What is the minimum number of swaps for K-Similar Strings?

The minimum number of swaps required is the smallest integer k such that s1 can be transformed into s2 using exactly k swaps of characters.

How does the BFS approach work for K-Similar Strings?

BFS explores all possible transformations level by level, ensuring the minimal swap count is found when s2 is reached.

Can K-Similar Strings be solved without BFS?

While BFS is the most efficient way to find the minimal number of swaps, other techniques like greedy methods or recursive backtracking could also be used, but they may not be as optimal.

What role does the hash table play in the solution?

The hash table stores previously seen states of s1, helping to avoid redundant calculations and ensuring efficient traversal of the string transformations.

What are the constraints of the K-Similar Strings problem?

The input strings s1 and s2 are anagrams of each other, with lengths between 1 and 20, and consist of lowercase letters from {'a', 'b', 'c', 'd', 'e', 'f'}.

#854

Hard

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LeetCode 题解工作台

相似度为 K 的字符串

对于某些非负整数 k ，如果交换 s1 中两个字母的位置恰好 k 次，能够使结果字符串等于 s2 ，则认为字符串 s1 和 s2 的相似度为 k 。给你两个字母异位词 s1 和 s2 ，返回 s1 和 s2 的相似度 k 的最小值。示例 1：输入： s1 = "ab", s2 = "ba" …

哈希表字符串广度优先搜索

题目描述

对于某些非负整数 k ，如果交换 s1 中两个字母的位置恰好 k 次，能够使结果字符串等于 s2 ，则认为字符串 s1 和 s2 的 相似度为 k 。

给你两个字母异位词 s1 和 s2 ，返回 s1 和 s2 的相似度 k 的最小值。

示例 1：

输入：s1 = "ab", s2 = "ba"
输出：1

示例 2：

输入：s1 = "abc", s2 = "bca"
输出：2

提示：

1 <= s1.length <= 20
s2.length == s1.length
s1 和 s2 只包含集合 {'a', 'b', 'c', 'd', 'e', 'f'} 中的小写字母
s2 是 s1 的一个字母异位词

lightbulb

解题思路

方法一：BFS

本题实际上是一类经典的问题：求解最小操作次数。从一个初始状态 $s_1$ ，经过最少 $k$ 次状态转换，变成目标状态 $s_2$ 。字符串长度不超过 $20$ ，我们考虑使用 BFS 搜索来求解。

首先将初始状态 $s_1$ 入队，用哈希表 vis 记录所有访问过的状态。

接下来每一轮，都是将队列中的所有状态转换到下一个状态，当遇到目标状态 $s_2$ 时，当前状态转换的轮数就是答案。

我们发现，题目的重点在于如何进行状态转换。对于本题，转换操作就是交换一个字符串中两个位置的字符。如果当前字符串 $s[i]$ 与 $s_2[i]$ 不相等，那么我们应该在 $s$ 中找到一个位置 $j$ ，满足 $s[j] = s_2[i]$ 并且 $s[j] \neq s_2[j]$ ，然后交换 $s[i]$ 和 $s[j]$ 。这样可以使得状态最接近于目标状态。这里的状态转换可以参考以下代码中的 next() 函数。

复杂度分析：BFS 剪枝不讨论时空复杂度。

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class Solution:
    def kSimilarity(self, s1: str, s2: str) -> int:
        def next(s):
            i = 0
            while s[i] == s2[i]:
                i += 1
            res = []
            for j in range(i + 1, n):
                if s[j] == s2[i] and s[j] != s2[j]:
                    res.append(s2[: i + 1] + s[i + 1 : j] + s[i] + s[j + 1 :])
            return res

        q = deque([s1])
        vis = {s1}
        ans, n = 0, len(s1)
        while 1:
            for _ in range(len(q)):
                s = q.popleft()
                if s == s2:
                    return ans
                for nxt in next(s):
                    if nxt not in vis:
                        vis.add(nxt)
                        q.append(nxt)
            ans += 1

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Candidate demonstrates an understanding of BFS for finding the shortest path in a graph.
question_mark
Candidate applies optimization techniques using hash tables to avoid redundant state processing.
question_mark
Candidate effectively uses string manipulation techniques to reduce unnecessary swaps.

warning

常见陷阱

外企场景

error
Overlooking the importance of BFS for level-based state exploration could lead to an inefficient solution.
error
Failing to use a hash table to track visited states will result in redundant calculations and higher time complexity.
error
Not considering early termination when the target string is reached can lead to unnecessary calculations and slow performance.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the two strings are already equal? In this case, the solution should immediately return 0 without further computation.
arrow_right_alt
Consider the case where only one swap is needed: this will test if the algorithm handles minimal cases effectively.
arrow_right_alt
What if there are repeated characters? Ensure that the algorithm handles string transformations correctly, even when characters appear multiple times.

help

常见问题

外企场景

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