CODEWITHSUNDEEP
c++algorithm
nz_21
nz_21

Reputation: 7353

Algorithm: Longest Approximate interval

I am trying to solve this question:

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?

You're given a sequence of 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.

A range [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 The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of data points.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).

Output Print a single number — the maximum length of an almost constant range of the given sequence.

https://codeforces.com/problemset/problem/602/B

I see a solution here but I don't understand the algorithm - specifically the body of the loop. I am familiar with C++ syntax and I understand what's happening. I just don't understand why the algorithm works. 

#include<cstdio>
#include<algorithm>
using namespace std;
int a[1000005];
int main()
{
int n,ans = 2,x;
scanf("%d",&n);
for(int i = 1; i <= n; i++)
{

    scanf("%d",&x);
    a[x] = i;
    if(a[x-1] > a[x+1])
        ans = max(ans,i-max(a[x+1],a[x-2]));
    else
        ans = max(ans,i-max(a[x+2],a[x-1]));
}
printf("%d\n",ans);
return 0;
}

Could someone explain it to me?

Upvotes: 2

Views: 353

Answers (1)

StillFantasy
StillFantasy

Reputation: 1740

Array a stands for the last position of value x.

Now let's calculate the left bound for each position(with value x), if a[x - 1] is more close to a[x] compare to a[x + 1], it means the position that will break the rule of almost constant is at a[x + 1](because there is a x - 1 in between) or a[x - 2](because there is a x - 1 in between).

Vice versa.

Upvotes: 2

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