The challenge

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 `` 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.

The solution in Java code

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class Solution {
    public boolean buddyStrings(String A, String B) {
        if (A.length() != B.length() || A.isEmpty()) return false;
        
        if (!A.equals(B)) {
            int count = 0;
            char first = 'a', second = 'a';
            for (int i = 0; i < A.length(); i++) {
                if (A.charAt(i) == B.charAt(i)) continue;
                if (count >= 2) return false;
                if (count == 0) {
                    first = A.charAt(i);
                    second = B.charAt(i);
                    count++;
                    continue;
                }
                if (A.charAt(i) != second || B.charAt(i) != first) return false;
                count++;
            }
            return count == 2;
        }

        if (A.length() > 26) return true;
        int [] fre = new int[26];
        for (char c : A.toCharArray()) {
            if (fre[c-'a'] == 1) return true;
            fre[c-'a']++;
        }
        return false;
    }
}