How do you solve Buddy Strings on LeetCode?

Buddy Strings (LeetCode #859) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Swapping requires exactly two differing characters to match the goal.

What is the brute force solution for Buddy Strings?

The brute force approach for Buddy Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of indices in the string `s` to see if swapping the characters at those indices results in the string `goal`. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Buddy Strings?

Buddy Strings uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Buddy Strings?

Always start by checking edge cases, like string length. Explain your thought process clearly while coding. Consider both the brute force and optimal solutions during interviews.

What are common mistakes when solving Buddy Strings?

Not checking for string length equality first. Overlooking cases where no swaps are needed but duplicates exist.

#859

Buddy Strings

Easy

Hash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach involves counting the differences between the two strings and checking specific conditions to determine if a valid swap can be made. This is more efficient than checking all pairs.

⚙️

Algorithm

6 steps

1Step 1: If the lengths of `s` and `goal` are not equal, return false.
2Step 2: Initialize a list to store the indices where `s` and `goal` differ.
3Step 3: Iterate through the strings and collect indices of differing characters.
4Step 4: If there are exactly two differing indices, check if swapping those characters results in equality.
5Step 5: If there are no differing indices and there are duplicate characters in `s`, return true (indicating a swap of identical characters).
6Step 6: Otherwise, return false.

solution.py9 lines

1def buddyStrings(s, goal):
2    if len(s) != len(goal): return False
3    diff = []
4    for i in range(len(s)):
5        if s[i] != goal[i]:
6            diff.append(i)
7    if len(diff) == 2:
8        return s[diff[0]] == goal[diff[1]] and s[diff[1]] == goal[diff[0]]
9    return len(diff) == 0 and len(set(s)) < len(s)

ℹ

Complexity note: The time complexity is O(n) because we only need to iterate through the strings once to find differences. The space complexity is O(n) due to the storage of differing indices.

1Swapping requires exactly two differing characters to match the goal.
2Identical characters can be swapped if they exist in the string.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.