Codeforces Round 683 (Div. 2, by Meet IT)
F2
Frequency Problem (Hard Version)
Codeforces
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原题
← F1→
翻译中…

This is the hard version of the problem. The difference between the versions is in the constraints on the array elements. You can make hacks only if all versions of the problem are solved.

翻译中…

You are given an array $[a_1, a_2, \dots, a_n]$.

翻译中…

Your goal is to find the length of the longest subarray of this array such that the most frequent value in it is not unique. In other words, you are looking for a subarray such that if the most frequent value occurs $f$ times in this subarray, then at least $2$ different values should occur exactly $f$ times.

翻译中…

An array $c$ is a subarray of an array $d$ if $c$ can be obtained from $d$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

翻译中…

Input

翻译中…

The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the length of the array.

翻译中…

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — elements of the array.

翻译中…

Output

翻译中…

You should output exactly one integer  — the length of the longest subarray of the array whose most frequent value is not unique. If there is no such subarray, output $0$.

翻译中…

Note

翻译中…

In the first sample, the subarray $[1, 1, 2, 2, 3, 3]$ is good, but $[1, 1, 2, 2, 3, 3, 3]$ isn't: in the latter there are $3$ occurrences of number $3$, and no other element appears $3$ times.

翻译中…

样例

Input #1
7
1 1 2 2 3 3 3
Output #1
6
InputCopy OutputCopy
Input #2
10
1 1 1 5 4 1 3 1 2 2
Output #2
7
InputCopy OutputCopy
Input #3
1
1
Output #3
0
InputCopy OutputCopy
题目信息
题号
1447F2
时间限制
2 s
内存限制
256 MB

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。