What is the time complexity of Find and Replace Pattern?

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

What data structures are used to solve Find and Replace Pattern?

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

What are the key insights for solving Find and Replace Pattern?

Insight 1: The core idea is to normalize both the pattern and the words. Normalization transforms each string into a sequence of integers representing the order of appearance of each unique character. For instance, 'abb' becomes [0, 1, 1] because 'a' is the first unique character (0), and 'b' is the second (1) which repeats. Insight 2: By normalizing, we effectively remove the specific characters themselves and focus solely on the structure of the string, represented by the order of unique character appearances.

Which companies ask LeetCode 890. Find and Replace Pattern in interviews?

This problem is frequently asked at Zomato.

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Find and Replace Pattern - LeetCode 890 Solution

Find and Replace Pattern - Complete Solution Guide

Find and Replace Pattern is LeetCode problem 890, a Medium 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 list of strings words and a string pattern , return a list of words[i] that match pattern . You may return the answer in any order . A word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x) , we get the desired word. Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter. Example 1: Input: words = ["abc","deq","m

Detailed Explanation

The problem asks us to find all words in a given list that match a specified pattern. A word matches the pattern if there exists a one-to-one mapping (bijection or permutation) between the letters of the pattern and the letters of the word. This means each letter in the pattern must map to a unique letter in the word, and vice-versa. The length of the pattern and each word is the same and consist of lowercase English letters. The goal is to return a list of words that satisfy this matching condition.

Solution Approach

The provided code implements a normalization technique to solve this problem. It normalizes both the input pattern and each word in the input list. It then compares the normalized representations. If the normalized representations match, the word is considered a match for the pattern and added to the result list. The normalization process involves creating a mapping from characters to indices based on their first appearance in the string. This allows for a direct comparison of string structures rather than character values.

Step-by-Step Algorithm

Step 1: Define a function `normalize(s)` that takes a string `s` as input.
Step 2: Inside `normalize(s)`, create an empty dictionary/map called `mapping` to store the character-to-index mapping.
Step 3: Iterate through the string `s` character by character.
Step 4: For each character `c`, check if it exists as a key in the `mapping`.
Step 5: If `c` is not in `mapping`, add it to `mapping` with a value equal to the current size of `mapping` (which is effectively the next available index).
Step 6: Append the index (value from `mapping`) corresponding to the character `c` to the normalized list.
Step 7: Return the normalized list.
Step 8: Normalize the input `pattern` using the `normalize` function and store the result as `pattern_key`.
Step 9: Initialize an empty list `result` to store the matching words.
Step 10: Iterate through the `words` list.
Step 11: For each `word` in `words`, normalize it using the `normalize` function.
Step 12: Compare the normalized representation of the `word` with `pattern_key`.
Step 13: If they are equal, append the `word` to the `result` list.
Step 14: Return the `result` list.

Key Insights

Insight 1: The core idea is to normalize both the pattern and the words. Normalization transforms each string into a sequence of integers representing the order of appearance of each unique character. For instance, 'abb' becomes [0, 1, 1] because 'a' is the first unique character (0), and 'b' is the second (1) which repeats.
Insight 2: By normalizing, we effectively remove the specific characters themselves and focus solely on the structure of the string, represented by the order of unique character appearances.
Insight 3: The one-to-one mapping requirement is crucial. If the mapping isn't a bijection (e.g., two different letters in the pattern map to the same letter in the word), the word doesn't match.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(m)

Topics

This problem involves: Array, Hash Table, String.

Companies

Asked at: Zomato.

Find and Replace Pattern | LeetCode 890

Medium

Array Hash Table String

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public List<String> findAndReplacePattern(String[] words, String pattern) {
        
        List<Integer> patternKey = normalize(pattern);
        
        List<String> result = new ArrayList<>();
        for (String word : words) {
            if (normalize(word).equals(patternKey)) {
                result.add(word);
            }
        }
        
        return result;
    }
    
    private List<Integer> normalize(String s) {
        Map<Character, Integer> mapping = new HashMap<>();
        List<Integer> normalized = new ArrayList<>();
        
        for (char c : s.toCharArray()) {
            if (!mapping.containsKey(c)) {
                mapping.put(c, mapping.size());
            }
            normalized.add(mapping.get(c));
        }
        
        return normalized;
    }
}

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Problem Info

Description

Problem 890: Find and Replace Pattern - Detailed explanation and solution approaches.

