【CodeForces 1257C --- Dominated Subarray】

【CodeForces 1257C --- Dominated Subarray】

Description

Let’s call an array t dominated by value v in the next situation.

At first, array t should have at least 2 elements. Now, let’s calculate number of occurrences of each number num in t and define it as occ(num). Then t is dominated (by v) if (and only if) occ(v)>occ(v′) for any other number v′. For example, arrays [1,2,3,4,5,2], [11,11] and [3,2,3,2,3] are dominated (by 2, 11 and 3 respectevitely) but arrays [3], [1,2] and [3,3,2,2,1] are not.

Small remark: since any array can be dominated only by one number, we can not specify this number and just say that array is either dominated or not.

You are given array a1,a2,…,an. Calculate its shortest dominated subarray or say that there are no such subarrays.

The subarray of a is a contiguous part of the array a, i. e. the array ai,ai+1,…,aj for some 1≤i≤j≤n.

Input

The first line contains single integer T (1≤T≤1000) — the number of test cases. Each test case consists of two lines.

The first line contains single integer n (1≤n≤2⋅105) — the length of the array a.

The second line contains n integers a1,a2,…,an (1≤ai≤n) — the corresponding values of the array a.

It’s guaranteed that the total length of all arrays in one test doesn’t exceed 2⋅105.

Output

Print T integers — one per test case. For each test case print the only integer — the length of the shortest dominated subarray, or −1 if there are no such subarrays.

Sample Input

4
1
1
6
1 2 3 4 5 1
9
4 1 2 4 5 4 3 2 1
4
3 3 3 3

Sample Output

-1
6
3
2

AC代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
const int MAXN = 2e5+5;
int arr[MAXN];

int main()
{
    SIS;
    int T;
    cin >> T;
    while(T--)
    {
        map<int,int> m;
        int n,ans=0,cnt=0x3f3f3f3f;
        cin >> n;
        for(int i=1;i<=n;i++)
        {
            cin >> arr[i];
            if(i>1 && m[arr[i]]!=0 && cnt>i-m[arr[i]])
            {
                cnt=i-m[arr[i]];
            }
            m[arr[i]]=i;
        } 
        if(n<2 || cnt<1 || cnt==0x3f3f3f3f)
            cout << -1 << endl;
        else
            cout << cnt+1 << endl;
    }
    return 0;
}

来源：CSDN

作者：私忆一秒钟

链接：https://blog.csdn.net/qq_41879343/article/details/103091376