangram(String sentence){\n        Set<Character> set=new HashSet<>();\n        for(int i=0;i<sentence.length();i++){\n            set.add(sentence.charAt(i));\n        }\n        if(set.size()==26){\n            return true;\n        }\n        return false;\n    }\n}\n```\n\n**3. using HashMap**\n\n```\nclass Solution{\n    public boolean checkIfPangram(String sentence){\n        HashMap<Character,Integer> hm=new HashMap<>();\n        for(int i=0;i<sentence.length();i++){\n            hm.put(sentence.charAt(i),0);\n        }\n        if(hm.size()==26){\n            return true;\n        }\n        return false;\n    }\n}\n```
['Ordered Set', 'Java']