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:

class Solution { public int solution(int[] 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.

// you can also use imports, for example:

// import java.util.*;

// you can write to stdout for debugging purposes, e.g.

// System.out.println("this is a debug message");

class Solution {

    public int solution(int[] A) {

        // write your code in Java SE 8

        int zeroCnt=0, oneCnt =0;

        for(int i=0; i<A.length; i++) {

            if(A[i] == 0) {

                zeroCnt += 1;

            } else if(A[i] == 1) {

                oneCnt += zeroCnt;

            }

            if(oneCnt > 1000000000)

            return -1;

        }

        return oneCnt;

    }

}

https://codility.com/demo/results/training8U5YGT-RJR/