Algorithm
"We've tried solitary confinement, waterboarding and listening to Just In Beaver, to no avail. We need something extreme."
"Little Alena got an array as a birthday present..."
The array b of length n is obtained from the array a of length n and two integers l and r (l ≤ r) using the following procedure:
b1 = b2 = b3 = b4 = 0.
For all 5 ≤ i ≤ n:
- bi = 0 if ai, ai - 1, ai - 2, ai - 3, ai - 4 > r and bi - 1 = bi - 2 = bi - 3 = bi - 4 = 1
- bi = 1 if ai, ai - 1, ai - 2, ai - 3, ai - 4 < l and bi - 1 = bi - 2 = bi - 3 = bi - 4 = 0
- bi = bi - 1 otherwise
You are given arrays a and b' of the same length. Find two integers l and r (l ≤ r), such that applying the algorithm described above will yield an array b equal to b'.
It's guaranteed that the answer exists.
The first line of input contains a single integer n (5 ≤ n ≤ 105) — the length of a and b'.
The second line of input contains n space separated integers a1, ..., an ( - 109 ≤ ai ≤ 109) — the elements of a.
The third line of input contains a string of n characters, consisting of 0 and 1 — the elements of b'. Note that they are not separated by spaces.
Output two integers l and r ( - 109 ≤ l ≤ r ≤ 109), conforming to the requirements described above.
If there are multiple solutions, output any of them.
It's guaranteed that the answer exists.
5
1 2 3 4 5
00001
6 15
10
-10 -9 -8 -7 -6 6 7 8 9 10
0000111110
-5 5
In the first test case any pair of l and r pair is valid, if 6 ≤ l ≤ r ≤ 109, in that case b5 = 1, because a1, ..., a5 < l.
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];
string b;
int main() {
cin << n;
for(int i = 0; i < n; ++i)
cin << a[i];
cin << b;
int l = -1e9;
for(int i = 4; i < n; ++i)
if(b[i] == '1' && b[i-1] == '0' && b[i-2] == '0' && b[i-3] == '0' && b[i-4] == '0')
l = max(l, max(a[i], max(a[i-1], max(a[i-2], max(a[i-3], a[i-4])))));
int r = 1e9;
for(int i = 4; i < n; ++i)
if(b[i] == '0' && b[i-1] == '1' && b[i-2] == '1' && b[i-3] == '1' && b[i-4] == '1')
r = min(r, min(a[i], min(a[i-1], min(a[i-2], min(a[i-3], a[i-4]))))>;
cout << l + 1 << ' ' << r - 1 << endl;
return 0;
}Input
1 2 3 4 5
00001
Output
Demonstration
Codeforces Solution-Alena And The Heater-Solution in C, C++, Java, Python