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 Python (3.4.3) (judge verdict: Accepted)

import sys

read= sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
MOD = 1000000007
sys.setrecursionlimit(10**7)

#N,*A = map(int,read().split())
N = int(input())
A = list(map(int,input().split()))

#現在作ってる数列の種類を記録
# 単調増加：1
# 単調減少 ：-1
# 初期値:0
State = 0

#答え入れる
res = 1

for i in range(1,N):
    
    #A[i]はA[i-1]と比べたときの大小を記録

    #X に現在の状態を記録
    # 増加：1
    # 減少 ：-1
    # 同じ値:0
    if A[i] > A[i-1]: #大きくなった時
        X = 1 
    elif A[i] == A[i-1]: #等しい時
        X = 0
    else :# それ以外(小さくなった時)
        X = -1

    #現在の状態と作っている数列の状態が同じとき
    if X == State:
        continue
    
    #配列の種類が0の時(どちらにもなりえる)
    if  State == 0: 
        State = X #Stateが決定されるため、更新

    #現在の状態と作っている数列の状態が違い、現在の状態が0出ないとき
    if X != State  and  X != 0: 
        res += 1 
        State = 0 #数列を初期化

print(res)