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

Category : Java | Sub Category : Java Programs from Coding Interviews | By Prasad Bonam Last updated: 2020-12-04 06:01:53 Viewed : 637

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. `    }`