What is the time complexity of Word Pattern?

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

What data structures are used to solve Word Pattern?

This problem uses Hash Table, String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Word Pattern?

Insight 1: Using two hash maps (or dictionaries) is crucial for efficiently checking the bijective mapping. One map tracks the character-to-word mapping, and the other tracks the word-to-character mapping. This ensures that both conditions (each character maps to exactly one word, and each word maps to exactly one character) are satisfied. Insight 2: The solution needs to handle the case where the number of words in `s` doesn't match the number of characters in `pattern`. This is a quick pre-check to avoid unnecessary processing.

Which companies ask LeetCode 290. Word Pattern in interviews?

This problem is frequently asked at Dropbox, Google, Zoho.

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Word Pattern - LeetCode 290 Solution

Word Pattern - Complete Solution Guide

Word Pattern is LeetCode problem 290, 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 pattern and a string s , find if s follows the same pattern. Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s . Specifically: Each letter in pattern maps to exactly one unique word in s . Each unique word in s maps to exactly one letter in pattern . No two letters map to the same word, and no two words map to the same letter. Example 1: Input: pattern = "abba", s = "dog cat cat dog" Output: true Explanation: The bijectio

Detailed Explanation

The problem asks whether a given string `s` follows the same pattern as a given pattern string `pattern`. 'Following the pattern' means there's a one-to-one mapping between each character in `pattern` and a non-empty word in `s`. This mapping must be consistent throughout the strings; no character can map to multiple words, and no word can map to multiple characters. The words in `s` are separated by single spaces. The input `pattern` contains only lowercase English letters, and `s` contains only lowercase English letters and spaces. The output is a boolean value: `true` if `s` follows the pattern, and `false` otherwise.

Solution Approach

The solution uses two hash maps (or dictionaries) to track the mappings between characters in `pattern` and words in `s`. It iterates through the input strings simultaneously. For each character and corresponding word, it checks if the mapping already exists. If it does, it verifies the consistency of the mapping. If not, it adds the new mapping to both maps. If any inconsistency is found (a character maps to multiple words or vice-versa), the function immediately returns `false`. Otherwise, after iterating through all characters and words, it returns `true`. The C solution uses arrays instead of hash tables due to C's limitations.

Step-by-Step Algorithm

Step 1: Split the input string `s` into an array of words using `s.split()`. Check if the lengths of the `pattern` and the word array are different. If they are, return `false`.
Step 2: Initialize two hash maps (or dictionaries), `p_map` and `w_map`, to store the character-to-word and word-to-character mappings, respectively.
Step 3: Iterate through both the `pattern` and the `words` array simultaneously using a loop. For each character `p` and word `w` at the same index:
Step 4: Check if `p` is already in `p_map`. If it is not, check if `w` is in `w_map`. If `w` is in `w_map`, there is a conflict (a word mapping to multiple characters), so return `false`. Otherwise, add the mapping (`p` to `w` and `w` to `p`) to both maps.
Step 5: If `p` is in `p_map`, check if the mapped word is equal to `w`. If not, there is a conflict (a character mapping to multiple words), so return `false`.
Step 6: After the loop completes, return `true` because all mappings are consistent.

Key Insights

Insight 1: Using two hash maps (or dictionaries) is crucial for efficiently checking the bijective mapping. One map tracks the character-to-word mapping, and the other tracks the word-to-character mapping. This ensures that both conditions (each character maps to exactly one word, and each word maps to exactly one character) are satisfied.
Insight 2: The solution needs to handle the case where the number of words in `s` doesn't match the number of characters in `pattern`. This is a quick pre-check to avoid unnecessary processing.
Insight 3: Empty words are not allowed, so the `s.split()` method conveniently handles this case, as it only generates non-empty words.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String.

Companies

Asked at: Dropbox, Google, Zoho.

