Passing Cars

Problem

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.

My Solution

failed O(N ** 2)

function solution(A) {
	let pairs = [];

	for (let i in A) {
		if (A[i] === 0) {
			pairs.push(...A.slice(i, A.length));
		}
	}

	let cnt = 0;
	for (let i of pairs) {
		if (i === 1) cnt++;
	}

	if (cnt > 1000000000) return -1;

	return cnt;
}

success O(N)

function solution(A) {
	let pairsCount = 0;
	let incCounter = 0;

	for (let i = 0; i < A.length; i++) {
		// If current car is east bound, increment the counter by 1
		// next time onwards, when westbound car comes, par count would be increased by increased counter
		if (A[i] === 0) {
			incCounter++;
		} else {
			pairsCount += incCounter;
		}
	}
	return pairsCount <= 1000000000 ? pairsCount : -1;
}