Simple Solution and easy to understand || [ Java] ✅✅
Intuition\n##### Store every charecter at string in HashMap and check at all english charecters if it exist or not\n\n# Approach\n### Step 1 : \n### Looping at 
# Intuition\n##### Store every charecter at string in HashMap and check at all english charecters if it exist or not\n\n# Approach\n### Step 1 : \n### Looping at all sentence charecter and store it at HashMap we used HashMap as it give us time O(1) when check the existence of the cherecter \n```\nmap.containsKey(chIndex)\n```\n### Step 2 : Looping at all 26 english charecters and check the existence of the charecter in the HashMap by the unique ASCII Code of each charecter . We Know small english charecter start with 97 at ASCII Code system\n##### Finally : If you like the method of solution and explanation, do not hesitate to vote up to encourage me to solve more problems and solve them in better ways.\n# Complexity\n- Time complexity: \nO(n) where n is the length of alphabet english characters\n- Space complexity:\nO(n) where n is the length of charecters at the Given String\n# Code\n```\nclass Solution {\n    public boolean checkIfPangram(String sentence) {\n        HashMap<Integer, Character> map = new HashMap<Integer, Character>();\n\n        for (int i = 0; i < sentence.length(); i++) {\n            int chIndex = (int) (sentence.charAt(i));\n            if (!map.containsKey(chIndex)) {\n                map.put( chIndex, sentence.charAt(i));\n            }\n        }\n        for (int i = 0; i < 26; i++) {\n       \n            if (!map.containsKey(97 + i)) {\n                return false;\n            }\n        }\n        return true;\n    }\n}\n\n```\n![upvote.png](https://assets.leetcode.com/users/images/1ca5fcca-7469-433d-b56a-dc1e1687d264_1713298771.5938907.png)\n
['Java']