# 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

BFS.

• class Solution {
public int kSimilarity(String s1, String s2) {
Deque<String> q = new ArrayDeque<>();
Set<String> vis = new HashSet<>();
q.offer(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)) {
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);
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;
}
}

• class Solution {
public:
int kSimilarity(string s1, string s2) {
queue<string> q{ {s1} };
unordered_set<string> vis{ {s1} };
int ans = 0;
while (1) {
for (int i = q.size(); i; --i) {
auto s = q.front();
q.pop();
if (s == s2) {
return ans;
}
for (auto& nxt : next(s, s2)) {
if (!vis.count(nxt)) {
vis.insert(nxt);
q.push(nxt);
}
}
}
++ans;
}
}

vector<string> next(string& s, string& s2) {
int i = 0, n = s.size();
for (; s[i] == s2[i]; ++i) {}
vector<string> res;
for (int j = i + 1; j < n; ++j) {
if (s[j] == s2[i] && s[j] != s2[j]) {
swap(s[i], s[j]);
res.push_back(s);
swap(s[i], s[j]);
}
}
return res;
}
};

• 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:
q.append(nxt)
ans += 1


• func kSimilarity(s1 string, s2 string) int {
next := func(s string) []string {
i := 0
res := []string{}
for ; s[i] == s2[i]; i++ {
}
for j := i + 1; j < len(s1); j++ {
if s[j] == s2[i] && s[j] != s2[j] {
res = append(res, s[:i]+string(s[j])+s[i+1:j]+string(s[i])+s[j+1:])
}
}
return res
}

q := []string{s1}
vis := map[string]bool{s1: true}
ans := 0
for {
for i := len(q); i > 0; i-- {
s := q[0]
q = q[1:]
if s == s2 {
return ans
}
for _, nxt := range next(s) {
if !vis[nxt] {
vis[nxt] = true
q = append(q, nxt)
}
}
}
ans++
}
}