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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 <= 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, $C = 26$.

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

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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;
    }
}

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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;
    }
};

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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
}

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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);
}

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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)
    }
}

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bool checkIfPangram(char* sentence) {
    int vis[26] = {0};
    for (int i = 0; sentence[i]; i++) {
        vis[sentence[i] - 'a'] = 1;
    }
    for (int i = 0; i < 26; i++) {
        if (!vis[i]) {
            return 0;
        }
    }
    return 1;
}

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:
    def checkIfPangram(self, sentence: str) -> bool:
        mask = 0
        for c in sentence:
            mask |= 1 << (ord(c) - ord('a'))
        return mask == (1 << 26) - 1

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