B. Approximating a Constant Range
time limit per test
memory limit per test
input
output
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
n data points a1, ..., an. There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.
[l, r] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let M be the maximum and m the minimum value of ai for l ≤ i ≤ r; the range [l, r] is almost constant if M - m ≤ 1.
Find the length of the longest almost constant range.
Input
n (2 ≤ n ≤ 100 000) — the number of data points.
n integers a1, a2, ..., an (1 ≤ ai).
Output
Print a single number — the maximum length of an almost constant range of the given sequence.
Sample test(s)
input
5 1 2 3 3 2
output
4
input
11 5 4 5 5 6 7 8 8 8 7 6
output
5
Note
[2, 5]; its length (the number of data points in it) is 4.
4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
#include <bits/stdc++.h>
using namespace std;
multiset <int> s;
int a[100005];
int main()
{
int n,ans=0;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",a+i);
int j=0,i=0;
while(j<n)
{
s.insert(a[j]);
if(abs(*s.begin()-*s.rbegin())>1)
{
s.erase(s.find(a[i]));
i++;
}
ans=max(ans,int(s.size()));
j++;
}
printf("%d\n",ans);
return 0;
}