### Passing Cars

January 17, 2020

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.

**0 ≤ P < Q < N** : the zeros will show before ones

Add the ones count from the end of the array, and add the previous ones count for each passing zero.

```
function solution(A) {
// write your code in JavaScript (Node.js 8.9.4)
// P < Q, P -east, 0; Q -west, 1
// the index of 0 shows before 1
// counting 1 from the end, add sub-total for each 0
var ones = 0, passing = 0;
for(var i=A.length-1; i>=0; i--) {
//console.log("i: ", i, " passing: ", passing, " ones: ", ones)
if (A[i] === 0){
passing += ones;
if (passing > 1000000000){
return -1;
}
} else {
ones ++;
}
}
return passing;
}
```