/*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:
class Solution { public int solution(int[] 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:
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.
____________________________________________________________________________________________________
Attampt Date : 2nd June 2022
Hetal Patel
onlinehet@gmail.com
(+1) 647 936 9867
https://www.linkedin.com/in/hetgammer/
_____________________________________________________________________________________________________*/
using System;
public class HPProgram
{
public static void Main(string[] args)
int[] A = {0,1,0,1,1};
var s = new Solution();
Console.WriteLine("Total Car Passed = "+s.solution(A));
Console.ReadLine();
}
class Solution
public int solution(int[] A)
Console.WriteLine("Given A = "+ string.Join(",", A)+"\n");
int count = 0;
int zero = 0;
foreach (int car in A)
zero = (car==0) ? zero = zero + 1 : zero;
if (zero > 0)
if (car == 1)
count = count + zero;
if (count > 1000000000)
return -1;
return count;
