# Leetcode - Buddy Strings Solution

Given two strings `a`

and `b`

, 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[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
```

**Constraints:**

`1 <= a.length, b.length <= 2 * 10`

^{4}`a`

and`b`

consist of lowercase letters.

## Solution in Python

```
class Solution:
def buddyStrings(self, a: str, b: str) -> bool:
diff = [(i,j) for i,j in zip(a,b) if i!=j]
freq = Counter(a)
if len(a) != len(b):
return False
if len(diff) == 2 and diff[0][0] == diff[1][1] and diff[0][1]==diff[1][0]:
return True
elif len(diff) == 0 and any(map(lambda x: x>1, freq.values())):
return True
else:
return False
```