B. Uniqueness

You are given an array a1,a2,…,ana1,a2,…,an. You can remove at most one subsegment from it. The remaining elements should be pairwise distinct.

In other words, at most one time you can choose two integers ll and rr (1≤l≤r≤n1≤l≤r≤n) and delete integers al,al+1,…,aral,al+1,…,ar from the array. Remaining elements should be pairwise distinct.

Find the minimum size of the subsegment you need to remove to make all remaining elements distinct.

The first line of the input contains a single integer nn (1≤n≤20001≤n≤2000) — the number of elements in the given array.

The next line contains nn spaced integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the elements of the array.

Print a single integer — the minimum size of the subsegment you need to remove to make all elements of the array pairwise distinct. If no subsegment needs to be removed, print 00.

In the first example all the elements are already distinct, therefore no subsegment needs to be removed.

In the second example you can remove the subsegment from index 22 to 33.

In the third example you can remove the subsegments from index 11 to 22, or from index 22 to 33, or from index 33 to 44.

题意：删除一段连续的区间，使得剩下的所有元素都是唯一的，求满足要求的最小区间长度

题解：逆向考虑，从左边开始[1，i ]最多可以取x个不同的数，然后从n到 i+1 最多可以连续取y个不同的数，不断枚举 i ，取cnt=max( cnt，x+y) 的最大值，n - cnt就是可以删除的最小长度

#include<iostream>

#include<algorithm>

#include<math.h>

#include<map>

using namespace std;

int a[];

map<int,int>mp;

int main()

{

    int n,ans=;

    cin>>n;

    for(int i=;i<=n;i++)

        cin>>a[i];

    for(int i=;i<=n;i++)

    {

        mp.clear();

        int cnt=;

        for(int j=;j<=i;j++)

        {

            if(mp[a[j]]==)

            {

                mp[a[j]]=;

                cnt++;

            }

            else

                break;

        }

        for(int j=n;j>i;j--)

        {

            if(mp[a[j]]==)

            {

                mp[a[j]]=;

                cnt++;

            }

            else

                break;

        }

        ans=max(ans,cnt);

    }

    cout<<n-ans<<endl;

    return ;

}