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

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

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

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

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

Insight 1: The swap operation only allows rearrangement of characters within the even indices and within the odd indices independently. Insight 2: The problem can be solved by checking if the characters at even positions in `s1` can be rearranged to match the characters at even positions in `s2`, and similarly for the odd positions.

Which companies ask LeetCode 2840. Check if Strings Can be Made Equal With Operations II in interviews?

This problem is frequently asked at Citrix.

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Check if Strings Can be Made Equal With Operations II - LeetCode 2840 Solution

Q: How to solve Check if Strings Can be Made Equal With Operations II?

The solution approach is based on the observation that characters at even indices can only be swapped with other characters at even indices, and similarly for odd indices. Therefore, we can separate the characters at even and odd positions into two groups for each string. We then compare the frequency of each character within these groups. If the frequencies match for both even and odd groups, the strings can be made equal by the allowed operations, and we return `true`. Otherwise, we return `false`.

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

Check if Strings Can be Made Equal With Operations II is LeetCode problem 2840, 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

You are given two strings s1 and s2 , both of length n , 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 i < j and the difference j - i is even , 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 = "abcdba", s2 = "cabdab" Output: true Explanation: We can apply the follo

Detailed Explanation

The problem asks us to determine if two strings, `s1` and `s2`, of the same length `n`, can be made equal by performing a series of swap operations. The swap operations are constrained: we can only swap characters at indices `i` and `j` if `i < j` and the difference `j - i` is even. The strings consist of lowercase English letters, and we can perform these swaps any number of times. We need to return `true` if `s1` can be transformed into `s2`, and `false` otherwise.

Solution Approach

The solution approach is based on the observation that characters at even indices can only be swapped with other characters at even indices, and similarly for odd indices. Therefore, we can separate the characters at even and odd positions into two groups for each string. We then compare the frequency of each character within these groups. If the frequencies match for both even and odd groups, the strings can be made equal by the allowed operations, and we return `true`. Otherwise, we return `false`.

Step-by-Step Algorithm

Step 1: Initialize two arrays, `even_diff` and `odd_diff`, of size 26 (for each lowercase English letter) to store the frequency difference of characters at even and odd indices between `s1` and `s2`, respectively.
Step 2: Iterate through the strings `s1` and `s2` from index 0 to n-1.
Step 3: In each iteration, check if the current index `i` is even or odd.
Step 4: If `i` is even, increment the frequency count of the character `s1[i]` in `even_diff` and decrement the frequency count of the character `s2[i]` in `even_diff`. This effectively calculates `frequency(s1[i]) - frequency(s2[i])` for characters at even indices.
Step 5: If `i` is odd, increment the frequency count of the character `s1[i]` in `odd_diff` and decrement the frequency count of the character `s2[i]` in `odd_diff`. This effectively calculates `frequency(s1[i]) - frequency(s2[i])` for characters at odd indices.
Step 6: After iterating through all characters, check if all the elements in `even_diff` and `odd_diff` are zero. If any element is not zero, it means the frequencies of characters at the corresponding indices do not match, and we return `false`.
Step 7: If all elements in both `even_diff` and `odd_diff` are zero, return `true`.

Key Insights

Insight 1: The swap operation only allows rearrangement of characters within the even indices and within the odd indices independently.
Insight 2: The problem can be solved by checking if the characters at even positions in `s1` can be rearranged to match the characters at even positions in `s2`, and similarly for the odd positions.
Insight 3: Instead of performing actual swaps, we can count the frequency of each character in even and odd positions for both strings and compare the frequencies.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Sorting.

Companies

Asked at: Citrix.

Check if Strings Can be Made Equal With Operations II | LeetCode 2840

Medium

Hash Table String Sorting

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

Solution Code

Available in:

class Solution {
    public boolean checkStrings(String s1, String s2) {
        int n = s1.length();
        
        int[] even_diff = new int[26];
        int[] odd_diff = new int[26];
        
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0) {
                even_diff[s1.charAt(i) - 'a'] += 1;
                even_diff[s2.charAt(i) - 'a'] -= 1;
            } else {
                odd_diff[s1.charAt(i) - 'a'] += 1;
                odd_diff[s2.charAt(i) - 'a'] -= 1;
            }
        }
        
        for (int i = 0; i < 26; i++) {
            if (even_diff[i] != 0 || odd_diff[i] != 0) {
                return false;
            }
        }
        
        return true;
    }
}

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Problem Info

Description

Problem 2840: Check if Strings Can be Made Equal With Operations II - Detailed explanation and solution approaches.

