Codility - PassingCars

A non-empty array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road.

Array A contains only 0s and/or 1s:

0 represents a car traveling east,
1 represents a car traveling west.
The goal is to count passing cars. We say that a pair of cars (P, Q), where 0 ≤ P < Q < N, is passing when P is traveling to the east and Q is traveling to the west.

For example, consider array A such that:

A[0] = 0
A[1] = 1
A[2] = 0
A[3] = 1
A[4] = 1
We have five pairs of passing cars: (0, 1), (0, 3), (0, 4), (2, 3), (2, 4).

Write a function:

function solution(A);

that, given a non-empty array A of N integers, returns the number of pairs of passing cars.

The function should return −1 if the number of pairs of passing cars exceeds 1,000,000,000.

For example, given:

A[0] = 0
A[1] = 1
A[2] = 0
A[3] = 1
A[4] = 1
the function should return 5, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..100,000];
each element of array A is an integer that can have one of the following values: 0, 1.
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1. East 방향의 자동차 수를 카운팅 해 둔 뒤, West 방향의 자동차를 만났을때,
   TotalCount 에 East 방향의 자동차 수를 더해주는 방식으로 풀이.

2. TotalCount 가 1,000,000,000를 넘어가는 경우에 대한 예외처리.

function solution(A) {
    // write your code in JavaScript (Node.js 8.9.4)
    let totalCount = 0, eastCar = 0;
    for (let i in A) {
        if (A[i] == 0) {
            eastCar += 1;
        } else {
            totalCount += eastCar;
        }

    }

    if (totalCount > 1000000000) 
        return -1;
        
    return totalCount;
}