2020-10-12

## 859. Buddy Strings

### Question:

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 `0` and `2` in `"abcd"` results in `"cbad"`.

#### Example 1:

``````Input: A = "ab", B = "ba"
Output: true
Explanation: You can swap A = 'a' and A = '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 = 'a' and A = 'b', which results in "ba" != B.
``````

#### Example 3:

``````Input: A = "aa", B = "aa"
Output: true
Explanation: You can swap A = 'a' and A = '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.

### Solution:

Find if two strings are the same first; if true, look for the duplicate character. Otherwise, check how many characters are different.

``````class Solution {
public boolean buddyStrings(String A, String B) {
if (A == null || B == null || A.length() != B.length()){
return false;
}
int pointer_a = -1;
int pointer_b = -1;
int diff = 0;
int[] counter = new int;

if (A.equals(B)) {
for (char c : A.toCharArray()) {
if (++counter[c - 'a'] >= 2) {
return true;
}
}
}
for (int i = 0; i < A.length(); i++)  {
if (A.charAt(i) != B.charAt(i)) {
if (pointer_a == -1) {
pointer_a = i;
} else if (pointer_b == -1) {
pointer_b = i;
}
diff++;
}
}
return (diff == 2 && A.charAt(pointer_a) == B.charAt(pointer_b) && A.charAt(pointer_b) == B.charAt(pointer_a));
}
}
``````

It can also do in a single pass.

``````class Solution {
public boolean buddyStrings(String A, String B) {
int pointer_a = -1;
int pointer_b = -1;
if (A == null || B == null || A.length() != B.length()){
return false;
}
int diff = 0;
boolean duplicate = false;
int[] counter = new int;
for (int i = 0; i < A.length(); i++)  {
if (A.charAt(i) != B.charAt(i)) {
if (pointer_a == -1) {
pointer_a = i;
} else if (pointer_b == -1) {
pointer_b = i;
}
diff++;
} else {
counter[A.charAt(i) - 'a']++;
if (counter[A.charAt(i) - 'a'] >= 2) {
duplicate = true;
}
}
}
return (diff == 0 && duplicate) || (diff == 2 && A.charAt(pointer_a) == B.charAt(pointer_b) && A.charAt(pointer_b) == B.charAt(pointer_a));
}
}
``````