LC 0859 [E] Buddy Strings - ALawliet/algorithms GitHub Wiki

1. if length differs or set of characters differ, return False directly
2. if A and B are equal, returns if we have at least 1 repetitive character in the list, so we just swap the duplicate
3. if two list have more than 2 indices with different characters (more than 2 differences), return False
4. In the end check if the swap can happen with the exactly 2 differences because this must be the 2 indices to swap

def buddyStrings(self, A, B):
        if len(A) != len(B): return False
        if A == B and len(set(A)) < len(A): return True
        dif = [(a, b) for a, b in zip(A, B) if a != b]
        return len(dif) == 2 and dif[0] == dif[1][::-1]

class Solution:
    def buddyStrings(self, A: str, B: str) -> bool:
        # check same length
        if len(A) != len(B): return False
        
        # if strings are equal - check if there is a double to swap
        if A == B:
            return True if len(A) - len(set(A)) >= 1 else False
        
        # count differences between strings
        diff = []
        for i in range(len(A)):
            if A[i] != B[i]:
                diff.append(i)
                if len(diff) > 2: return False
                
        # not exactly two differences
        if len(diff) != 2: return False
        
        # check if can be swapped
        if A[diff[0]] == B[diff[1]] and A[diff[1]] == B[diff[0]]:
            return True
        
        return False

class Solution:
    def buddyStrings(self, A, B):
        if len(A) != len(B) or set(A) != set(B):
            return False       
        
        if A == B:
            return len(A) - len(set(A)) >= 1
        
        indices = []
        counter = 0
        for i in range(len(A)):
            if A[i] != B[i]:
                counter += 1
                indices.append(i)       
            if counter > 2:
                return False       
            
        return A[indices[0]] == B[indices[1]] and A[indices[1]] == B[indices[0]]