Problem:
Score : 300 points

Problem Statement

You are given an array A of length N.
Your task is to divide it into several contiguous subarrays.
Here, all subarrays obtained must be sorted in either non-decreasing or non-increasing order.
At least how many subarrays do you need to divide A into?

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

Each A_i is an integer.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the minimum possible number of subarrays after division of A.

Sample Input 1

6
1 2 3 2 2 1

Sample Output 1

2

One optimal solution is to divide the array into [1,2,3] and [2,2,1].

Sample Input 2

9
1 2 1 2 1 2 1 2 1

Sample Output 2

5

Sample Input 3

7
1 2 3 2 1 999999999 1000000000

Sample Output 3

3

Solution in Rust (1.15.1) (judge verdict: Accepted)

#![allow(non_snake_case)]

use std::io::{stdin, Read};
use std::str::FromStr;
fn read_option<T: FromStr>() -> Option<T> {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok()
}
fn read<T: FromStr>() -> T {
    let opt = read_option();
    opt.expect("failed to parse token")
}

fn main() {
    let n: usize = read();
    let vec: Vec<i32> = (0..n).map(|_| read()).collect();
    let mut unique = vec![vec[0]];
    for v in vec.windows(2) {
        if v[0] != v[1] {
            unique.push(v[1]);
        }
    }
    let mut ans = 0;
    let mut seq = vec![];
    for v in unique {
        if seq.is_empty() {
            seq.push(v);
            ans += 1;
            continue;
        }
        if seq.len() == 1 {
            seq.push(v);
            continue;
        }
        let (a, b, x) = (seq[seq.len() - 2], seq[seq.len() - 1], v);
        if (a > b && b > x) || (a < b && b < x) {
            seq.push(x);
        } else {
            seq = vec![x];
            ans += 1;
        }
    }
    println!("{}", ans);
}