What is the time complexity of Check if One String Swap Can Make Strings Equal?

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

What data structures are used to solve Check if One String Swap Can Make Strings Equal?

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 if One String Swap Can Make Strings Equal?

Insight 1: The problem only requires checking for at most one swap. This means we only need to consider the differences between the two strings. If there are more than two differences, it's impossible to make them equal with one swap. Insight 2: Focusing on the indices where the characters differ is crucial. The solution efficiently tracks these differences without unnecessary iterations.

Which companies ask LeetCode 1790. Check if One String Swap Can Make Strings Equal in interviews?

This problem is frequently asked at DoorDash.

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Check if One String Swap Can Make Strings Equal - LeetCode 1790 Solution

Check if One String Swap Can Make Strings Equal - Complete Solution Guide

Check if One String Swap Can Make Strings Equal is LeetCode problem 1790, 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 of equal length. A string swap is an operation where you choose two indices in a string (not necessarily different) and swap the characters at these indices. Return true if it is possible to make both strings equal by performing at most one string swap on exactly one of the strings. Otherwise, return false . Example 1: Input: s1 = "bank", s2 = "kanb" Output: true Explanation: For example, swap the first character with the last character of s2 to make "bank". E

Detailed Explanation

The problem asks whether two strings of equal length can be made identical by performing at most one swap of characters within *one* of the strings. The swap can involve swapping two different characters or the same character with itself (in essence, doing nothing). The input consists of two strings, `s1` and `s2`, and the output is a boolean value: `true` if a single swap (or no swap) can make the strings equal, and `false` otherwise.

Solution Approach

The solution iterates through both strings simultaneously, comparing characters at corresponding positions. It maintains a counter for the number of differences and records the indices where differences occur. If the number of differing indices is 0 (strings are equal), or 2 (a single swap can potentially fix), it proceeds to check if a single swap is sufficient. If the number of differing indices is greater than 2, it's impossible to achieve equality with a single swap.

Step-by-Step Algorithm

Step 1: Check if the strings are already equal. If so, return `true`.
Step 2: Iterate through the strings, counting the number of differing character positions and storing the indices of these positions.
Step 3: If the difference count is greater than 2, return `false`.
Step 4: If the difference count is 2, check if swapping the characters at the stored indices makes the strings equal. If so, return `true`; otherwise, return `false`.
Step 5: If the difference count is 0, return `true` (strings are already equal).

Key Insights

Insight 1: The problem only requires checking for at most one swap. This means we only need to consider the differences between the two strings. If there are more than two differences, it's impossible to make them equal with one swap.
Insight 2: Focusing on the indices where the characters differ is crucial. The solution efficiently tracks these differences without unnecessary iterations.
Insight 3: Handling the edge case where the strings are already equal is important for efficiency. The early return improves the overall time complexity.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: DoorDash.

Check if One String Swap Can Make Strings Equal | LeetCode 1790

Easy

Hash Table String Counting

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

Solution Code

Available in:

class Solution {
    public boolean areAlmostEqual(String s1, String s2) {
        if (s1.equals(s2)) {
            return true;
        }
        int first = -1, second = -1;
        int diffCount = 0;
        for (int i = 0; i < s1.length(); i++) {
            if (s1.charAt(i) != s2.charAt(i)) {
                diffCount++;
                if (first == -1) {
                    first = i;
                } else if (second == -1) {
                    second = i;
                }
            }
        }
        if (diffCount == 2 && s1.charAt(first) == s2.charAt(second) && s1.charAt(second) == s2.charAt(first)) {
            return true;
        }
        return false;
    }
}

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Problem Info

Description

Problem 1790: Check if One String Swap Can Make Strings Equal - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

DoorDash

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1790

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: Check if the strings are already equal. If so, return `true`.
2
Step 2: Iterate through the strings, counting the number of differing character positions and storing the indices of these positions.
3
Step 3: If the difference count is greater than 2, return `false`.
4
Step 4: If the difference count is 2, check if swapping the characters at the stored indices makes the strings equal. If so, return `true`; otherwise, return `false`.
5
Step 5: If the difference count is 0, return `true` (strings are already equal).

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The solution iterates through the strings once, using only a constant amount of extra space to store the indices of differences and the difference count. The space used is independent of the input string length.

📖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 Hash Maps

We could use a hash map to count the frequency of each character in both strings. If the character frequencies are identical except for at most two positions, a single swap may be possible. This approach would involve more overhead.

Time: O(n)Space: O(1) - since there are only 26 possible characters in the alphabet

Trade-offs: This approach is less efficient than the provided solution because of the extra overhead of creating and managing the hash maps. The current approach is more direct and optimal.

✨

Key Insights

1Insight 1: The problem only requires checking for at most one swap. This means we only need to consider the differences between the two strings. If there are more than two differences, it's impossible to make them equal with one swap.
2Insight 2: Focusing on the indices where the characters differ is crucial. The solution efficiently tracks these differences without unnecessary iterations.
3Insight 3: Handling the edge case where the strings are already equal is important for efficiency. The early return improves the overall time complexity.

📚

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: s1 and s2 are already equal. The solution correctly handles this by immediately returning `true`.
Edge case 2: s1 and s2 differ in more than two positions. The solution correctly returns `false` in this scenario.
Edge case 3: The two different characters are the same (e.g., s1 = 'ab', s2 = 'ba'). The solution correctly handles this, recognizing that a swap would equate the strings.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the strings are already identical. This leads to incorrect results.

✗Common mistake 2: Not properly handling the case where a single swap is needed but not sufficient to make the strings identical. This often leads to a false positive.

✗Common mistake 3: Not considering that the two differing characters must be swapped for the strings to be equal. Checking only that there are two differing characters is insufficient.

✅

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: Checking if two data entries are almost identical, allowing for a single typographical error.

🏦Real-world scenario 2: Network communication: Detecting and correcting minor errors in transmitted data packets (though more sophisticated algorithms would be used for larger-scale error correction).

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Hash Table. It's a great foundation for more complex problems.

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 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 if One String Swap Can Make Strings Equal"?

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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