Circular Sentence - Practice Coding Problems

Circular Sentence - Problem

String Easy

Imagine you're creating a word game where sentences must form a perfect circle! A circular sentence is one where each word connects to the next like links in a chain, and the last word connects back to the first.

A sentence is a list of words separated by single spaces with no leading or trailing spaces. For example: "Hello World", "HELLO", or "hello world hello world".

Circular Rules:

The last character of each word must equal the first character of the next word
The last character of the final word must equal the first character of the first word
Case matters: 'A' ≠ 'a'

Examples:

✅ "leetcode exercises sound delightful" → e→e, s→s, d→d, l→l
✅ "eetcode" → single word where first 'e' = last 'e'
❌ "Leetcode is cool" → 'e' ≠ 'i', 's' ≠ 'c', 'l' ≠ 'L'

Given a string sentence, return true if it forms a circular sentence, false otherwise.

Input & Output

example_1.py — Circular Sentence

$ Input: "leetcode exercises sound delightful"

› Output: true

💡 Note: All words connect properly: leetcode ends with 'e' and exercises starts with 'e', exercises ends with 's' and sound starts with 's', sound ends with 'd' and delightful starts with 'd', and delightful ends with 'l' which matches the first character of leetcode.

example_2.py — Non-Circular Sentence

$ Input: "Leetcode is cool"

› Output: false

💡 Note: The connection fails at the first word boundary: Leetcode ends with 'e' but is starts with 'i'. Since 'e' ≠ 'i', the sentence is not circular.

example_3.py — Single Word

$ Input: "eetcode"

› Output: true

💡 Note: Single word case: we only need to check if the first character equals the last character. Here 'e' == 'e', so it's circular.

Constraints

1 ≤ sentence.length ≤ 500
sentence consists 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

Visualization

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Understanding the Visualization

Identify Links

Each word is a chain link with connection points at first and last characters

Check Connections

Verify each link connects properly to the next one

Complete Circle

Ensure the last link connects back to the first link

Key Takeaway

🎯 Key Insight: We can check circular connectivity in O(1) space by scanning for spaces and comparing adjacent characters, plus checking first/last characters match.

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal solution uses a one-pass string scan approach with O(1) space complexity. Instead of splitting the sentence into words, we scan through the string and check character connections at each space boundary, plus verify the first and last characters match.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Split and Check)	O(n)	O(n)	Split sentence into words array, then check each adjacent pair separately
One-Pass String Scan (Optimal)	O(n)	O(1)	Scan the string once, checking connections at space boundaries without splitting

Brute Force (Split and Check) — Algorithm Steps

Split the sentence into an array of words
Loop through each word from 0 to n-2
For each word i, check if word[i][-1] == word[i+1][0]
Separately check if last_word[-1] == first_word[0]
Return false if any connection fails

Visualization

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Step-by-Step Walkthrough

Split Sentence

Break sentence into individual words

Check Adjacent

Verify each word connects to the next

Check Loop

Verify last word connects to first

Code -

solution.c — C

bool isCircularSentence(char* sentence) {
    char* words[1000];
    int wordCount = 0;
    char* token = strtok(sentence, " ");
    
    // Split sentence into words
    while (token != NULL && wordCount < 1000) {
        words[wordCount++] = token;
        token = strtok(NULL, " ");
    }
    
    // Check connections between adjacent words
    for (int i = 0; i < wordCount - 1; i++) {
        int len1 = strlen(words[i]);
        if (words[i][len1 - 1] != words[i + 1][0]) {
            return false;
        }
    }
    
    // Check if last word connects to first word
    int lastLen = strlen(words[wordCount - 1]);
    if (words[wordCount - 1][lastLen - 1] != words[0][0]) {
        return false;
    }
    
    return true;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

O(n) to split string + O(w) to check w words = O(n) where n is string length

✓ Linear Growth

Space Complexity

O(n)

O(n) space to store the array of words split from the sentence

⚡ Linearithmic Space

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Medium Frequency

~8 min Avg. Time

845 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Split and Check) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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