# 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 $O(1)$.

• 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)
}
}