Buddy Strings, by LeetCode

Problem is here: https://leetcode.com/problems/buddy-strings/description/

Given two strings A and B of lowercase letters, return true if and only if we can swap two letters in A so that the result equals B.

Example 1:

Input: A = "ab", B = "ba"
Output: true

Example 2:

Input: A = "ab", B = "ab"
Output: false

Example 3:

Input: A = "aa", B = "aa"
Output: true

Example 4:

Input: A = "aaaaaaabc", B = "aaaaaaacb"
Output: true

Example 5:

Input: A = "", B = "aa"
Output: false

Note:

1. 0 <= A.length <= 20000
2. 0 <= B.length <= 20000
3. A and B consist only of lowercase letters.

Problem is deceivingly more obnoxious than it looks. Not hard in terms of "algorithms", but rather the different "special" cases need to be thought thru properly. Here they are:

• Conditions where you know there won't be a solution:
• String have different lengths, OR
• At least one of them is null or empty, OR
• They are equal with unique characters only, OR
• They differ by more or less than 2 characters
• If the above passes, the next step is to check:
• If they are equal but at least one character is duplicated, then return true (the dupe chars can be swapped)
• If they differ by two characters, make sure that swapping them makes the strings equal

The code is a little more concise than the explanation above (below).
The interesting thing to notice here is that sometimes in real life problems, they are actually more like this nature where you're dealing mostly with the "special cases", trying to cover all the logical bases, rather than dealing with complex algorithms. Hope you enjoy, cheers, Marcelo

public class Solution
{
 private bool OneDupe(string s)
 {
  Hashtable count = new Hashtable();
  for (int i = 0; i < s.Length; i++)
  {
   char c = s[i];
   if (!count.ContainsKey(c)) count.Add(c, 0);
   count[c] = (int)count[c] + 1;

   if ((int)count[c] > 1) return true;
  }
  return false;
 }
 public bool BuddyStrings(string A, string B)
 {
  if (String.IsNullOrEmpty(A) || String.IsNullOrEmpty(B)) return false;
  if (A.Length != B.Length) return false;

  if (A.Equals(B))
   return OneDupe(A);

  int indexDiff1 = -1;
  int indexDiff2 = -1;
  int countDiff = 0;
  for (int i = 0; i < A.Length; i++)
  {
   if (A[i] != B[i])
   {
    countDiff++;
    if (countDiff == 1)
     indexDiff1 = i;
    else if (countDiff == 2)
     indexDiff2 = i;
    else return false;
   }
  }

  return countDiff == 2 && A[indexDiff1] == B[indexDiff2] && A[indexDiff2] == B[indexDiff1];
 }
}