Given a string check if it is Pangram or not. A pangram is a sentence containing every letter in the English Alphabet. Lowercase and Uppercase are considered the same. Examples:
Input : str = 'The quick brown fox jumps over
the lazy dog'
Output : Yes
// Contains all the characters from āaā to āzā
Input : str='The quick brown fox jumps over the dog'
Output : No
// Doesnāt contains all the characters from āaā
// to āzā, as ālā, āzā, āyā are missing
This problem has existing solution please refer Pangram Checking link. We will solve this in Python using Set() data structure and List() comprehension. Approach is very simple,
- Convert complete sentence in lower case using lower() method of string data type in python.
- Now pass this sentence into Set(str) so that we could have list of all unique characters present in given string.
- Now separate out list of all alphabets (a-z), if length of list is 26 that means all characters are present and sentence is Pangram otherwise not.
Implementation:
Python3
# function to check pangramĀ
def pangram(input):Ā Ā Ā Ā Ā Ā Ā Ā Ā # convert input string into lower caseĀ Ā Ā Ā input = input.lower()Ā Ā Ā Ā Ā Ā Ā Ā Ā # convert input string into Set() so that we willĀ Ā Ā Ā # list of all unique characters present in sentenceĀ Ā Ā Ā input = set(input)Ā
Ā Ā Ā Ā # separate out all alphabetsĀ Ā Ā Ā # ord(ch) returns ascii value of characterĀ Ā Ā Ā alpha = [ ch for ch in input if ord(ch) in range(ord('a'), ord('z')+1)]Ā
Ā Ā Ā Ā if len(alpha) == 26:Ā Ā Ā Ā Ā Ā Ā Ā return 'true'Ā Ā Ā Ā else:Ā Ā Ā Ā Ā Ā Ā Ā return 'false'Ā
# Driver programif __name__ == "__main__":Ā Ā Ā Ā input = 'The quick brown fox jumps over the lazy dog'Ā Ā Ā Ā print (pangram(input)) |
Java
import java.util.*;Ā
public class PangramChecker {Ā Ā Ā Ā public static void main(String[] args) {Ā Ā Ā Ā Ā Ā Ā Ā // Define the input stringĀ Ā Ā Ā Ā Ā Ā Ā String input = "The quick brown fox jumps over the lazy dog";Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Call the pangram function and print the resultĀ Ā Ā Ā Ā Ā Ā Ā System.out.println(pangram(input));Ā Ā Ā Ā }Ā
Ā Ā Ā Ā public static String pangram(String input) {Ā Ā Ā Ā Ā Ā Ā Ā // Convert the input string to lowercaseĀ Ā Ā Ā Ā Ā Ā Ā input = input.toLowerCase();Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Create a set to store unique characters in the input stringĀ Ā Ā Ā Ā Ā Ā Ā Set<Character> inputSet = new HashSet<>();Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Loop through each character in the input stringĀ Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < input.length(); i++) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā char ch = input.charAt(i);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Check if the character is an alphabetĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (ch >= 'a' && ch <= 'z') {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Add the character to the input setĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā inputSet.add(ch);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Convert the input set to a list and sort itĀ Ā Ā Ā Ā Ā Ā Ā List<Character> alpha = new ArrayList<>(inputSet);Ā Ā Ā Ā Ā Ā Ā Ā Collections.sort(alpha);Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Create a StringBuilder to concatenate the sorted characters into a stringĀ Ā Ā Ā Ā Ā Ā Ā StringBuilder sb = new StringBuilder();Ā Ā Ā Ā Ā Ā Ā Ā for (char ch : alpha) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā sb.append(ch);Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā String uniqueChars = sb.toString();Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Check if the length of the resulting string is 26 (i.e., contains all alphabets)Ā Ā Ā Ā Ā Ā Ā Ā if (uniqueChars.length() == 26) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return "true";Ā Ā Ā Ā Ā Ā Ā Ā } else {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return "false";Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā }} |
Output
true
Time complexity:Ā
- Lowercasing the input string takes O(n) time, where n is the length of the string.
- Converting the input string to a set takes O(n) time.
- The list comprehension that separates out all alphabets takes O(n) time.Ā
- The len() function to count the number of alphabets takes O(1) time.Ā
- The overall time complexity of the function is therefore O(n).
Auxiliary space:
- Creating the set to store unique characters in the input string requires O(n) space, where n is the length of the string.Ā
- Creating the list of alphabets requires O(k) space, where k is the number of alphabets in the string (which is at most 26).Ā
- The overall auxiliary space complexity of the function is therefore O(n).
