2490. Circular Sentence

Description

A sentence is a list of words that are separated by a single space with no leading or trailing spaces.

For example, "Hello World", "HELLO", "hello world hello world" are all sentences.

Words consist of only uppercase and lowercase English letters. Uppercase and lowercase English letters are considered different.

A sentence is circular if:

The last character of each word in the sentence is equal to the first character of its next word.
The last character of the last word is equal to the first character of the first word.

For example, "leetcode exercises sound delightful", "eetcode", "leetcode eats soul" are all circular sentences. However, "Leetcode is cool", "happy Leetcode", "Leetcode" and "I like Leetcode" are not circular sentences.

Given a string sentence, return true if it is circular. Otherwise, return false.

Example 1:

Input: sentence = "leetcode exercises sound delightful"
Output: true
Explanation: The words in sentence are ["leetcode", "exercises", "sound", "delightful"].
- leetcode's last character is equal to exercises's first character.
- exercises's last character is equal to sound's first character.
- sound's last character is equal to delightful's first character.
- delightful's last character is equal to leetcode's first character.
The sentence is circular.

Example 2:

Input: sentence = "eetcode"
Output: true
Explanation: The words in sentence are ["eetcode"].
- eetcode's last character is equal to eetcode's first character.
The sentence is circular.

Example 3:

Input: sentence = "Leetcode is cool"
Output: false
Explanation: The words in sentence are ["Leetcode", "is", "cool"].
- Leetcode's last character is not equal to is's first character.
The sentence is not circular.

Constraints:

1 <= sentence.length <= 500
sentence consist of only lowercase and uppercase English letters and spaces.
The words in sentence are separated by a single space.
There are no leading or trailing spaces.

Solutions

Solution 1: Simulation

We split the string into words by spaces, then check whether the last character of each word is equal to the first character of the next word. If they are not equal, return false. Otherwise, return true after traversing all the words.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string.

Python3JavaC++GoTypeScriptRustJavaScript

class Solution:
    def isCircularSentence(self, sentence: str) -> bool:
        ss = sentence.split()
        n = len(ss)
        return all(s[-1] == ss[(i + 1) % n][0] for i, s in enumerate(ss))

class Solution {
    public boolean isCircularSentence(String sentence) {
        var ss = sentence.split(" ");
        int n = ss.length;
        for (int i = 0; i < n; ++i) {
            if (ss[i].charAt(ss[i].length() - 1) != ss[(i + 1) % n].charAt(0)) {
                return false;
            }
        }
        return true;
    }
}

class Solution {
public:
    bool isCircularSentence(string sentence) {
        auto ss = split(sentence, ' ');
        int n = ss.size();
        for (int i = 0; i < n; ++i) {
            if (ss[i].back() != ss[(i + 1) % n][0]) {
                return false;
            }
        }
        return true;
    }

    vector<string> split(string& s, char delim) {
        stringstream ss(s);
        string item;
        vector<string> res;
        while (getline(ss, item, delim)) {
            res.emplace_back(item);
        }
        return res;
    }
};

func isCircularSentence(sentence string) bool {
    ss := strings.Split(sentence, " ")
    n := len(ss)
    for i, s := range ss {
        if s[len(s)-1] != ss[(i+1)%n][0] {
            return false
        }
    }
    return true
}

function isCircularSentence(sentence: string): boolean {
    const ss = sentence.split(' ');
    const n = ss.length;
    for (let i = 0; i < n; ++i) {
        if (ss[i][ss[i].length - 1] !== ss[(i + 1) % n][0]) {
            return false;
        }
    }
    return true;
}

impl Solution {
    pub fn is_circular_sentence(sentence: String) -> bool {
        let ss: Vec<String> = sentence.split(' ').map(String::from).collect();
        let n = ss.len();
        for i in 0..n {
            if ss[i].as_bytes()[ss[i].len() - 1] != ss[(i + 1) % n].as_bytes()[0] {
                return false;
            }
        }
        return true;
    }
}

/**
 * @param {string} sentence
 * @return {boolean}
 */
var isCircularSentence = function (sentence) {
    const ss = sentence.split(' ');
    const n = ss.length;
    for (let i = 0; i < n; ++i) {
        if (ss[i][ss[i].length - 1] !== ss[(i + 1) % n][0]) {
            return false;
        }
    }
    return true;
};

Solution 2: Simulation (Space Optimization)

We can first check whether the first and last characters of the string are equal. If they are not equal, return false. Otherwise, traverse the string. If the current character is a space, check whether the previous character and the next character are equal. If they are not equal, return false. Otherwise, return true after traversing all the characters.

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