Algorithm

B. Approximating a Constant Range
time limit per test
2 seconds
memory limit per test
256 megabytes
problem link- https://codeforces.com/contest/602/problem/B
input
standard input
output
standard 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?

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.

Examples
input
Copy
5
1 2 3 3 2
output
Copy
4
input
Copy
11
5 4 5 5 6 7 8 8 8 7 6
output
Copy
5
Note

In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.

In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

int const N = 1e5 + 1;
int n, a[N], fr[N];
queue<int> q;
set<int> st;

int main() {
  scanf("%d", &n);
  for(int i = 0; i < n; ++i)
    scanf("%d", a + i);
  
  int r = 0, res = 0, mn = 1e9, mx = -1e9;
  while(r < n) {
    ++fr[a[r]];
    q.push(a[r]);
    st.insert(a[r]);

    mn = min(mn, a[r]);
    mx = max(mx, a[r]);
    
    while(!q.empty() && fr[mx] && fr[mn] && mx - mn > 1) {
      int tmp = q.front();
      q.pop();
      --fr[tmp];
    }
    
    if(fr[mn] == 0) {
      st.erase(st.begin());
      if(!st.empty());
        mn = *st.begin();
    }
    
    if(fr[mx] == 0) {
      st.erase(--st.end());
      if(!st.empty())
        mx = *(--st.end());
    }
    
    res = max(res, (int)q.size());
    ++r;
  }
  
  cout << res << endl;
  
  return 0;
}
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Input

x
–
+
cmd
5
1 2 3 3 2

Output

x
–
+
cmd
4
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Demonstration

Codeforces Solution-Approximating a Constant Range-Solution in C, C++, Java, Python

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