B.Ternary String(欲張り)

題意説明
You are given a string s such that each its character is either 1, 2, or 3. You have to choose the shortest contiguous substring of s such that it contains each of these three characters at least once.
A contiguous substring of string s is a string that can be obtained from s by removing some (possibly zero) characters from the beginning of s and some (possibly zero) characters from the end of s.
Input The first line contains one integer t (1≤t≤20000) — the number of test cases.
Each test case consists of one line containing the string s (1≤|s|≤200000). It is guaranteed that each character of s is either 1, 2, or 3.
The sum of lengths of all strings in all test cases does not exceed 200000.
Output For each test case, print one integer — the length of the shortest contiguous substring of s containing all three types of characters at least once. If there is no such substring, print 0 instead.
構想
タイトルに規定されているサブストリングは削除ヘッダまたは削除末尾であるため,まず各数字の出現回数を統計し,1つの数字が繰り返されると最初から削除し,最後に1,2,3が出現すると最小出現回数を統計すればよい.
ACコード

#include
#define x first
#define y second
#define pb push_back
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long LL;
const int N=1e5+10;
const int M=1e6;
const int INF=0x3f3f3f3f;
const int MOD=1000000007;
int a[N];
string str[6]={"123","132","213","231","312","321"};
int ans=INF;
int main()
{
    IOS;
    int T;cin>>T;
    while(T--){
        int cnt[4];
        memset(cnt,0,sizeof cnt);
        string s;cin>>s;
        ans=INF;
        int l=0;
        for(int i=0;i<s.size();i++){
            cnt[s[i]-'0']++;
            while(cnt[s[l]-'0']>=2){
                cnt[s[l]-'0']--;
                l++;
            }
            if(cnt[1] && cnt[3] && cnt[2]) ans=min(ans,i-l+1);
        }
        if(ans==INF) cout<<0<<endl;
        else cout<<ans<<endl;
    }
    return 0;
}