What is the time complexity of Check if Strings Can be Made Equal With Operations I?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Check if Strings Can be Made Equal With Operations I?

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

What are the key insights for solving Check if Strings Can be Made Equal With Operations I?

Insight 1: The allowed swaps only affect pairs of characters at even and odd indices separately. Characters at index 0 and 2 can be swapped amongst themselves, and characters at index 1 and 3 can be swapped amongst themselves. They cannot be mixed. Insight 2: The problem boils down to checking if the sorted pairs of even-indexed characters and odd-indexed characters are the same in both strings.

Which companies ask LeetCode 2839. Check if Strings Can be Made Equal With Operations I in interviews?

This problem is frequently asked at Citrix.

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Check if Strings Can be Made Equal With Operations I - LeetCode 2839 Solution

Check if Strings Can be Made Equal With Operations I - Complete Solution Guide

Check if Strings Can be Made Equal With Operations I is LeetCode problem 2839, 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

You are given two strings s1 and s2 , both of length 4 , consisting of lowercase English letters. You can apply the following operation on any of the two strings any number of times: Choose any two indices i and j such that j - i = 2 , then swap the two characters at those indices in the string. Return true if you can make the strings s1 and s2 equal, and false otherwise . Example 1: Input: s1 = "abcd", s2 = "cdab" Output: true Explanation: We can do the following operations on s1: - Choose the

Detailed Explanation

The problem asks whether two strings, `s1` and `s2`, both of length 4, can be made equal using a specific operation. The operation allows swapping two characters at indices `i` and `j` where `j - i = 2`. This means you can swap characters at indices (0, 2) and (1, 3). The goal is to determine if, after applying this operation any number of times to either string, `s1` and `s2` can become identical. The strings consist only of lowercase English letters.

Solution Approach

The solution efficiently separates even and odd indexed characters from both strings. It then sorts these character pairs independently for both strings and compares the sorted results. If the sorted even-indexed characters are identical and the sorted odd-indexed characters are identical, the strings can be made equal; otherwise, they cannot.

Step-by-Step Algorithm

Step 1: Extract even-indexed characters (indices 0 and 2) into separate strings (`s1_even`, `s2_even`) and odd-indexed characters (indices 1 and 3) into separate strings (`s1_odd`, `s2_odd`).
Step 2: Sort the characters within `s1_even`, `s1_odd`, `s2_even`, and `s2_odd`.
Step 3: Compare the sorted even-indexed character strings (`s1_even` and `s2_even`) and the sorted odd-indexed character strings (`s1_odd` and `s2_odd`).
Step 4: Return `true` if both comparisons result in equality; otherwise, return `false`.

Key Insights

Insight 1: The allowed swaps only affect pairs of characters at even and odd indices separately. Characters at index 0 and 2 can be swapped amongst themselves, and characters at index 1 and 3 can be swapped amongst themselves. They cannot be mixed.
Insight 2: The problem boils down to checking if the sorted pairs of even-indexed characters and odd-indexed characters are the same in both strings.
Insight 3: The strings are of fixed length 4, making the algorithm's time complexity constant. No complex data structures are needed.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: String.

Companies

Asked at: Citrix.

Check if Strings Can be Made Equal With Operations I | LeetCode 2839

Easy

String

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

Solution Code

Available in:

class Solution {
    public boolean canBeEqual(String s1, String s2) {
        String s1_even = "";
        String s1_odd = "";
        String s2_even = "";
        String s2_odd = "";

        for (int i = 0; i < 4; i++) {
            if (i % 2 == 0) {
                s1_even += s1.charAt(i);
                s2_even += s2.charAt(i);
            } else {
                s1_odd += s1.charAt(i);
                s2_odd += s2.charAt(i);
            }
        }

        char[] even1 = s1_even.toCharArray();
        char[] even2 = s2_even.toCharArray();
        char[] odd1 = s1_odd.toCharArray();
        char[] odd2 = s2_odd.toCharArray();

        java.util.Arrays.sort(even1);
        java.util.Arrays.sort(even2);
        java.util.Arrays.sort(odd1);
        java.util.Arrays.sort(odd2);

        return java.util.Arrays.equals(even1, even2) && java.util.Arrays.equals(odd1, odd2);
    }
}

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Problem Info

Description

Problem 2839: Check if Strings Can be Made Equal With Operations I - Detailed explanation and solution approaches.

Topics

String

Companies

Citrix

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2839

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

4 steps

1
Step 1: Extract even-indexed characters (indices 0 and 2) into separate strings (`s1_even`, `s2_even`) and odd-indexed characters (indices 1 and 3) into separate strings (`s1_odd`, `s2_odd`).
2
Step 2: Sort the characters within `s1_even`, `s1_odd`, `s2_even`, and `s2_odd`.
3
Step 3: Compare the sorted even-indexed character strings (`s1_even` and `s2_even`) and the sorted odd-indexed character strings (`s1_odd` and `s2_odd`).
4
Step 4: Return `true` if both comparisons result in equality; 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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The input size is fixed (strings of length 4). Sorting a constant number of characters takes constant time. All other operations (string manipulation, comparison) also take constant time. Therefore, both time and space complexity are constant.

📖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

Bit manipulation

Represent the characters as integers and use bitwise operations to check for equality after potential swaps. This may be slightly faster in some cases but is less readable and more complex to implement correctly for all characters.

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

Trade-offs: Slightly faster in some scenarios but less readable and potentially harder to debug. The current solution is preferred due to readability and maintainability.

✨

Key Insights

1Insight 1: The allowed swaps only affect pairs of characters at even and odd indices separately. Characters at index 0 and 2 can be swapped amongst themselves, and characters at index 1 and 3 can be swapped amongst themselves. They cannot be mixed.
2Insight 2: The problem boils down to checking if the sorted pairs of even-indexed characters and odd-indexed characters are the same in both strings.
3Insight 3: The strings are of fixed length 4, making the algorithm's time complexity constant. No complex data structures are needed.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Empty strings are not considered (problem constraints specify length 4).
Edge case 2: Strings with repeated characters. The solution correctly handles this since sorting will still result in the same ordered sequence for identical characters.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the character pairs before comparison. The allowed swap operation implies that the order of characters within even and odd indices does not matter, only their values.

✗Common mistake 2: Assuming that the allowed swaps can mix even and odd indexed characters. This is incorrect; the swaps are only allowed between pairs of even or pairs of odd indices.

✅

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 or manipulation where certain rearrangements are allowed within specific data subsets (e.g., rearranging bytes in a specific protocol).

🏦Real-world scenario 2: Simplified representation of more complex puzzle-solving or state-space search algorithms where movement restrictions impose constraints similar to the allowed swaps in this problem.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: October 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

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 Strings Can be Made Equal With Operations I"?

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

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