PassingCars.md

Task description A non-empty zero-indexed 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 zero-indexed 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.

Assume that:

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. Complexity:

expected worst-case time complexity is O(N); expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments). Elements of input arrays can be modified.

function solution(A) {
    var result = 0;

    var aLen = A.length;
    var zeroFactor = 0;

    for (var i = 0; i < aLen; i++)
    {
        if (A[i] == 0)
        {
            zeroFactor += 1;
        }
        
        if (A[i] == 1)
        {
            result += zeroFactor * 1;

            if (result > 1000000000)
            {
                result = -1;
                break;
            }
        }
    }

    return result;
}