Local Max Version 2

#include <bits/stdc++.h>

using namespace std;

#pragma GCC optimize("-Ofast")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-funroll-all-loops,-fpeel-loops,-funswitch-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")

#define ll long long
#define IO ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);

const int N = 2e6 + 5, M = 1e2 + 5, MM = 55, OO = 1e9;
const double PI = acos(-1), EPS = 1e-6;
const int dx[8] = {0, 0, 1, -1, 1, 1, -1, -1};
const int dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
const long long MOD = 1e9 + 7, MOD2 = 998244353;

int t = 1, n, a[M], ans, mem[M][M][M][MM], k;

int dp(int first, int mid, int i, int k)
{
    if (k < 0)return OO;
    if (i == n + 1)
    {
        if (k == 0)return 0;
        return OO;
    }

    int &ret = mem[first][mid][i][k];
    if (~ret)return ret;

    int c1 = 1 + dp(first, mid, i + 1, k), c2;
    if (first == 0)c2 = dp(i, 0, i + 1, k);
    else if (mid == 0)c2 = dp(first, i, i + 1, k);
    else c2 = dp(mid, i, i + 1, k - (a[mid] > a[first] && a[mid] > a[i]));

    c1 = min(c1, OO);
    c2 = min(c2, OO);

    return ret = min(c1, c2);
}

void solve()
{
    memset(mem, -1, sizeof mem);
    cin >> n >> k;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    cout << (dp(0, 0, 1, k) != OO ? mem[0][0][1][k] : -1);
    return;
}

int main()
{
    IO

    // cin >> t;
    while (t--)
    {
        solve();
    }

    return 0;
}