C. 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 , [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⋅1e5) — 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⋅1e5.

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.

Example
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
output
-1
6
3
2
Note
In the first test case, there are no subarrays of length at least 2, so the answer is −1.

In the second test case, the whole array is dominated (by 1) and it’s the only dominated subarray.

In the third test case, the subarray a4,a5,a6 is the shortest dominated subarray.

In the fourth test case, all subarrays of length more than one are dominated.

Problem solving:

Code:

#include <bits/stdc++.h>
using namespace std;
int main()
{
int t, n;
cin >> t;
while (t--)
{
int         ans = 0x3f3f3f3f;
cin >> n;
vector<int> a(n +1); vector<int> pos(n+1);
for (int i = 1; i <= n; i++)
{
cin >> a[i];
if (!pos[a[i]])
pos[a[i]] = i;
else
ans = min(ans, i - pos[a[i]]+1), pos[a[i]] = i;
}
if (ans == 0x3f3f3f3f)
cout << "-1\n";
else
cout << ans << endl;
}
}
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