#include <iostream>
#include <cstdio>
#include <cassert>
#include <cstring>
#include <vector>
#include <valarray>
#include <array>
#include <queue>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <algorithm>
#include <cmath>
#include <complex>
#include <random>

using namespace std;
typedef long long ll;
typedef pair<int, int> P;

/**
 * RangeMinQuery
 *
 * 区間系データ構造の中ではFenwick Treeの次あたりに有名？
 * 値の変更と区間の最小値がO(logn)で処理できる
 *
 * template引数のint Sは要素数が2^Sであることを示す
 */
template <int S>
struct RangeMinQuery {
    static const int N = 1<<S;
    typedef ll D;
    static const D INF = 1LL<<62;
    D seg[2*N];

    void init() {
        fill_n(seg, 2*N, -INF);
    }

    void init(int n, D x[]) {
        init();
        for (int i = 0; i < n; i++) {
            seg[i+N] = x[i];
        }
        for (int i = N-1; i >= 1; i--) {
            seg[i] = max(seg[i*2], seg[i*2+1]);
        }
    }

    /// i番目の要素をxにする
    void set(int i, D x) {
        i += N;
        seg[i] = x;
        while (i) {
            i /= 2;
            seg[i] = max(seg[i*2], seg[i*2+1]);
        }
    }

    /// [a, b)での最小値を求める
    D get(int a, int b, int k = 1, int l = 0, int r = N) {
        if (r <= a || b <= l) return -INF;
        if (a <= l && r <= b) return seg[k];
        int md = (l+r)/2;
        D dl = get(a, b, k*2, l, md);
        D dr = get(a, b, k*2+1, md, r);
        return max(dl, dr);
    }
};


const int MN = 100100;
int n;
int a[MN];

RangeMinQuery<17> rm1, rm2;

int solve() {
	rm1.init(); rm2.init();
	rm1.set(a[0], 0);
	rm2.set(a[0], 0);
	// printf("te %d %lld \n", a[0], rm1.get(1, 2));
	// for (int i = 0; i < n; i++) {
	// 	printf("%d ", a[i]);
	// } printf("\n");
	for (int i = 1; i < n; i++) {
		rm1.set(a[i], max(rm1.get(a[i], a[i]+1), 1+rm2.get(0, a[i])));
		rm2.set(a[i], max(rm2.get(a[i], a[i]+1), 1+rm1.get(a[i]+1, MN)));
		// printf("rm1 ");
		// for (int j = 0; j <= n; j++) {
		// 	printf("%lld ", rm1.get(j, j+1));
		// } printf("\n");
		// printf("rm2 ");
		// for (int j = 0; j <= n; j++) {
		// 	printf("%lld ", rm2.get(j, j+1));
		// } printf("\n");
	}

	int re = max(rm1.get(0, MN), rm2.get(0, MN));

	if (re <= 2) re = 0;

	return n-re;
}


int main() {
	cin >> n;
	set<int> s;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		s.insert(a[i]);
	}
	vector<int> v(s.begin(), s.end());
	for (int i = 0; i < n; i++) {
		a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
	}
	cout << solve() << endl;

    return 0;
}