# Solving Leetcode Interviews in Seconds with AI: Buddy Strings


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "859" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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: s = "ab", goal = "ab" Output: false Explanation: The only letters you can swap are s[0] = 'a' and s[1] = 'b', which results in "ba" != goal.  Example 3:  Input: s = "aa", goal = "aa" Output: true Explanation: You can swap s[0] = 'a' and s[1] = 'a' to get "aa", which is equal to goal.    Constraints:  1 <= s.length, goal.length <= 2 * 104 s and goal consist of lowercase letters.  

	# Explanation
	Here's the breakdown:

*   **High-Level Approach:**
    *   First, check if the lengths of the two strings are equal. If not, they cannot be made equal by swapping.
    *   Next, identify the indices where the characters in `s` and `goal` differ. If there are exactly two such indices, check if swapping the characters at those indices in `s` makes it equal to `goal`.
    *   If the strings are identical, check if there are duplicate characters in `s`. If there are, then swapping them won't change the string and will make the two strings the same. Otherwise, return false.

*   **Complexity:**
    *   Runtime: O(n), where n is the length of the strings. Storage: O(1) - constant extra space.

	
	# Code
	```python
	def buddyStrings(s: str, goal: str) -> bool:
    if len(s) != len(goal):
        return False

    if s == goal:
        seen = set()
        for char in s:
            if char in seen:
                return True
            seen.add(char)
        return False

    diff_indices = []
    for i in range(len(s)):
        if s[i] != goal[i]:
            diff_indices.append(i)

    if len(diff_indices) == 2:
        i, j = diff_indices[0], diff_indices[1]
        return s[i] == goal[j] and s[j] == goal[i]

    return False
	```