What is the time complexity of Determine if Two Strings Are Close?

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

What data structures are used to solve Determine if Two Strings Are Close?

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

What are the key insights for solving Determine if Two Strings Are Close?

Insight 1: The exact order of characters doesn't matter due to the swap operation. Thus, sorting is a viable approach to compare character frequencies. Insight 2: The presence of characters *does* matter. If a character exists in one string but not the other, they cannot be close.

Which companies ask LeetCode 1657. Determine if Two Strings Are Close in interviews?

This problem is frequently asked at Postmates.

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Determine if Two Strings Are Close - LeetCode 1657 Solution

Determine if Two Strings Are Close - Complete Solution Guide

Determine if Two Strings Are Close is LeetCode problem 1657, 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

Two strings are considered close if you can attain one from the other using the following operations: Operation 1: Swap any two existing characters. For example, a b cd e -> a e cd b Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character. For example, aa c abb -> bb c baa (all a 's turn into b 's, and all b 's turn into a 's) You can use the operations on either string as many times as necessary. Given two strin

Detailed Explanation

The problem asks us to determine if two strings are "close". Two strings are considered close if one can be transformed into the other using two operations: 1) swapping any two characters, and 2) transforming every occurrence of one character into another character and vice-versa (e.g., 'a' becomes 'b' and 'b' becomes 'a' throughout the string). The length of the strings is up to 10^5, and they contain only lowercase English letters.

Solution Approach

The solution checks if two strings are close based on the following criteria: 1) They have the same length. 2) They have the same set of characters. 3) The sorted frequency counts of the characters are the same. If all these conditions are met, the strings are considered close. The core idea is that the operations allowed don't change the set of characters present, nor do they change the multiset of frequencies of the characters. They only allow rearranging and renaming.

Step-by-Step Algorithm

Step 1: Check if the lengths of the two strings are equal. If not, return `false`.
Step 2: Count the frequency of each character in both strings using a `Counter` (in Python) or a `HashMap` (in Java/C++).
Step 3: Check if the sets of characters in both strings are equal (i.e., they contain the same characters). If not, return `false`.
Step 4: Extract the frequency counts from the frequency maps into arrays/lists.
Step 5: Sort the frequency count arrays/lists.
Step 6: Compare the sorted frequency count arrays/lists. If they are equal, return `true`; otherwise, return `false`.

Key Insights

Insight 1: The exact order of characters doesn't matter due to the swap operation. Thus, sorting is a viable approach to compare character frequencies.
Insight 2: The presence of characters *does* matter. If a character exists in one string but not the other, they cannot be close.
Insight 3: The *frequency counts* of the characters are important. The 'transform' operation preserves the set of frequency counts. For example, 'aabbc' can become 'ccaab' in frequency terms.

Complexity Analysis

Time Complexity: O(nlogn)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Sorting, Counting.

Companies

Asked at: Postmates.

Determine if Two Strings Are Close | LeetCode 1657

Medium

Hash Table String Sorting Counting

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

class Solution {
    public boolean closeStrings(String word1, String word2) {
        if (word1.length() != word2.length()) {
            return false;
        }

        Map<Character, Integer> count1 = new HashMap<>();
        Map<Character, Integer> count2 = new HashMap<>();

        for (char c : word1.toCharArray()) {
            count1.put(c, count1.getOrDefault(c, 0) + 1);
        }

        for (char c : word2.toCharArray()) {
            count2.put(c, count2.getOrDefault(c, 0) + 1);
        }

        Set<Character> keys1 = count1.keySet();
        Set<Character> keys2 = count2.keySet();

        if (!keys1.equals(keys2)) {
            return false;
        }

        int[] values1 = count1.values().stream().mapToInt(Integer::intValue).toArray();
        int[] values2 = count2.values().stream().mapToInt(Integer::intValue).toArray();

        Arrays.sort(values1);
        Arrays.sort(values2);

        if (!Arrays.equals(values1, values2)) {
            return false;
        }

        return true;
    }
}

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Problem Info

Description

Problem 1657: Determine if Two Strings Are Close - Detailed explanation and solution approaches.

Topics

Hash Table String Sorting Counting

Companies

Postmates

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1657

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: Check if the lengths of the two strings are equal. If not, return `false`.
2
Step 2: Count the frequency of each character in both strings using a `Counter` (in Python) or a `HashMap` (in Java/C++).
3
Step 3: Check if the sets of characters in both strings are equal (i.e., they contain the same characters). If not, return `false`.
4
Step 4: Extract the frequency counts from the frequency maps into arrays/lists.
5
Step 5: Sort the frequency count arrays/lists.
6
Step 6: Compare the sorted frequency count arrays/lists. If they are equal, return `true`; otherwise, return `false`.

💡

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(nlogn)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the sorting operation. The space complexity is dominated by the character frequency counts, which are bounded by the number of possible characters (26).

📖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

Frequency Counting without Sorting

Instead of sorting the frequencies, we could use an array of size n (where n is the length of the strings) to count the occurrences of each frequency. If two strings are close, they should have the same distribution of frequencies. This can improve the time complexity, but the space complexity might increase.

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

Trade-offs: This approach avoids sorting, potentially improving performance if the length is very long. But it requires more space. This approach will be a better choice for large `n`.

✨

Key Insights

1Insight 1: The exact order of characters doesn't matter due to the swap operation. Thus, sorting is a viable approach to compare character frequencies.
2Insight 2: The presence of characters *does* matter. If a character exists in one string but not the other, they cannot be close.
3Insight 3: The *frequency counts* of the characters are important. The 'transform' operation preserves the set of frequency counts. For example, 'aabbc' can become 'ccaab' in frequency terms.

📚

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!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty strings. The code implicitly handles this as the length check will immediately fail if one string is empty and the other is not.
Edge case 2: Strings of different lengths. The code explicitly checks for this condition and returns `false`.
Edge case 3: Strings with different character sets. The code explicitly checks if the sets of characters are different and returns `false`.
Edge case 4: Strings with same character sets but different frequency counts. The code checks this by comparing the sorted frequency counts.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if the strings have the same set of characters. Two strings can have the same frequency counts but different characters, which are not close.

✗Common mistake 2: Incorrectly assuming that the strings need to be exactly the same after just swapping characters. The transform operation is crucial.

✗Common mistake 3: Not handling the edge case where string lengths differ. This is a crucial first step.

✅

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 compression. Identifying patterns in character frequencies can help design efficient compression algorithms.

🏦Real-world scenario 2: Anagram detection. While this problem isn't directly about anagrams, the character counting and comparison techniques are useful in anagram solvers and related word games.

🎮Real-world scenario 3: Bioinformatics. Analyzing DNA or protein sequences involves similar techniques for understanding the frequency and distribution of different bases or amino acids.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: July 8, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Determine if Two Strings Are Close"?

For this medium-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.

🤖 Need help with this problem?