O(1) Java Solution Easy to Understand.
\n# Complexity\n- Time complexity:\nO(1)\n\n- Space complexity:\nO(N)\n\n# Code\n\nclass Solution {\n    public boolean checkIfPangram(String sentence) {\n\n   
\n# Complexity\n- Time complexity:\nO(1)\n\n- Space complexity:\nO(N)\n\n# Code\n```\nclass Solution {\n    public boolean checkIfPangram(String sentence) {\n\n    Character[] charArray = \n    sentence.chars().mapToObj(c -> (char)c).toArray(Character[]::new); //don\'t get confused about it because i also copied it from stack overflow , this function basically create the Character[] type array from the string.\n\n    List<Character>check=Arrays.asList(charArray);//now we convert the Character type of array into List<Character>\n\n    Set<Character>st=new HashSet<>(check);//now we store that list into set\n    \n    return st.size()==26;//as we all know that the set contains unique character so it is confirmed that if the set size is 26 which means that set contains all the characters from a to z.\n\n    }\n}\n```
['Hash Table', 'String', 'Java']