Problem:
Score : 300 points

Problem Statement

You are given strings S and T consisting of lowercase English letters.

You can perform the following operation on S any number of times:

Operation: Choose two distinct lowercase English letters c_1 and c_2, then replace every occurrence of c_1 with c_2, and every occurrence of c_2 with c_1.

Determine if S and T can be made equal by performing the operation zero or more times.

Constraints

1 \leq |S| \leq 2 \times 10^5

|S| = |T|

S and T consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S
T

Output

If S and T can be made equal, print Yes; otherwise, print No.

Sample Input 1

azzel
apple

Sample Output 1

Yes

azzel can be changed to apple, as follows:

Choose e as c_1 and l as c_2. azzel becomes azzle.

Choose z as c_1 and p as c_2. azzle becomes apple.

Sample Input 2

chokudai
redcoder

Sample Output 2

No

No sequences of operation can change chokudai to redcoder.

Sample Input 3

abcdefghijklmnopqrstuvwxyz
ibyhqfrekavclxjstdwgpzmonu

Sample Output 3

Yes

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)


#include<bits/stdc++.h>
#include<utility>

using namespace std;
#define int long long
#define br cout<<"\n"
#define all(V) V.begin(),V.end()
#define rall(V) V.rbegin(),V.rend()
#define umap unordered_map
#define uset unordered_set

typedef vector<int> vi;
typedef vector<vi> vvi;

bool isPwrOf2(int x) { return x && (!(x&(x-1))); }
/*------------------------------------------------*/

/*------------------------------------------------*/

signed main()
{
    ios_base::sync_with_stdio(false); cin.tie(nullptr);
    // #ifndef ONLINE_JUDGE
    //     freopen("./input.txt", "r", stdin);
    //     freopen("./output.txt", "w", stdout);
    // #endif
    /*------------------------------------------------*/
    string s,t;
    cin>>s>>t;
    int sarr[26]={0}, tarr[26]={0};
    for(auto c:s){
        sarr[c-'a']++;
    }
    for(auto c:t){
        tarr[c-'a']++;
    }
    int sod=0,sev=0,tod=0,tev=0;
    for(int i=0;i<26;i++){
        (sarr[i]&1) ? sod++ : sev++ ;
        (tarr[i]&1) ? tod++ : tev++ ;
    }
    if(sod==tod and tev==sev){
        cout<<"Yes";
    }
    else{
        cout<<"No";
    }
    /*------------------------------------------------*/
    return 0;
}