#include "sequence.h"
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <typename T>
using o_set = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
int sequence(int N, vector<int> A) {
// subtask 4
// binary search for the answer for the case of 1 and 3
// the case of 2, sort all the array, and remove elements greedily until you get 2 as the median
// this is if there was 3 distinct numbers
// in case of only 2 distinct, return the maximum of them
// in case of only 1 distinct number, return N
map<int, int> mp;
for(int i = 0; i < N; i++)
mp[A[i]]++;
if(mp.size() == 1)
{
return mp.begin()->second;
}
if(mp.size() == 2)
{
return max(mp.begin()->second, mp.rbegin()->second);
}
// we got 1, 2, 3 in our sequence
vector idx(4, deque<int>());
for(int i = 0; i < N; i++)
idx[A[i]].push_back(i);
vector pref(N + 1, vector<int>(4));
for(int i = 0; i < N; i++)
{
if(i)
pref[i] = pref[i - 1];
pref[i][A[i]]++;
}
int mx = 1;
int l = 1, r = mp[1], ans = 1;
while(l <= r)
{
int md = (l + r) / 2;
bool can = false;
for(int i = md - 1; i < idx[1].size(); i++)
{
int j = idx[1][i];
int k = idx[1][i - md + 1];
int p = pref[j][1] - (k ? pref[k - 1][1] : 0);
int others = j - k + 1 - p;
int all = p + others;
can |= p >= (all + 1) / 2;
}
if(can)
{
l = md + 1;
ans = md;
}
else
{
r = md - 1;
}
}
mx = max(mx, ans);
l = 1, r = mp[3], ans = 1;
while(l <= r)
{
int md = (l + r) / 2;
bool can = false;
for(int i = md - 1; i < idx[3].size(); i++)
{
int j = idx[3][i];
int k = idx[3][i - md + 1];
int p = pref[j][3] - (k ? pref[k - 1][3] : 0);
int others = j - k + 1 - p;
int all = p + others;
can |= p >= (all + 1) / 2;
}
if(can)
{
l = md + 1;
ans = md;
}
else
{
r = md - 1;
}
}
mx = max(mx, ans);
l = 1, r = mp[2], ans = 1;
while(l <= r)
{
int md = (l + r) / 2;
bool can = false;
for(int i = md - 1; i < idx[2].size(); i++)
{
int j = idx[2][i];
int k = idx[2][i - md + 1];
int p = pref[j][2] - (k ? pref[k - 1][2] : 0);
int others = j - k + 1 - p;
int all = p + others;
can |= p >= (all + 1) / 2;
}
if(can)
{
l = md + 1;
ans = md;
}
else
{
r = md - 1;
}
}
mx = max(mx, ans);
// the answer for 2 needs more, but let's submit this
return mx;
}
