Strings `s1` and `s2` are `k`**\-similar** (for some non-negative integer `k`) if we can swap the positions of two letters in `s1` exactly `k` times so that the resulting string equals `s2`.

Given two anagrams `s1` and `s2`, return the smallest `k` for which `s1` and `s2` are `k`**\-similar**.

**Example 1:**

**Input:** s1 =  "ab ", s2 =  "ba "
**Output:** 1
**Explanation:** The two string are 1-similar because we can use one swap to change s1 to s2:  "ab " -->  "ba ".

**Example 2:**

**Input:** s1 =  "abc ", s2 =  "bca "
**Output:** 2
**Explanation:** The two strings are 2-similar because we can use two swaps to change s1 to s2:  "abc " -->  "bac " -->  "bca ".

**Constraints:**

*   `1 <= s1.length <= 20`
*   `s2.length == s1.length`
*   `s1` and `s2` contain only lowercase letters from the set `{'a', 'b', 'c', 'd', 'e', 'f'}`.
*   `s2` is an anagram of `s1`.


def kSimilarity(s1: str, s2: str) -> int:
    k = 0
    temp = list(s1)

    for i in range(len(temp)):
        if temp[i] != s2[i]:
            j = i + 1
            while temp[j] != s2[i] or s2[j] == temp[j]:
                j += 1
            temp[i], temp[j] = temp[j], temp[i]
            k += 1

    return k
