What is the time complexity of Buddy Strings?

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

What data structures are used to solve Buddy Strings?

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

What are the key insights for solving Buddy Strings?

Insight 1: The lengths of the two strings must be equal for a swap to potentially make them identical. Insight 2: If the strings are equal, we need to check if there are duplicate characters in `s`. If there is at least one duplicate, swapping those identical characters will result in the same string.

Which companies ask LeetCode 859. Buddy Strings in interviews?

This problem is frequently asked at DoorDash.

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Buddy Strings - LeetCode 859 Solution

Buddy Strings - Complete Solution Guide

Buddy Strings is LeetCode problem 859, 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

Given two strings s and goal , return true if you can swap two letters in s so the result is equal to goal , otherwise, return false . Swapping letters is defined as taking two indices i and j (0-indexed) such that i != j and swapping the characters at s[i] and s[j] . For example, swapping at indices 0 and 2 in "abcd" results in "cbad" . Example 1: Input: s = "ab", goal = "ba" Output: true Explanation: You can swap s[0] = 'a' and s[1] = 'b' to get "ba", which is equal to goal. Example 2: Input:

Detailed Explanation

The problem asks us to determine if two strings, `s` and `goal`, can become equal if we swap exactly two characters within string `s`. We must return `true` if such a swap is possible, and `false` otherwise. The constraints are that the strings contain only lowercase letters and have a maximum length of 2 * 10^4.

Solution Approach

The solution first checks if the lengths of the strings are different. If they are, it immediately returns `false`. If the strings are equal, it checks if string `s` contains duplicate characters. If so, it returns `true` (because swapping the duplicate characters will result in `s` itself). If the strings are not equal, it finds the indices where `s` and `goal` differ. It checks if there are exactly two such indices. If there are, it swaps the characters at those indices in `s` conceptually, and then returns `true` if the swapped string is equal to the `goal` string, otherwise `false`.

Step-by-Step Algorithm

Step 1: Check if the lengths of `s` and `goal` are different. If they are, return `false`.
Step 2: If `s` and `goal` are equal, check for duplicate characters in `s`. Iterate through the characters of `s` and use a `set` or frequency array to determine if any character appears more than once. If a duplicate is found, return `true`. If no duplicates are found after iterating through all characters, return `false`.
Step 3: If `s` and `goal` are not equal, find the indices where they differ. Store these indices in a list or array.
Step 4: Check the number of differing indices. If the number of differing indices is not equal to 2, return `false`.
Step 5: If there are exactly two differing indices, `i` and `j`, check if `s[i] == goal[j]` and `s[j] == goal[i]`. If this condition is true, return `true`; otherwise, return `false`.

Key Insights

Insight 1: The lengths of the two strings must be equal for a swap to potentially make them identical.
Insight 2: If the strings are equal, we need to check if there are duplicate characters in `s`. If there is at least one duplicate, swapping those identical characters will result in the same string.
Insight 3: If the strings are not equal, we need to find the indices where the characters differ. There should be exactly two such indices. If there are more than two differences, no single swap can fix it. Finally, we must verify that swapping the characters at these two indices in `s` results in `goal`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String.

Companies

Asked at: DoorDash.

Buddy Strings | LeetCode 859

Easy

Hash Table String

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

Solution Code

Available in:

class Solution {
    public boolean buddyStrings(String s, String goal) {
        if (s.length() != goal.length()) return false;
        if (s.equals(goal)) {
            int[] count = new int[26];
            for (char c : s.toCharArray()) {
                count[c - 'a']++;
                if (count[c - 'a'] > 1) return true;
            }
            return false;
        }
        int diff = 0;
        int i = 0;
        int j = 0;
        for (; i < s.length(); i++) {
            if (s.charAt(i) != goal.charAt(i)) {
                diff++;
                j = i;
            }
            if (diff > 2) return false;
        }
        
        if (diff == 2) {
            return s.charAt(i - 1) == goal.charAt(j) && s.charAt(j) == goal.charAt(i -1);
        }
        return false;

    }
}

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Problem Info

Description

Problem 859: Buddy Strings - Detailed explanation and solution approaches.

Topics

Hash Table String

Companies

DoorDash

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:859

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 lengths of `s` and `goal` are different. If they are, return `false`.
2
Step 2: If `s` and `goal` are equal, check for duplicate characters in `s`. Iterate through the characters of `s` and use a `set` or frequency array to determine if any character appears more than once. If a duplicate is found, return `true`. If no duplicates are found after iterating through all characters, return `false`.
3
Step 3: If `s` and `goal` are not equal, find the indices where they differ. Store these indices in a list or array.
4
Step 4: Check the number of differing indices. If the number of differing indices is not equal to 2, return `false`.
5
Step 5: If there are exactly two differing indices, `i` and `j`, check if `s[i] == goal[j]` and `s[j] == goal[i]`. If this condition is true, 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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because the loops iterate over each element of the input string(s) at most a constant number of times. The space complexity is dominated by the duplicate check for equal strings, and storing indices of differing characters, which is O(1) or O(n) depending on implementation detail.

📖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 and Difference Analysis

Instead of directly comparing the strings and finding differing indices, count the frequency of each character in both strings. If the frequencies are different, then it's impossible to make them equal with a single swap. Further analyze the differences in character frequencies to determine if a valid swap is possible.

Time: O(n)Space: O(1), using a fixed-size array of size 26 to count the frequency of lowercase English letters.

Trade-offs: This approach might be slightly more complex to implement initially, but it can offer a different perspective on the problem and might be beneficial if you need to analyze character frequencies for other purposes as well. It can be useful in situations where further character-based analyses are needed beyond simply identifying swappable strings.

✨

Key Insights

1Insight 1: The lengths of the two strings must be equal for a swap to potentially make them identical.
2Insight 2: If the strings are equal, we need to check if there are duplicate characters in `s`. If there is at least one duplicate, swapping those identical characters will result in the same string.
3Insight 3: If the strings are not equal, we need to find the indices where the characters differ. There should be exactly two such indices. If there are more than two differences, no single swap can fix it. Finally, we must verify that swapping the characters at these two indices in `s` results in `goal`.

📚

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: `s` and `goal` have different lengths. The solution correctly returns `false` immediately.
Edge case 2: `s` and `goal` are equal, but `s` has no duplicate characters (e.g., `s` = 'abcde', `goal` = 'abcde'). The solution correctly identifies that no swap will result in a different string, so it returns `false`.
Edge case 3: `s` and `goal` are equal, and `s` has duplicate characters (e.g., `s` = 'aabcde', `goal` = 'aabcde'). The solution correctly identifies that swapping the 'a's will result in the same string, so it returns `true`.
Edge case 4: `s` and `goal` are only different by one character or by more than 2 characters. The solution correctly returns `false`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where `s` and `goal` are equal correctly. It is crucial to check for duplicate characters in this case.

✗Common mistake 2: Incorrectly assuming that if two characters are different, any swap will work. It's important to check that swapping the differing characters in `s` actually results in `goal`.

✗Common mistake 3: Forgetting to check if the lengths of `s` and `goal` are equal before doing any other comparisons.

✅

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: Spell checkers can use a similar approach to suggest corrections for misspelled words. If a word is not found in the dictionary, the algorithm can determine if swapping two letters will result in a valid word in the dictionary.

🏦Real-world scenario 2: In bioinformatics, this algorithm could be adapted to identify similar DNA sequences. If two DNA sequences differ by only two base pairs, this algorithm could be used to determine if a simple mutation (swapping two base pairs) could have transformed one sequence into the other.

💡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: 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

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.

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 "Buddy Strings"?

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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