PassingCars - Count the number of passing cars on the road.

Category : Java | Sub Category : Java Programs from Coding Interviews | By Runner Dev Last updated: 2020-12-04 11:31:53 Viewed : 200


1. PassingCars - Count the number of passing cars on the road.

 

  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.

 

2. Solution

 

PassingCars.java

 

 package javaprograms;

 

public class PassingCars {

     public static void main(String ar[]) {

         int[] a = {0,1,0,1,1,0};

         int result = solution(a);

         System.out.println(result);

     }

     public static int solution(int[] A) {

         int[] travellingToWestPassingCars = new int[A.length];

         travellingToWestPassingCars[0] = A[0];

         int totalCarsTravelingToWest = A[0];

         for (int i = 1; i < A.length; i++) {

              travellingToWestPassingCars[i] = travellingToWestPassingCars[i - 1] + A[i];

              if (A[i] == 1) {

                  totalCarsTravelingToWest += 1;

              }

         }

 

         int numberOfPairsOfPassingCars = 0;

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

              if (A[i] == 0) {

                  numberOfPairsOfPassingCars += totalCarsTravelingToWest - travellingToWestPassingCars[i];

                  if (numberOfPairsOfPassingCars > 1000000000) {

                       numberOfPairsOfPassingCars = -1;

                       break;

                  }

              }

         }

 

         return numberOfPairsOfPassingCars;

     }

}

 Output:  5

Explanation: {0,1,0,1,1,0}; => (0,1)(0,1),(0,1),(0,1),(0,1)

(0,1) =>index[0],index[1]

(0,1)=>index[0],index[3]

(0,1)=>index[0],index[4]

(0,1)=>index[2],index[3]

(0,1)=>index[2],index[4]

So, total 5 pairs

Another way of implementation :

  1.   static int passingPairs(int[] array){
    //{0,1,0,1,1,0};
  2.         int pairs=0; 
  3.         int count=0; 
  4.         int i;
  5.         for(i = array.length-1; i>=0; i--){
  6.             if(array[i] == 1)
  7.             {
  8.                 count++;
  9.             }
  10.             else{
  11.                 pairs+=count;
  12.             }
  13.         }
  14.         return pairs;
  15.     }




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