What is the time complexity of Check Whether Two Strings are Almost Equivalent?

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

What data structures are used to solve Check Whether Two Strings are Almost Equivalent?

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

What are the key insights for solving Check Whether Two Strings are Almost Equivalent?

Insight 1: Using a frequency array (or hash map) is efficient to count character occurrences in strings. This avoids iterating through the string repeatedly for each character. Insight 2: The problem can be solved by calculating the difference in frequencies of each character between the two strings. No need to store the frequencies of each string separately, only their difference.

Which companies ask LeetCode 2068. Check Whether Two Strings are Almost Equivalent in interviews?

This problem is frequently asked at J.P. Morgan, Salesforce, Vanguard.

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Check Whether Two Strings are Almost Equivalent - LeetCode 2068 Solution

Check Whether Two Strings are Almost Equivalent - Complete Solution Guide

Check Whether Two Strings are Almost Equivalent is LeetCode problem 2068, 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

Two strings word1 and word2 are considered almost equivalent if the differences between the frequencies of each letter from 'a' to 'z' between word1 and word2 is at most 3 . Given two strings word1 and word2 , each of length n , return true if word1 and word2 are almost equivalent , or false otherwise . The frequency of a letter x is the number of times it occurs in the string. Example 1: Input: word1 = "aaaa", word2 = "bccb" Output: false Explanation: There are 4 'a's in "aaaa" but 0 'a's in "b

Detailed Explanation

The problem asks you to determine if two strings, `word1` and `word2`, are 'almost equivalent'. Two strings are considered almost equivalent if the absolute difference in the frequency of each letter (a-z) between the two strings is at most 3. The input consists of two strings of equal length containing only lowercase English letters. The output is a boolean value: `true` if they are almost equivalent, and `false` otherwise.

Solution Approach

The solution uses a frequency array (or hash map in Python) to efficiently count the differences in character frequencies between the two input strings. It iterates through each string once, updating the frequency of each character accordingly. Then, it iterates through the frequency array to check if the absolute difference for any character exceeds 3. If it does, the strings are not almost equivalent, and `false` is returned; otherwise, `true` is returned.

Step-by-Step Algorithm

Step 1: Initialize a frequency array `freq` of size 26 (for a-z), with all elements set to 0. This array will store the difference in frequency of each character.
Step 2: Iterate through `word1`. For each character `c`, increment `freq[c - 'a']`.
Step 3: Iterate through `word2`. For each character `c`, decrement `freq[c - 'a']`.
Step 4: Iterate through the `freq` array. If the absolute value of any element `abs(freq[i])` is greater than 3, return `false`.
Step 5: If the loop in Step 4 completes without finding any differences greater than 3, return `true`.

Key Insights

Insight 1: Using a frequency array (or hash map) is efficient to count character occurrences in strings. This avoids iterating through the string repeatedly for each character.
Insight 2: The problem can be solved by calculating the difference in frequencies of each character between the two strings. No need to store the frequencies of each string separately, only their difference.
Insight 3: The code needs to handle cases where a letter appears in one string but not the other. The solution correctly handles this implicitly by initializing the frequency array to 0.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: J.P. Morgan, Salesforce, Vanguard.

Check Whether Two Strings are Almost Equivalent | LeetCode 2068

Easy

Hash Table String Counting

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean checkAlmostEquivalent(String word1, String word2) {
        int[] freq = new int[26];

        for (char c : word1.toCharArray()) {
            freq[c - 'a']++;
        }

        for (char c : word2.toCharArray()) {
            freq[c - 'a']--;
        }

        for (int i = 0; i < 26; i++) {
            if (Math.abs(freq[i]) > 3) {
                return false;
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 2068: Check Whether Two Strings are Almost Equivalent - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

J.P. MorganSalesforceVanguard

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:2068

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: Initialize a frequency array `freq` of size 26 (for a-z), with all elements set to 0. This array will store the difference in frequency of each character.
2
Step 2: Iterate through `word1`. For each character `c`, increment `freq[c - 'a']`.
3
Step 3: Iterate through `word2`. For each character `c`, decrement `freq[c - 'a']`.
4
Step 4: Iterate through the `freq` array. If the absolute value of any element `abs(freq[i])` is greater than 3, return `false`.
5
Step 5: If the loop in Step 4 completes without finding any differences greater than 3, return `true`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through each string once (O(n) for each string) and then iterates through the constant-sized frequency array (O(1)). Therefore, the overall time complexity is dominated by the string iterations, resulting in O(n). The space used by the frequency array is constant regardless of the input string length, resulting in O(1) space complexity.

📖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 two frequency maps

Create separate frequency maps for `word1` and `word2`, then iterate through the alphabet and compare the absolute difference of frequencies for each letter.

Time: O(n)Space: O(1)

Trade-offs: This approach is less efficient than using a single frequency array because it involves more memory allocation and iterations. The single array approach is more concise and memory-efficient.

✨

Key Insights

1Insight 1: Using a frequency array (or hash map) is efficient to count character occurrences in strings. This avoids iterating through the string repeatedly for each character.
2Insight 2: The problem can be solved by calculating the difference in frequencies of each character between the two strings. No need to store the frequencies of each string separately, only their difference.
3Insight 3: The code needs to handle cases where a letter appears in one string but not the other. The solution correctly handles this implicitly by initializing the frequency array to 0.

📚

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: One or both strings are empty. The solution correctly handles this because the loops will not execute, and the frequency array will remain initialized to zeros, resulting in a true return.
Edge case 2: A character appears in one string but not the other. This is correctly handled because the frequency array element will either be positive (if only in word1) or negative (if only in word2), and the absolute value check in step 4 will detect a difference.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to take the absolute difference of frequencies. The problem requires checking `abs(count) <= 3`, not just `count <=3`.

✗Common mistake 2: Inefficiently checking frequency differences using nested loops instead of using an array/hash map for efficient lookups. This would lead to an O(n^2) solution.

✅

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: Comparing DNA sequences for similarity. Minor variations (like SNPs – single nucleotide polymorphisms) can be considered 'almost equivalent' in biological contexts.

🏦Real-world scenario 2: Document comparison in plagiarism detection. Identifying documents that are similar but not identical, considering minor word or character changes.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

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

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: August 28, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 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 "Check Whether Two Strings are Almost Equivalent"?

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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