codeforces 616D Longest k-Good Segment(two pointer)

D. Longest k-Good Segment

 

The array a with n integers is given. Let's call the sequence of one or more consecutive elements in a segment. Also let's call the segment k-good if it contains no more than k different values.

Find any longest k-good segment.

As the input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printfinstead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 106) — the elements of the array a.

Output

Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from 1 to n from left to right.

Sample test(s)

input

5 5
1 2 3 4 5

output

1 5

input

9 3
6 5 1 2 3 2 1 4 5

output

3 7

input

3 1
1 2 3

output

1 1

//从左到右维护一个最长的不同元素个数<=k的区间
#include<cstdio>
#include<cstring>
#include<stack>
#include<iterator>
#include<queue>
#include<set>
#include<vector>
#include<iostream>
#include<map>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF=0x3f3f3f3f;
const int maxn=5e5+5;
int cnt[1000005];
int a[maxn];
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    int p1=0,p2=-1,l=0,r=0,tot=0;
    while(p2<n)
    {
        while(++p2<n)
        {
            if(!cnt[a[p2]])tot++;
            cnt[a[p2]]++;
            if(tot>k)break;
            if(p2-p1>=r-l)r=p2,l=p1;
        }
        while(tot>k)
        {
            cnt[a[p1]]--;
            if(cnt[a[p1]]==0)
            {
                tot--;p1++;
                break;
            }
            p1++;
        }
    }
    printf("%d %d
",l+1,r+1);
    return 0;
}