#include <vector>
#include <cstdlib>

#define DECREASING -1
#define INCREASING  1
#define CONSTANT    0

// Since each element is within the range [-1000000, 1000000], there could be problems with
// functions like factorial and combination

// calculates factorial of a number (n!)
double factorial(int num)
{
    double sum = 1;

    for(int i = 2 ; i <= num ; i++)
    {
        sum *= i;
    }

    return sum;
}

// calculates number of combinations (2,n)
int combination(int n)
{
    return ( factorial(n) ) / ( factorial(2) * factorial(n-2) );
}

// Finds number of monotonic segments, then uses combinations to count every segments combination
int solution(std::vector<int> &A)
{
    int segment_cnt = 0;
    int segment_length = 2;
    int prev_state;
    int current_state;
    int state_finder;

    // find first state
    state_finder = A[1] - A[0];

    // Could use switch-case (for maybe a little more efficiency)
    // or get prev_state directly with
    // state_finder = (A[1] - A[0])/ abs(A[1] - A[0])
    // but CONSTANT will cause a 0/0 division
    if(state_finder > 0)
    {
        prev_state = INCREASING;
    }

    if(state_finder == 0)
    {
        prev_state = CONSTANT;
    }

    if(state_finder < 0)
    {
        prev_state = DECREASING;
    }

    // find different monotonic segment
    for(int i = 2 ; i < A.size() ; i++)
    {
        state_finder = A[i] - A[i-1];
        if(state_finder > 0)
        {
            current_state = INCREASING;
        }

        if(state_finder == 0)
        {
            current_state = CONSTANT;
        }

        if(state_finder < 0)
        {
            current_state = DECREASING;
        }

        // if state changes, then we have a new segment
        if(current_state != prev_state)
        {
            segment_cnt += combination(segment_length);
            prev_state = current_state;
            segment_length = 2;
            continue;
        }

        segment_length++;

    }

    // add the very last segment's count
    segment_cnt += combination(segment_length);


    return segment_cnt;
}