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Formatted question description: https://leetcode.ca/all/2490.html

# 2490. Circular Sentence

• Difficulty: Easy.
• Related Topics: .
• Similar Questions: Defuse the Bomb.

## Problem

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 a word is equal to the first character of the 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.

## Solution (Java, C++, Python)

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

• class Solution {
public:
bool isCircularSentence(string sentence) {
if (sentence[0] != sentence[sentence.size() - 1]) return false;
istringstream is(sentence);
vector<string> ss;
string s;
while (is >> s) ss.emplace_back(s);
for (int i = 1; i < ss.size(); ++i) {
if (ss[i][0] != ss[i - 1][ss[i - 1].size() - 1]) {
return false;
}
}
return true;
}
};

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


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

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


• var isCircularSentence = function (sentence) {
const words = sentence.split(' ');
const post = words[0].charCodeAt(0);
let prev = words[0].charCodeAt(words[0].length - 1);
const n = words.length;
for (let i = 1; i < n; i++) {
let cur = words[i];
if (cur.charCodeAt(0) !== prev) {
return false;
}
prev = cur.charCodeAt(cur.length - 1);
}
return post === prev;
};


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



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).