human
Solve in Python:
There are $n$ beautiful skyscrapers in New York, the height of the $i$-th one is $h_i$. Today some villains have set on fire first $n - 1$ of them, and now the only safety building is $n$-th skyscraper.

Let's call a jump from $i$-th skyscraper to $j$-th ($i < j$) discrete, if all skyscrapers between are strictly lower or higher than both of them. Formally, jump is discrete, if $i < j$ and one of the following conditions satisfied:   $i + 1 = j$  $\max(h_{i + 1}, \ldots, h_{j - 1}) < \min(h_i, h_j)$  $\max(h_i, h_j) < \min(h_{i + 1}, \ldots, h_{j - 1})$. 

At the moment, Vasya is staying on the first skyscraper and wants to live a little longer, so his goal is to reach $n$-th skyscraper with minimal count of discrete jumps. Help him with calcualting this number.

-----Input-----

The first line contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — total amount of skyscrapers.

The second line contains $n$ integers $h_1, h_2, \ldots, h_n$ ($1 \le h_i \le 10^9$) — heights of skyscrapers.

-----Output-----

Print single number $k$ — minimal amount of discrete jumps. We can show that an answer always exists.

-----Examples-----
Input
5
1 3 1 4 5

Output
3
Input
4
4 2 2 4

Output
1
Input
2
1 1

Output
1
Input
5
100 1 100 1 100

Output
2

-----Note-----

In the first testcase, Vasya can jump in the following way: $1 \rightarrow 2 \rightarrow 4 \rightarrow 5$.

In the second and third testcases, we can reach last skyscraper in one jump.

Sequence of jumps in the fourth testcase: $1 \rightarrow 3 \rightarrow 5$.

Give me a Python program to print single number $k$ — minimal amount of discrete jumps?
assistant
#Here is Python program to print single number $k$ — minimal amount of discrete jumps.

class SegmentTree:
    def __init__(self, data, default=0, func=max):
        """initialize the segment tree with data"""
        self._default = default
        self._func = func
        self._len = len(data)
        self._size = _size = 1 << (self._len - 1).bit_length()

        self.data = [default] * (2 * _size)
        self.data[_size:_size + self._len] = data
        for i in reversed(list(range(_size))):
            self.data[i] = func(self.data[i + i], self.data[i + i + 1])

    def __delitem__(self, idx):
        self[idx] = self._default

    def __getitem__(self, idx):
        return self.data[idx + self._size]

    def __setitem__(self, idx, value):
        idx += self._size
        self.data[idx] = value
        idx >>= 1
        while idx:
            self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
            idx >>= 1

    def __len__(self):
        return self._len

    def query(self, start, stop):
        """func of data[start, stop)"""
        start += self._size
        stop += self._size

        res_left = res_right = self._default
        while start < stop:
            if start & 1:
                res_left = self._func(res_left, self.data[start])
                start += 1
            if stop & 1:
                stop -= 1
                res_right = self._func(self.data[stop], res_right)
            start >>= 1
            stop >>= 1

        return self._func(res_left, res_right)

    def __repr__(self):
        return "SegmentTree({0})".format(self.data)

n = int(input())
l = list(map(int, input().split()))

stMax = SegmentTree(l)
stMin = SegmentTree(l, default = 10**10,func=min)

nexSmol = [-1] * n
nexTol = [-1] * n

smolS = [n-1]
tolS = [n-1]

for i in range(n - 2, -1 ,-1):
    while smolS and l[smolS[-1]] >= l[i]:
        smolS.pop()
    if smolS:
        nexSmol[i] = smolS[-1]
    smolS.append(i)

    while tolS and l[tolS[-1]] <= l[i]:
        tolS.pop()
    if tolS:
        nexTol[i] = tolS[-1]
    tolS.append(i)

best = [n] * n
best[0] = 0

for i in...