Codeforces 1389c good string

Let’s call left cyclic shift of some string t1t2t3…tn−1tn as string t2t3…tn−1tnt1

.

Analogically, let’s call right cyclic shift of string t
as string tnt1t2t3…tn−1

.

Let’s say string t

is good if its left cyclic shift is equal to its right cyclic shift.

You are given string s

which consists of digits 0–9.

What is the minimum number of characters you need to erase from s

to make it good?
Input

The first line contains single integer t
(1≤t≤1000

) — the number of test cases.

Next t
lines contains test cases — one per line. The first and only line of each test case contains string s (2≤|s|≤2⋅105). Each character si

is digit 0–9.

It’s guaranteed that the total length of strings doesn’t exceed 2⋅105 .
Output

For each test case, print the minimum number of characters you need to erase from s

to make it good.

Example
Input

3
95831
100120013
252525252525

Output

3
5
0

Note

In the first test case, you can erase any 3
characters, for example, the 1-st, the 3-rd, and the 4

-th. You’ll get string 51 and it is good.

In the second test case, we can erase all characters except 0: the remaining string is 0000 and it’s good.

In the third test case, the given string s
is already good.

题意：
给出一段字符串，判断要让其变为good最少要删除多少个字符。经分析可得一个字符串如果要是good需要所有的字符串一样，或者是一直重复两个字符。于是我们就枚举这两个字符所有的可能，包括是相同的情况，求出所有情况中的最小删除量即可。
代码：

#include 
using namespace std;
int main()
{
    int t;
    cin>>t;
    char m[15]={'0','1','2','3','4','5','6','7','8','9'};

    while(t--){
        int minn=100000000;
        int ch=0;
        string s;
        cin>>s;
        for(int i=0;i<10;i++){
            for(int j=0;j<10;j++){
                ch=0;
                int flag=1;
                for(int k=0;k<s.length();k++){
                    if(s[k]==m[i]&&flag==1){
                        flag=0;
                        if(i==j) ch++;
                    }
                    else if(s[k]==m[j]&&flag==0){
                        flag=1;
                        if(i==j) ch++;
                        else ch=ch+2;
                    }
                }
                int len=s.length();
                minn=min(minn,len-ch);
            }
        }
        cout<<minn<<endl;
    }
}