Not simply beatiful strings (判断)

最新推荐文章于 2020-03-27 13:04:47 发布

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本文探讨了一个有趣的问题：如何判断一个字符串是否能被划分为两个可爱的子串，即这两个子串都可以重新排列形成两个连续的相同字符组。文章提供了一种算法实现思路，并给出了具体的代码示例。

Not simply beatiful strings

Let's call a string adorable if its letters can be realigned in such a way that they form two consequent groups of equal symbols (note that different groups must contain different symbols). For example, ababa is adorable (you can transform it to aaabb, where the first three letters form a group of a-s and others — a group of b-s), but cccc is not since in each possible consequent partition letters in these two groups coincide.

You're given a string s. Check whether it can be split into two non-empty subsequences such that the strings formed by these subsequences are adorable. Here a subsequence is an arbitrary set of indexes of the string.

Input

The only line contains s (1 ≤ |s| ≤ 10⁵) consisting of lowercase latin letters.

Output

Print «Yes» if the string can be split according to the criteria above or «No» otherwise.

Each letter can be printed in arbitrary case.

Examples

Input

ababa

Output

Yes

Input

zzcxx

Output

Yes

Input

yeee

Output

No

Note

In sample case two zzcxx can be split into subsequences zc and zxx each of which is adorable.

There's no suitable partition in sample case three.

这道题的意思是让我们把一个串分成两部分，使得这两部分子串的每个子串都仅由两种字母组成（这里的子串可以随意选字母组不一定按照原串的顺序）

我没有想到简单的思路，就是一步一步直接判断的，首先我们可以先统计一下整个串的字母种类数，如果只有1种或大于四种一定是不满足的，如aaaaaa再怎么分也不可能满足，而abcde，怎么分都会使得一个子串的中含有三个即以上字母种类

所以答案就在234中，字母种类为2时，必须保证每种字母个数大于等于2，因为要是每种字母在两个子串中都出现，而字母种类为3时，只要有一个种类的字母个数大于等于2即可，即作为那个公共的字母，另外两种字母一个串一种，当四个的时候一定符合，不管字母多少个。

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1e5+10;
int main(){
    char s[maxn];
    scanf("%s",s);
    int vis[27];
    memset(vis,0,sizeof(vis));
    int cnt = 0;
    for(int i = 0; s[i]; i++){
        vis[s[i]-'a']++;
    }
    for(int i = 0; i < 26; i++){
        if(vis[i]) cnt++;
    }
    if(cnt >= 5 || cnt < 2) printf("No\n");
    else{
        if(cnt == 2){
            int flag = 1;
            for(int i = 0; i < 26; i++){
                if(vis[i] == 1){
                    flag = 0;
                    break;
                }
            }
            if(flag) printf("Yes\n");
            else printf("No\n");
        }
        else if(cnt == 3){
            int flag = 0;
            for(int i = 0; i < 26; i++){
                if(vis[i] > 1){
                    flag = 1;
                    break;
                }
            }
            if(flag) printf("Yes\n");
            else printf("No\n");
        }
        else{
            printf("Yes\n");
        }
    }

    return 0;
}