What is the time complexity of Check If a Word Occurs As a Prefix of Any Word in a Sentence?

The optimal solution has a time complexity of O(n*m).

What data structures are used to solve Check If a Word Occurs As a Prefix of Any Word in a Sentence?

This problem uses Two Pointers, String, String Matching. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Check If a Word Occurs As a Prefix of Any Word in a Sentence?

Insight 1: The problem can be efficiently solved by iterating through the words of the sentence and checking if each word starts with the `searchWord` using the `startsWith()` method (or equivalent). Insight 2: String splitting is a fundamental operation for processing text. Efficient string splitting is crucial for performance.

Which companies ask LeetCode 1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence in interviews?

This problem is frequently asked at Yelp.

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Check If a Word Occurs As a Prefix of Any Word in a Sentence - LeetCode 1455 Solution

Check If a Word Occurs As a Prefix of Any Word in a Sentence - Complete Solution Guide

Check If a Word Occurs As a Prefix of Any Word in a Sentence is LeetCode problem 1455, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a sentence that consists of some words separated by a single space , and a searchWord , check if searchWord is a prefix of any word in sentence . Return the index of the word in sentence ( 1-indexed ) where searchWord is a prefix of this word . If searchWord is a prefix of more than one word, return the index of the first word (minimum index) . If there is no such word return -1 . A prefix of a string s is any leading contiguous substring of s . Example 1: Input: sentence = "i love eating

Detailed Explanation

The problem asks you to find the first word in a sentence that starts with a given `searchWord`. The input is a sentence (string) and the `searchWord` (string). The output is the 1-indexed position of the word in the sentence if found; otherwise, -1. The sentence consists of words separated by single spaces. The `searchWord` must be a prefix (beginning substring) of a word, not necessarily the entire word itself. Constraints limit the length of the sentence and `searchWord` to relatively small values.

Solution Approach

The provided solutions all follow a similar approach: they first split the sentence into individual words using the space character as a delimiter. Then, they iterate through the words, checking if each word starts with the given `searchWord`. The index of the first word that satisfies this condition is returned (plus 1 for 1-based indexing). If no word matches, -1 is returned.

Step-by-Step Algorithm

Step 1: Split the input sentence string into an array of individual words using the space character (' ') as the delimiter.
Step 2: Iterate through the array of words using a loop (e.g., `for` loop).
Step 3: For each word, check if it starts with the `searchWord` using a `startsWith()` function or string comparison (e.g., `strncmp` in C).
Step 4: If a word starts with `searchWord`, return its index (plus 1 for 1-based indexing).
Step 5: If the loop completes without finding a matching word, return -1.

Key Insights

Insight 1: The problem can be efficiently solved by iterating through the words of the sentence and checking if each word starts with the `searchWord` using the `startsWith()` method (or equivalent).
Insight 2: String splitting is a fundamental operation for processing text. Efficient string splitting is crucial for performance.
Insight 3: Handling edge cases like an empty sentence or `searchWord` and a sentence without words matching the prefix are essential to prevent unexpected behavior or errors.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(n)

Topics

This problem involves: Two Pointers, String, String Matching.

Companies

Asked at: Yelp.

Check If a Word Occurs As a Prefix of Any Word in a Sentence | LeetCode 1455

Easy

Two Pointers String String Matching

Time: O(n*m)Space: O(n)

Solution Code

Available in:

class Solution {
    public int isPrefixOfWord(String sentence, String searchWord) {
        String[] words = sentence.split(" ");
        for (int i = 0; i < words.length; i++) {
            if (words[i].startsWith(searchWord)) {
                return i + 1;
            }
        }
        return -1;
    }
}

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Problem Info

Description

Problem 1455: Check If a Word Occurs As a Prefix of Any Word in a Sentence - Detailed explanation and solution approaches.

Topics

Two Pointers String String Matching

Companies

Yelp

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1455

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Split the input sentence string into an array of individual words using the space character (' ') as the delimiter.
2
Step 2: Iterate through the array of words using a loop (e.g., `for` loop).
3
Step 3: For each word, check if it starts with the `searchWord` using a `startsWith()` function or string comparison (e.g., `strncmp` in C).
4
Step 4: If a word starts with `searchWord`, return its index (plus 1 for 1-based indexing).
5
Step 5: If the loop completes without finding a matching word, return -1.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*m)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity comes from the nested iteration (implied in startsWith, which iterates through characters) The space complexity is dominated by the space needed to store the list of words generated after splitting the sentence.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Regular Expression

Use a regular expression to search for the pattern `\bsearchWord\w*\b` (word boundary, search word, any word characters, word boundary). This approach directly searches for the prefix within the words, avoiding explicit word splitting.

Time: O(n), where n is the length of the sentence. The complexity of regex matching is dependent on implementation, but usually linear in the length of the input string. Space: O(1) - Constant extra space is used.

Trade-offs: Regular expressions can be less readable but offer a potentially more concise solution. It is usually faster for larger sentences as it avoids splitting the string, but the overhead of regex compilation needs to be taken into account. For this specific problem, this isn't necessarily a significant advantage given the size constraints.

✨

Key Insights

1Insight 1: The problem can be efficiently solved by iterating through the words of the sentence and checking if each word starts with the `searchWord` using the `startsWith()` method (or equivalent).
2Insight 2: String splitting is a fundamental operation for processing text. Efficient string splitting is crucial for performance.
3Insight 3: Handling edge cases like an empty sentence or `searchWord` and a sentence without words matching the prefix are essential to prevent unexpected behavior or errors.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty sentence. The solutions correctly handle this by returning -1 because there are no words to check.
Edge case 2: Empty searchWord. While not explicitly mentioned in constraints, the solutions will return the index of the first word if the `searchWord` is empty as any word starts with an empty string.
Edge case 3: Search word not found. The solutions correctly return -1 in this case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect indexing. Forgetting to add 1 to the index to account for 1-based indexing in the problem statement.

✗Common mistake 2: Not handling the case where the `searchWord` is not found. Failing to return -1 when no word starts with the given prefix.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Autocomplete systems. Finding words that start with a given prefix in a dictionary for autocomplete suggestions.

🏦Real-world scenario 2: Search engines. Quickly identifying documents containing words starting with a given search term to improve search speed.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

👨‍💻 From Experience

This is one of the first problems I solved when learning Two Pointers. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 16, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Check If a Word Occurs As a Prefix of Any Word in a Sentence"?

For this easy-level Two Pointers problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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