Given two strings A and B of lowercase letters, return true if you can swap two letters in A so the result is equal to B, 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 A[i] and A[j]. For example, swapping at indices 0 and 2 in "abcd" results in "cbad".
Example 1:
Input: A = "ab", B = "ba"
Output: true
Explanation: You can swap A[0] = 'a' and A[1] = 'b' to get "ba", which is equal to B.
Example 2:
Input: A = "ab", B = "ab"
Output: false
Explanation: The only letters you can swap are A[0] = 'a' and A[1] = 'b', which results in "ba" != B.
Example 3:
Input: A = "aa", B = "aa"
Output: true
Explanation: You can swap A[0] = 'a' and A[1] = 'a' to get "aa", which is equal to B.
Example 4:
Input: A = "aaaaaaabc", B = "aaaaaaacb"
Output: true
Example 5:
Input: A = "", B = "aa"
Output: false
Constraints:
0 <= A.length <= 20000
0 <= B.length <= 20000
A and B consist of lowercase letters.
class Solution:
    def buddyStrings(self, A: str, B: str) -> bool:
        
        if A == B and len(set(A)) < len(B): return True
        
        if len(A) != len(B): return False
        
        diff_A = []
        diff_B = []
        
        for i in range(len(A)):
            if A[i] != B[i]:
                diff_A.append(A[i])
                diff_B.append(B[i])
            
        return len(diff_A) == len(diff_B) == 2 and diff_A[::-1] == diff_B
class Solution:
    def buddyStrings(self, A: str, B: str) -> bool:
        
        if A == B and len(set(A)) < len(B): return True
        
        if len(A) != len(B): return False
        
        diff_A = []
        diff_B = []
        
        for i in range(len(A)):
            if A[i] != B[i]:
                diff_A.append(A[i])
                diff_B.append(B[i])
            
        return len(diff_A) == len(diff_B) == 2 and diff_A[::-1] == diff_B