LeetCode 859 - Buddy Strings

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

https://leetcode.com/problems/buddy-strings/discuss/141780/Easy-Understood

If A.length() != B.length(): no possible swap

If A == B, we need swap two same characters. Check is duplicated char in A.

In other cases, we find index for A[i] != B[i]. There should be only 2 diffs and it's our one swap.

    public boolean buddyStrings(String A, String B) {
        if (A.length() != B.length()) return false;
        if (A.equals(B)) {
            Set<Character> s = new HashSet<Character>();
            for (char c : A.toCharArray()) s.add(c);
            return s.size() < A.length();
        }
        List<Integer> dif = new ArrayList<>();
        for (int i = 0; i < A.length(); ++i) if (A.charAt(i) != B.charAt(i)) dif.add(i);
        return dif.size() == 2 && A.charAt(dif.get(0)) == B.charAt(dif.get(1)) && A.charAt(dif.get(1)) == B.charAt(dif.get(0));
    }

  public boolean buddyStrings(String A, String B) {

    if (A.length() != B.length())

      return false;

    if (A.equals(B)) {

      int[] count = new int[26];

      for (int i = 0; i < A.length(); ++i)

        count[A.charAt(i) - 'a']++;

      for (int c : count)

        if (c > 1)

          return true;

      return false;

    } else {

      int first = -1, second = -1;

      for (int i = 0; i < A.length(); ++i) {

        if (A.charAt(i) != B.charAt(i)) {

          if (first == -1)

            first = i;

          else if (second == -1)

            second = i;

          else

            return false;

        }

      }

      return (second != -1 && A.charAt(first) == B.charAt(second) && A.charAt(second) == B.charAt(first));

    }

  }

https://leetcode.com/problems/buddy-strings/discuss/141794/Java-O(1)-space-O(n)-time

public boolean buddyStrings(String A, String B) {
    if (A == null || B == null || A.length() != B.length())  return false;
    int a = -1, b = -1, diff = 0;
    int[] count = new int[26];
    // check if able to switch with the same character.
    boolean canSwitch = false;
    for (int i = 0; i < A.length(); i++) {
      if (++count[A.charAt(i) - 'a'] >= 2)  canSwitch = true;
      if (A.charAt(i) != B.charAt(i)) {
        diff++;
        if (a == -1)  a = i;
        else if (b == -1)  b = i;
      }
    }
    return (diff == 0 && canSwitch) || (diff == 2 && A.charAt(a) == B.charAt(b) && A.charAt(b) == B.charAt(a));
  }