Topics

Hash Table String Sorting

Companies

Citrix

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2840

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

7 steps

1
Step 1: Initialize two arrays, `even_diff` and `odd_diff`, of size 26 (for each lowercase English letter) to store the frequency difference of characters at even and odd indices between `s1` and `s2`, respectively.
2
Step 2: Iterate through the strings `s1` and `s2` from index 0 to n-1.
3
Step 3: In each iteration, check if the current index `i` is even or odd.
4
Step 4: If `i` is even, increment the frequency count of the character `s1[i]` in `even_diff` and decrement the frequency count of the character `s2[i]` in `even_diff`. This effectively calculates `frequency(s1[i]) - frequency(s2[i])` for characters at even indices.
5
Step 5: If `i` is odd, increment the frequency count of the character `s1[i]` in `odd_diff` and decrement the frequency count of the character `s2[i]` in `odd_diff`. This effectively calculates `frequency(s1[i]) - frequency(s2[i])` for characters at odd indices.
6
Step 6: After iterating through all characters, check if all the elements in `even_diff` and `odd_diff` are zero. If any element is not zero, it means the frequencies of characters at the corresponding indices do not match, and we return `false`.
7
Step 7: If all elements in both `even_diff` and `odd_diff` are zero, 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 time complexity is dominated by the single iteration through the input strings. The space complexity is constant because the space used by the `even_diff` and `odd_diff` arrays does not depend on the input size, but rather on the number of possible characters (lowercase English letters).

📖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

Sorting approach

Separate even and odd indexed characters into separate arrays. Sort each array. Then compare the sorted arrays. If they are equal, return true, otherwise false.

Time: O(n log n) due to the sorting operations, where n is the length of the string.Space: O(n) as we create new arrays to store characters at even and odd indices.

Trade-offs: This approach is less efficient than the main solution (frequency counting) in terms of time complexity but might be easier to understand for some.

✨

Key Insights

1Insight 1: The swap operation only allows rearrangement of characters within the even indices and within the odd indices independently.
2Insight 2: The problem can be solved by checking if the characters at even positions in `s1` can be rearranged to match the characters at even positions in `s2`, and similarly for the odd positions.
3Insight 3: Instead of performing actual swaps, we can count the frequency of each character in even and odd positions for both strings and compare the frequencies.

📚

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 solution works correctly for empty strings because the loop will not execute, and both `even_diff` and `odd_diff` will remain all zeros, resulting in a `true` return value.
Edge case 2: Strings of length 1. The solution correctly handles strings of length 1 by checking only the even position (index 0).
Edge case 3: Strings that are already equal. The solution correctly returns true in this case because the diff arrays will be all zeroes.
Edge case 4: Strings that are permutations of each other but cannot be made equal due to even/odd constraint. Example: s1="abc", s2="bac". Even chars in s1 are 'a' and 'c'. Even chars in s2 are 'b' and 'c'. s1 won't be convertible to s2 in this example. The solution correctly identifies this.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Ignoring the even/odd index constraint and trying to check if the strings are simply permutations of each other. This would lead to incorrect results as not all permutations are achievable with the given operation.

✗Common mistake 2: Using a hashmap instead of a fixed-size array to store the character frequencies. While it would work, it's less efficient due to the overhead of hashmap operations and unnecessary as we know the character set beforehand.

✗Common mistake 3: Not handling the edge case of empty strings or strings of length 1 correctly.

✅

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 where specific elements at even/odd positions must adhere to a particular order or arrangement.

🏦Real-world scenario 2: Encryption/Decryption algorithms that utilize similar swap operations based on index parity to encode or decode messages.

💡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

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

⚠️ Watch Out

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

👨‍💻 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: October 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 13

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

@AlgoMasterJul 6

❓

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.

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 "Check if Strings Can be Made Equal With Operations II"?

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.

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