Welcome to Subscribe On Youtube

1832. Check if the Sentence Is Pangram

Description

A pangram is a sentence where every letter of the English alphabet appears at least once.

Given a string sentence containing only lowercase English letters, return true if sentence is a pangram, or false otherwise.

 

Example 1:

Input: sentence = "thequickbrownfoxjumpsoverthelazydog"
Output: true
Explanation: sentence contains at least one of every letter of the English alphabet.

Example 2:

Input: sentence = "leetcode"
Output: false

 

Constraints:

• 1 <= sentence.length <= 1000
• sentence consists of lowercase English letters.

Solutions

Solution 1: Array or Hash Table

Traverse the string sentence, use an array or hash table to record the letters that have appeared, and finally check whether there are $26$ letters in the array or hash table.

The time complexity is $O(n)$, and the space complexity is $O(C)$. Where $n$ is the length of the string sentence, and $C$ is the size of the character set. In this problem, $C = 26$.

Solution 2: Bit Manipulation

We can also use an integer $mask$ to record the letters that have appeared, where the $i$-th bit of $mask$ indicates whether the $i$-th letter has appeared.

Finally, check whether there are $26$ $1$s in the binary representation of $mask$, that is, check whether $mask$ is equal to $2^{26} - 1$. If so, return true, otherwise return false.

The time complexity is $O(n)$, where $n$ is the length of the string sentence. The space complexity is $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public boolean checkIfPangram(String sentence) {
          boolean[] vis = new boolean[26];
          for (int i = 0; i < sentence.length(); ++i) {
              vis[sentence.charAt(i) - 'a'] = true;
          }
          for (boolean v : vis) {
              if (!v) {
                  return false;
              }
          }
          return true;
      }
  }
  

• class Solution {
  public:
      bool checkIfPangram(string sentence) {
          int vis[26] = {0};
          for (char& c : sentence) vis[c - 'a'] = 1;
          for (int& v : vis)
              if (!v) return false;
          return true;
      }
  };
  

• class Solution:
      def checkIfPangram(self, sentence: str) -> bool:
          return len(set(sentence)) == 26
  
  

• func checkIfPangram(sentence string) bool {
  	vis := [26]bool{}
  	for _, c := range sentence {
  		vis[c-'a'] = true
  	}
  	for _, v := range vis {
  		if !v {
  			return false
  		}
  	}
  	return true
  }
  

• function checkIfPangram(sentence: string): boolean {
      const vis = new Array(26).fill(false);
      for (const c of sentence) {
          vis[c.charCodeAt(0) - 'a'.charCodeAt(0)] = true;
      }
      return vis.every(v => v);
  }
  
  

• impl Solution {
      pub fn check_if_pangram(sentence: String) -> bool {
          let mut vis = [false; 26];
          for c in sentence.as_bytes() {
              vis[(*c - b'a') as usize] = true;
          }
          vis.iter().all(|v| *v)
      }
  }
  
  

All Problems

All Solutions