Codeforces Round #598 (Div. 3) F. Equalizing Two Strings 构造

F. Equalizing Two Strings

You are given two strings s and t both of length n and both consisting of lowercase Latin letters.

In one move, you can choose any length len from 1 to n and perform the following operation:

Choose any contiguous substring of the string s of length len and reverse it;
at the same time choose any contiguous substring of the string t of length len and reverse it as well.
Note that during one move you reverse exactly one substring of the string s and exactly one substring of the string t.

Also note that borders of substrings you reverse in s and in t can be different, the only restriction is that you reverse the substrings of equal length. For example, if len=3 and n=5, you can reverse s[1…3] and t[3…5], s[2…4] and t[2…4], but not s[1…3] and t[1…2].

Your task is to say if it is possible to make strings s and t equal after some (possibly, empty) sequence of moves.

You have to answer q independent test cases.

Input

The first line of the input contains one integer q (1≤q≤104) — the number of test cases. Then q test cases follow.

The first line of the test case contains one integer n (1≤n≤2⋅105) — the length of s and t.

The second line of the test case contains one string s consisting of n lowercase Latin letters.

The third line of the test case contains one string t consisting of n lowercase Latin letters.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 (∑n≤2⋅105).

Output

For each test case, print the answer on it — "YES" (without quotes) if it is possible to make strings s and t equal after some (possibly, empty) sequence of moves and "NO" otherwise.

Example

input
4
4
abcd
abdc
5
ababa
baaba
4
asdf
asdg
4
abcd
badc
output
NO
YES
NO
YES

题意

如今给你两个字符串，你能够进行若干次操做。

每次操做须要在每一个字符串都选择出长度为len的一个区间，而后将这个区间的字符都进行翻转。

问你进行若干次操做后，这俩字符串能变成同样的吗？

题解

按照这个顺序进行判断：

1. 若是两个字符串存在不一样的字符，那么确定是NO
2. 若是某个字符串存在两个相同的字符，那么必定是YES，由于能够就在这两个字符中进行无限次的翻转
3. 若是两个字符串的逆的奇偶性相同，那么必定是YES

第三个怎么理解呢？在判断1和2以后，咱们获得的必定是一个排列，问题就变成你能够翻转若干次，两个排列可否相同。

咱们考虑咱们同时翻转相同长度的，咱们排列的逆必定会发生奇偶性的变化，那么若是一开始奇偶性就不一样，那么无论怎么翻转，都不会相同。

代码

#include<bits/stdc++.h>
using namespace std;

int n;
string s1,s2,S1,S2;
int Count(string s){
    int num=0;
    for(int i=0;i<s.size();i++){
        for(int j=0;j<i;j++){
            if(s[j]>s[i])
                num++;
        }
    }
    return num;
}
void solve(){
    cin>>n>>S1>>S2;
    s1=S1;s2=S2;
    sort(s1.begin(),s1.end());
    sort(s2.begin(),s2.end());
    for(int i=0;i<n;i++){
        if(s1[i]!=s2[i]){
            puts("NO");
            return;
        }
    }
    for(int i=1;i<n;i++){
        if(s1[i]==s1[i-1]){
            puts("YES");
            return;
        }
        if(s2[i]==s2[i-1]){
            puts("YES");
            return;
        }
    }

    if(Count(S1)%2==Count(S2)%2){
        puts("YES");
    }else{
        puts("NO");
    }
    return;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--)solve();
}