PassingCars

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Task description

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:

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.

Algorithm

/**
 * @author Ravi Yadav
 * blog address: www.meritmix.com 
 */
public class PassingCars {

 public static void main(String[] args) {
  int[] A ={0,1,0,1,1};
  System.out.println(new PassingCars().solution(A));
 }
 public int solution(int[] A) {
  
  if(A.length == 0) {
   return 0;
  }

  long max = 1_000_000_000;
  long pairs = 0;
  
  int currentMultiplier=0;
  for (int i = 0; i < A.length; i++) {
   if(A[i] == 0) {
    currentMultiplier++;
   }else if(A[i] ==1) {
    pairs+=currentMultiplier;
   }
  }
  return pairs > max? -1: (int)pairs;
    }
}

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