Topics

Array Hash Table String

Companies

Zomato

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:890

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

14 steps

1
Step 1: Define a function `normalize(s)` that takes a string `s` as input.
2
Step 2: Inside `normalize(s)`, create an empty dictionary/map called `mapping` to store the character-to-index mapping.
3
Step 3: Iterate through the string `s` character by character.
4
Step 4: For each character `c`, check if it exists as a key in the `mapping`.
5
Step 5: If `c` is not in `mapping`, add it to `mapping` with a value equal to the current size of `mapping` (which is effectively the next available index).
6
Step 6: Append the index (value from `mapping`) corresponding to the character `c` to the normalized list.
7
Step 7: Return the normalized list.
8
Step 8: Normalize the input `pattern` using the `normalize` function and store the result as `pattern_key`.
9
Step 9: Initialize an empty list `result` to store the matching words.
10
Step 10: Iterate through the `words` list.
11
Step 11: For each `word` in `words`, normalize it using the `normalize` function.
12
Step 12: Compare the normalized representation of the `word` with `pattern_key`.
13
Step 13: If they are equal, append the `word` to the `result` list.
14
Step 14: Return the `result` list.

💡

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(m)

📝

Why?

The time complexity is determined by the nested loops: iterating through the words and then normalizing each word. The space complexity is determined primarily by the storage needed for the normalized pattern and the mappings within the normalize function. The size of these mappings is bounded by the length of the pattern.

📖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

Two Hash Maps for Direct Mapping Check

Instead of normalizing, we could directly check if a word matches the pattern by using two hash maps. One hash map would map pattern characters to word characters, and the other would map word characters to pattern characters. We'd iterate through the strings and check if the mappings are consistent. This directly enforces the bijective constraint.

Time: O(n*m), where n is the number of words and m is the length of the pattern.Space: O(m), where m is the length of the pattern (in the worst case, where all characters in the pattern are unique).

Trade-offs: This approach may be slightly more intuitive for some, but the normalization approach might be more efficient in practice due to the simpler comparison after normalization.

✨

Key Insights

1Insight 1: The core idea is to normalize both the pattern and the words. Normalization transforms each string into a sequence of integers representing the order of appearance of each unique character. For instance, 'abb' becomes [0, 1, 1] because 'a' is the first unique character (0), and 'b' is the second (1) which repeats.
2Insight 2: By normalizing, we effectively remove the specific characters themselves and focus solely on the structure of the string, represented by the order of unique character appearances.
3Insight 3: The one-to-one mapping requirement is crucial. If the mapping isn't a bijection (e.g., two different letters in the pattern map to the same letter in the word), the word doesn't match.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

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 word list. The algorithm will correctly return an empty list in this case.
Edge case 2: Single-character pattern and word (e.g., pattern = 'a', word = 'b'). The algorithm will correctly identify matches in these cases because the normalized representations will be equal (e.g. [0] == [0]).
Edge case 3: Words with repeating letters that don't match the pattern's repetitions. The normalization function ensures that this is handled correctly because it captures the structure of repetitions, not just the isolated presence of letters.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not enforcing the bijective (one-to-one) mapping. For example, allowing 'abb' to map to 'mee' while also allowing 'a' to map to 'c', which violates the one-to-one constraint. The normalization ensures this doesn't happen.

✗Common mistake 2: Incorrectly handling edge cases where the word list is empty or contains words of different lengths. The problem statement specifies words should have the same length as the pattern, but real-world code should ideally handle discrepancies.

✅

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: Anonymization of data where the structure is important but the exact values are not. For example, if you want to maintain the relationship between events but don't want to reveal the actual event types, you can normalize the data.

🏦Real-world scenario 2: Detecting structural similarity in text analysis. For example, identifying if two code snippets or sentences have the same structure, even if the variable names or words are different. This is similar to code refactoring where variable names are changed.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: November 18, 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 13

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

@AlgoMasterJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Find and Replace Pattern"?

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

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