# 859. Buddy Strings

## 1. Question

Given two strings s and goal, return true if you can swap two letters in s so the result is equal to goal, 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 s[i] and s[j].

• For example, swapping at indices 0 and 2 in "abcd" results in "cbad".

## 2. Examples

Example 1:

Input: s = "ab", goal = "ba"
Output: true
Explanation: You can swap s[0] = 'a' and s[1] = 'b' to get "ba", which is equal to goal.


Example 2:

Input: s = "ab", goal = "ab"
Output: false
Explanation: The only letters you can swap are s[0] = 'a' and s[1] = 'b', which results in "ba" != goal.


Example 3:

Input: s = "aa", goal = "aa"
Output: true
Explanation: You can swap s[0] = 'a' and s[1] = 'a' to get "aa", which is equal to goal.


Example 4:

Input: s = "aaaaaaabc", goal = "aaaaaaacb"
Output: true


## 3. Constraints

• 1 <= s.length, goal.length <= 2 * 104
• s and goal consist of lowercase letters.

## 5. Solutions

class Solution {
public boolean buddyStrings(String s, String goal) {
if (s.length() != goal.length()) {
return false;
}

int[] arr1 = new int[26];
int[] arr2 = new int[26];
int diff = 0;
boolean flag = false;

for(int i = 0; i < s.length(); i++) {
int a = s.charAt(i) - 'a';
int b = goal.charAt(i) - 'a';

arr1[a]++;
arr2[b]++;

// 标记同一位置上字符不同的数量
if(a != b) {
diff++;
}
}

for(int i = 0; i < arr1.length; i++) {
// 如果所含字符数量本身不同，那么不可能构成亲密字符串
if (arr1[i] != arr2[i]) {
return false;
}
// 如果所有字符的数量相等，但有两个及以上相同的字符串，则也可构成亲密字符串
if (arr1[i] >= 2) {
flag = true;
}
}

return diff == 2 || (diff == 0 && flag);
}
}