854. K-Similar Strings

Description

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.

Solutions

Solution 1

Python3JavaC++Go

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

class Solution {
    public int kSimilarity(String s1, String s2) {
        Deque<String> q = new ArrayDeque<>();
        Set<String> vis = new HashSet<>();
        q.offer(s1);
        vis.add(s1);
        int ans = 0;
        while (true) {
            for (int i = q.size(); i > 0; --i) {
                String s = q.pollFirst();
                if (s.equals(s2)) {
                    return ans;
                }
                for (String nxt : next(s, s2)) {
                    if (!vis.contains(nxt)) {
                        vis.add(nxt);
                        q.offer(nxt);
                    }
                }
            }
            ++ans;
        }
    }

    private List<String> next(String s, String s2) {
        int i = 0, n = s.length();
        char[] cs = s.toCharArray();
        for (; cs[i] == s2.charAt(i); ++i) {
        }

        List<String> res = new ArrayList<>();
        for (int j = i + 1; j < n; ++j) {
            if (cs[j] == s2.charAt(i) && cs[j] != s2.charAt(j)) {
                swap(cs, i, j);
                res.add(new String(cs));
                swap(cs, i, j);
            }
        }
        return res;
    }

    private void swap(char[] cs, int i, int j) {
        char t = cs[i];
        cs[i] = cs[j];
        cs[j] = t;
    }
}