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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 <= 1000sentenceconsists 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 $O(1)$.
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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) } }