Strings A and B are K-similar (for some non-negative integer K) if we can swap the positions of two letters in A exactly K times so that the resulting string equals B.
Given two anagrams A and B, return the smallest K for which A and B are K-similar.
Example 1:
Input: A = "ab", B = "ba"
Output: 1


Example 2:
Input: A = "abc", B = "bca"
Output: 2


Example 3:
Input: A = "abac", B = "baca"
Output: 2


Example 4:
Input: A = "aabc", B = "abca"
Output: 2



Note:

1 <= A.length == B.length <= 20
A and B contain only lowercase letters from the set {'a', 'b', 'c', 'd', 'e', 'f'}

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

def _disabled_input(*args, **kwargs):
    raise RuntimeError("input() is disabled in callable mode")
builtins.input = _disabled_input

ns = {"__name__": "__judge__"}  # prevent if __name__ == "__main__" blocks
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = "class Solution:\n    def kSimilarity(self, A: str, B: str) -> int:\n        # 根据最后一位是否相等进行贪心的求解。dfs深搜\n        memo = {}\n        def helper(a, b):\n            if a == b:\n                return 0\n            if (a, b) in memo:\n                return memo[(a, b)]\n            res = float('inf')\n            if a[-1] == b[-1]:\n                res = min(res, helper(a[:-1], b[:-1]))\n            else:\n                for i in range(len(a) - 1):\n                    if a[i] == b[-1] and a[i] != b[i]:   \n                        # 切记当前位置的ab不能对应相等，要不然肯定多操作了。\n                        a_new = a[:i] + a[-1] + a[i + 1:-1]  #把a[i]与a[-1]交换了\n                        res = min(res, 1 + helper(a_new, b[:-1]))\n            memo[(a, b)] = res\n            return res\n        return helper(A, B)\n        '''\n        # 这道题虽然是hard但是通过你理解题意，很容易找到潜在的解答方法。\n        # 第一重思路就是看到最小。那么只要比最小的大都OK，那么很容易想到用二分法去猜测答案。但是如何check存疑\n        # 第二重思路就是看到两个node。第一个node通过交换变成第二个。交换的可能性很多。因此可以看乘bfs\n        # BFS就是给你一个起点一个终点，问你最近的距离，最少的方式等等是多少！\n        # by build directed edges bfs to find k\n        def neighbor(x):\n            i = 0\n            while x[i] == B[i]:  # 更换的种类很多种，但是仍然需要剪枝无效的。因此相同部分就不改变了。\n                i += 1\n            for j in range(i + 1, len(x)): # 找到第一个不同的地方，然后去找可以过来填坑的letter\n                if x[j] == B[i]:\n                    yield x[:i] + x[j] + x[i + 1:j] + x[i] + x[j + 1:]\n        q, seen = [(A, 0)], {A}\n        for x, d in q:\n            if x == B:\n                return d\n            for y in neighbor(x):\n                if y not in seen:\n                    seen.add(y)\n                    q.append((y, d + 1))\n        '''"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

_target = None
_fn_name = 'kSimilarity'
if _fn_name:
    if "Solution" in ns:
        try:
            _obj = ns["Solution"]()
            if hasattr(_obj, _fn_name):
                _target = getattr(_obj, _fn_name)
        except Exception:
            _target = None
    if _target is None and _fn_name in ns and callable(ns[_fn_name]):
        _target = ns[_fn_name]
# Fallback: if there's exactly one user-defined function, use it
if _target is None:
    _cands = [v for k, v in ns.items() if callable(v) and getattr(v, "__module__", "") in ("__judge__", "__main__")]
    _user_funcs = [c for c in _cands if getattr(c, "__name__", "").startswith("_") is False]
    if len(_user_funcs) == 1:
        _target = _user_funcs[0]
if _target is None:
    raise RuntimeError(f"Could not resolve callable '{_fn_name}'")

_raw = sys.stdin.read().strip()
if _raw:
    _parsed = json.loads(_raw)
    if isinstance(_parsed, dict):
        _kwargs = _parsed.get("kwargs", {})
        _args = _parsed.get("args", [])
    else:
        _args, _kwargs = _parsed, {}
else:
    _args, _kwargs = [], {}
if not isinstance(_args, (list, tuple)):
    _args = [_args]
_res = _target(*_args, **_kwargs)
try:
    sys.stdout.write(repr(_res) + "\n")
except Exception:
    pass

```

Measured memory_bytes: 6734