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Formatted question description: https://leetcode.ca/all/1832.html

# 1832. Check if the Sentence Is Pangram

Easy

## 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.

## Solution

Use a set to store all the lowercase letters that exist in sentence. Return true if and only if the set’s size is 26.

• class Solution {
public boolean checkIfPangram(String sentence) {
Set<Character> set = new HashSet<Character>();
int length = sentence.length();
for (int i = 0; i < length; i++) {
char c = sentence.charAt(i);
if (c >= 'a' && c <= 'z')
}
return set.size() == 26;
}
}

############

class Solution {
public boolean checkIfPangram(String sentence) {
for (int i = 0; i < sentence.length(); ++i) {
mask |= 1 << (sentence.charAt(i) - 'a');
}
return mask == (1 << 26) - 1;
}
}

• // OJ: https://leetcode.com/problems/check-if-the-sentence-is-pangram/
// Time: O(N)
// Space: O(C) where C is the length of the character set
class Solution {
public:
bool checkIfPangram(string s) {
unordered_set<char> ss(begin(s), end(s));
return ss.size() == 26;
}
};

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

############

# 1832. Check if the Sentence Is Pangram
# https://leetcode.com/problems/check-if-the-sentence-is-pangram/

class Solution:
def checkIfPangram(self, sentence: str) -> bool:
c = collections.Counter(sentence)

return len(c) >= 26


• func checkIfPangram(sentence string) bool {
for _, c := range sentence {
}
}

• function checkIfPangram(sentence: string): boolean {
let mark = 0;
for (const c of sentence) {
mark |= 1 << (c.charCodeAt(0) - 'a'.charCodeAt(0));
}
return mark === (1 << 26) - 1;
}


• impl Solution {
pub fn check_if_pangram(sentence: String) -> bool {
let mut mark = 0;
for c in sentence.as_bytes() {
mark |= 1 << *c - b'a';
}
mark == (1 << 26) - 1
}
}