Word Pattern | LeetCode 290

Easy

Hash Table String

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public boolean wordPattern(String pattern, String s) {
        String[] words = s.split(" ");
        if (words.length != pattern.length()) return false;
        Map<Character, String> map = new HashMap<>();
        Set<String> set = new HashSet<>();
        for (int i = 0; i < pattern.length(); i++) {
            char c = pattern.charAt(i);
            if (map.containsKey(c)) {
                if (!map.get(c).equals(words[i])) return false;
            } else {
                if (set.contains(words[i])) return false;
                map.put(c, words[i]);
                set.add(words[i]);
            }
        }
        return true;
    }
}

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Problem Info

Description

Problem 290: Word Pattern - Detailed explanation and solution approaches.

Topics

Hash Table String

Companies

DropboxGoogleZoho

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:290

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

6 steps

1
Step 1: Split the input string `s` into an array of words using `s.split()`. Check if the lengths of the `pattern` and the word array are different. If they are, return `false`.
2
Step 2: Initialize two hash maps (or dictionaries), `p_map` and `w_map`, to store the character-to-word and word-to-character mappings, respectively.
3
Step 3: Iterate through both the `pattern` and the `words` array simultaneously using a loop. For each character `p` and word `w` at the same index:
4
Step 4: Check if `p` is already in `p_map`. If it is not, check if `w` is in `w_map`. If `w` is in `w_map`, there is a conflict (a word mapping to multiple characters), so return `false`. Otherwise, add the mapping (`p` to `w` and `w` to `p`) to both maps.
5
Step 5: If `p` is in `p_map`, check if the mapped word is equal to `w`. If not, there is a conflict (a character mapping to multiple words), so return `false`.
6
Step 6: After the loop completes, return `true` because all mappings are consistent.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant factor in both time and space complexity is the iteration through the input string and the size of the hash maps (which are proportional to the input size). The hash map lookups are considered O(1) on average.

📖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

Using a single hash map

It's possible to use only one hash map by storing a combined key (character + word) or word + character as a string, and checking for collisions within this single map. This approach slightly simplifies the code but might slightly impact performance due to string concatenation overhead.

Time: O(n)Space: O(n)

Trade-offs: While slightly less code, the single hash map approach might be less readable and slightly less efficient due to string operations.

✨

Key Insights

1Insight 1: Using two hash maps (or dictionaries) is crucial for efficiently checking the bijective mapping. One map tracks the character-to-word mapping, and the other tracks the word-to-character mapping. This ensures that both conditions (each character maps to exactly one word, and each word maps to exactly one character) are satisfied.
2Insight 2: The solution needs to handle the case where the number of words in `s` doesn't match the number of characters in `pattern`. This is a quick pre-check to avoid unnecessary processing.
3Insight 3: Empty words are not allowed, so the `s.split()` method conveniently handles this case, as it only generates non-empty words.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

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

Edge Cases

Edge case 1: Empty input pattern or input string. The solution implicitly handles this because the length check in Step 1 will return `false`.
Edge case 2: Patterns with repeated characters mapping to different words or the same word multiple times. The solution explicitly handles this case in Steps 4 and 5 using the hash maps to detect inconsistencies.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check for the case where the number of words in `s` is not equal to the number of characters in `pattern`. This can lead to index out-of-bounds errors or incorrect results.

✗Common mistake 2: Not using both hash maps (or equivalent structure) to ensure the one-to-one mapping in both directions (from character to word and from word to character). This would lead to an incomplete check and inaccurate results.

✅

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: Data validation and parsing, where you need to verify if a data stream conforms to a predefined pattern (e.g., validating log file formats).

🏦Real-world scenario 2: Natural language processing (NLP) tasks, such as checking if a sentence follows a grammatical pattern or structure, or verifying code style consistency against coding conventions.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

👨‍💻 From Experience

This is one of the first problems I solved when learning Hash Table. 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: September 19, 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

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Word Pattern"?

For this easy-level Hash Table 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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