You are given an array a1, a2, ..., an consisting of n integers, and an integer k. You have to split the array into exactly k non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the k obtained minimums. What is the maximum possible integer you can get?
Definitions of subsegment and array splitting are given in notes.
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 105) — the size of the array a and the number of subsegments you have to split the array to.
The second line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109).
Print single integer — the maximum possible integer you can get if you split the array into k non-empty subsegments and take maximum of minimums on the subsegments.
5 2 1 2 3 4 5
5
5 1 -4 -5 -3 -2 -1
-5
A subsegment [l, r] (l ≤ r) of array a is the sequence al, al + 1, ..., ar.
Splitting of array a of n elements into k subsegments [l1, r1], [l2, r2], ..., [lk, rk] (l1 = 1, rk = n, li = ri - 1 + 1 for all i > 1) is k sequences (al1, ..., ar1), ..., (alk, ..., ark).
In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are min(1, 2, 3, 4) = 1 and min(5) = 5. The resulting maximum is max(1, 5) = 5. It is obvious that you can't reach greater result.
In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence( - 4, - 5, - 3, - 2, - 1). The only minimum is min( - 4, - 5, - 3, - 2, - 1) = - 5. The resulting maximum is - 5.
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5;
int a[N+10],n,k,premi[N+10],aftermi[N+10];
int main()
{
//freopen("F:\\rush.txt","r",stdin);
scanf("%d%d",&n,&k);
for (int i = 1;i <= n;i++)
scanf("%d",&a[i]);
if (k==1)
{
int ans = a[1];
for (int i = 2;i <= n;i++)
ans = min(ans,a[i]);
printf("%d\n",ans);
}else if (k==2)
{
premi[1] = a[1];
for (int i = 2;i <= n;i++)
premi[i] = min(premi[i-1],a[i]);
aftermi[n] = a[n];
for (int i = n-1;i >= 1;i--)
aftermi[i] = min(aftermi[i+1],a[i]);
int ans = max(premi[1],aftermi[2]);
for (int i = 2;i <= n-1;i++)
ans = max(ans,max(premi[i],aftermi[i+1]));
printf("%d\n",ans);
}else {
int ans = a[1];
for (int i = 1;i <= n;i++)
ans = max(ans,a[i]);
printf("%d\n",ans);
}
return 0;
}