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This approach leverages a set data structure to track the unique characters. The idea is to iterate through the sentence and add each letter to a set. Finally, we check if the size of the set is 26, which indicates that all letters of the alphabet are present.
Time Complexity: O(n), where n is the length of the sentence.
Space Complexity: O(1), since we use a fixed-size array of 26.
1def checkIfPangram(sentence: str) -> bool:
2 return len(set(sentence)) == 26
3The Python solution takes advantage of the set data type to filter out unique characters in the sentence. If the set length equals 26 after processing the sentence, it indicates a pangram.
This method uses a boolean array of size 26 to directly map each alphabet character to an index. By iterating over each character in the sentence, we update its corresponding index in the array to true. The sentence is confirmed as a pangram if all indices in the boolean array are true.
Time Complexity: O(n), with n denoting sentence length; we check each letter.
Space Complexity: O(1), constant size boolean array.
1
This approach relies on a 26-element boolean array, where each index corresponds to a letter of the alphabet (0 for 'a', 1 for 'b', etc.). As we parse the sentence, each letter sets its spot to true. A final loop checks that all positions in the array are true, affirming the pangram status.