Java

/**

 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:

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

 */
public class Buddy_Strings {
    class Solution {
        public boolean buddyStrings(String A, String B) {
            if (A == null || B == null) {
                return false;
            }
            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;
            }

            // abcd, badc => false
            int p1 = -1;
            int p2 = -1;

            for (int i = 0; i < A.length(); i++) {
                if (A.charAt(i) != B.charAt(i)) {
                    // record this position
                    if (p1 < 0) {
                        p1 = i;
                    } else if (p2 < 0) {
                        p2 = i;
                    } else {
                        return false; // found the 2 positions, for abcd and badc, also need to check the rest strings

                    }
                }
            }

            if (p1 < 0 | p2 < 0) {
                return false;
            }

            return A.charAt(p1) == B.charAt(p2) && A.charAt(p2) == B.charAt(p1) && A.substring(p2 + 1).equals(B.substring(p2 + 1));
        }
    }
}

// OJ: https://leetcode.com/problems/buddy-strings/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool buddyStrings(string A, string B) {
        if (A.size() != B.size()) return false;
        int cnts[26] = {0}, first = -1, second = -1;
        bool hasDup = false;
        for (int i = 0; i < A.size(); ++i) {
            if (++cnts[A[i] - 'a'] == 2) hasDup = true;
            if (A[i] == B[i]) continue;
            if (first == -1) first = i;
            else if (second == -1) second = i;
            else return false;
        }
        return (first != -1 && second != -1 && A[first] == B[second] && A[second] == B[first])
            || (first == -1 && second == -1 && hasDup);
    }
};

class Solution:
    def buddyStrings(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: bool
        """
        if len(A) != len(B):
            return False
        diff = 0
        idxs = []
        for i, a in enumerate(A):
            if B[i] != a:
                diff += 1
                idxs.append(i)
        counter = dict()
        if diff == 0:
            for a in A:
                if a in counter and counter[a]:
                    return True
                else:
                    counter[a] = True
        if diff != 2:
            return False
        return A[idxs[0]] == B[idxs[1]] and A[idxs[1]] == B[idxs[0]]

Java

class Solution {
    public boolean buddyStrings(String A, String B) {
        if (A == null || B == null || A.length() != B.length())
            return false;
        if (A.equals(B)) {
            char[] array = A.toCharArray();
            Arrays.sort(array);
            int length = array.length;
            for (int i = 1; i < length; i++) {
                if (array[i] == array[i - 1])
                    return true;
            }
            return false;
        }
        List<Integer> differentIndices = new ArrayList<Integer>();
        int length = A.length();
        for (int i = 0; i < length; i++) {
            char cA = A.charAt(i), cB = B.charAt(i);
            if (cA != cB)
                differentIndices.add(i);
        }
        if (differentIndices.size() != 2)
            return false;
        int index1 = differentIndices.get(0), index2 = differentIndices.get(1);
        return A.charAt(index1) == B.charAt(index2) && A.charAt(index2) == B.charAt(index1);
    }
}

// OJ: https://leetcode.com/problems/buddy-strings/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool buddyStrings(string A, string B) {
        if (A.size() != B.size()) return false;
        int cnts[26] = {0}, first = -1, second = -1;
        bool hasDup = false;
        for (int i = 0; i < A.size(); ++i) {
            if (++cnts[A[i] - 'a'] == 2) hasDup = true;
            if (A[i] == B[i]) continue;
            if (first == -1) first = i;
            else if (second == -1) second = i;
            else return false;
        }
        return (first != -1 && second != -1 && A[first] == B[second] && A[second] == B[first])
            || (first == -1 && second == -1 && hasDup);
    }
};

class Solution:
    def buddyStrings(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: bool
        """
        if len(A) != len(B):
            return False
        diff = 0
        idxs = []
        for i, a in enumerate(A):
            if B[i] != a:
                diff += 1
                idxs.append(i)
        counter = dict()
        if diff == 0:
            for a in A:
                if a in counter and counter[a]:
                    return True
                else:
                    counter[a] = True
        if diff != 2:
            return False
        return A[idxs[0]] == B[idxs[1]] and A[idxs[1]] == B[idxs[0]]

859 - Buddy Strings

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