JAVA  Solution || Beats 100% || Easy Approach
Intuition\n1. Length Check: It first checks if the sentence length is less than 26. If so, it cannot be a pangram as it wouldn\'t have enough characters to incl
# Intuition\n1. Length Check: It first checks if the sentence length is less than 26. If so, it cannot be a pangram as it wouldn\'t have enough characters to include all letters.\nLetter Iteration: It then iterates through all lowercase letters from \'a\' to \'z\'.\n2. Letter Search: For each letter, it checks if it exists in the sentence using the indexOf method. If the letter is not found, it returns false as the sentence is missing that letter.\n3. Pangram Determination: If the loop completes without finding a missing letter, it means the sentence contains all letters and therefore is a pangram.\n \nThis solution is efficient as it avoids unnecessary iterations if a missing letter is found early on. The indexOf method is generally optimized for string searching, making it a suitable choice for this task.\n\n# Approach\n1. check if (the length is less than 26) then return false\n2. iterate from a to z \n3. check that the given string contains all the elements from a to z using indexOf() method in string\n\n# Complexity\n- Time complexity:O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n    public boolean checkIfPangram(String sentence) {\n        if(sentence.length()<26) return false;\n        for(char ch=\'a\';ch<=\'z\';ch++){\n            if(sentence.indexOf(ch)<0){\n                return false;\n            }\n         }\n        return true;\n    }\n}\n```
['Java']