Easy and simple approach
Intuition\nThe goal is to check if the given string s contains all the letters of the alphabet at least once, making it a pangram. The provided code appears to 
# Intuition\nThe goal is to check if the given string `s` contains all the letters of the alphabet at least once, making it a pangram. The provided code appears to iterate through the alphabet and checks if each letter is present in the input string. The count of letters found is then compared to the total number of letters in the alphabet.\n\n# Approach\nThe approach taken in the code is to iterate through each letter of the alphabet and check if it exists in the input string. The count of letters found is then compared to the total number of letters in the alphabet. If the count is equal to or greater than 26 (the total number of letters in the English alphabet), the function returns true, indicating that the input string is a pangram.\n\n# Complexity\n- Time complexity: O(n * m), where n is the length of the input string and m is the length of the alphabet. The `contains` method takes O(n) time in the worst case, and it is called for each letter in the alphabet.\n- Space complexity: O(1), as the space used is constant (for the alphabet string and integer variables).\n\n# Code\n```java\nclass Solution {\n    public boolean checkIfPangram(String s) {\n        int count = 0;\n        String alphabet = "qwertyuiopasdfghjklzxcvbnm";\n\n        for (int i = 0; i < alphabet.length(); i++) {\n            if (s.contains(String.valueOf(alphabet.charAt(i)))) {\n                count++;\n            }\n        }\n        \n        return count >= 26;\n    }\n}\n```
['Java']