Count passing car pairs

A non-empty binary array A consisting of size N is given where,

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.

Examples:

Input  : arr[] = {0, 1, 0, 1, 1}
Output : 5
The 5 pairs are (A, A), (A, A), (A, A),
(A, A) and (A, A). Note that in all pairs
first element is 0, second element is 1 and index of
first element is smaller than index of second element.

Input  : arr[] = {1, 0, 0, 0, 1}
Output : 3

Input : arr[] = {0, 0, 1, 0, 0}
Output : 2

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to use two nested loops. The outer loop searches for a 0, the inner loop counts number of 1s after 0. Finally we return total count.

C++

 // Simple C++ program to count passing cars #include using namespace std;    // Returns count of passing cars int getPassingCars(int A[], int n) {     int result = 0;     for (int i=0; i

Java

 // Simple Java program to count passing cars class GFG {    // Returns count of passing cars static int getPassingCars(int[] A, int n) {     int result = 0;     for (int i = 0; i < n - 1; i++)     {     if (A[i] == 0)     {         for (int j = i + 1; j < n; j++)              if (A[j] == 1)                 result++;      }      }     return result; }    // Driver Code public static void main(String[] args) {     int[] A = {0, 1, 0, 1, 1};     int n = A.length;     System.out.println(getPassingCars(A, n)); } }    // This code is contributed // by Code_Mech

Python3

 # Simple Python 3 program to  # count passing cars    # Returns count of passing cars def getPassingCars(A, n):     result = 0     for i in range(0, n - 1, 1):         if (A[i] == 0):             for j in range(i + 1, n, 1):                 if (A[j]):                     result += 1                return result    # Driver Code if __name__ == '__main__':     A = [0, 1, 0, 1, 1]     n = len(A)     print(getPassingCars(A, n))        # This code is contributed by # Sanjit_Prasad

C#

 // Simple C# program to count passing cars using System;    class GFG {    // Returns count of passing cars static int getPassingCars(int[] A, int n) {     int result = 0;     for (int i = 0; i < n - 1; i++)     {     if (A[i] == 0)     {         for (int j = i + 1; j < n; j++)              if (A[j] == 1)                 result++;      }      }     return result; }    // Driver Code public static void Main() {     int[] A = {0, 1, 0, 1, 1};     int n = A.Length;     Console.WriteLine(getPassingCars(A, n)); } }    // This code is contributed // by Akanksha Rai

PHP



Output:

5

Time Complexity : O(n2)
Auxiliary Space : O(1)

An efficient solution to traverse array from right, keep track of counts of 1s from right. Whenever we see a 0, we increment the result by count of 1s.

C++

 // Efficient C++ program to count passing cars #include using namespace std;    // Returns count of passing cars int getPassingCars(int A[], int n) {     // Initialize count of 1s from right     // and result     int countOne = 0, result = 0;     while (n >= 1)     {         if (A[n-1] == 1)             countOne++;         else             result += countOne;         n--;     }        return result; }    // Driver program int main() {     int A[] = {0, 1, 0, 1, 1};     int n = sizeof(A)/sizeof(A);     cout << getPassingCars(A, n);     return 0; }

Java

 // Efficient Java program to count passing cars class GFG {        // Returns count of passing cars static int getPassingCars(int A[], int n) {     // Initialize count of 1s from right     // and result     int countOne = 0, result = 0;     while (n >= 1)     {         if (A[n-1] == 1)             countOne++;         else             result += countOne;         n--;     }        return result; }    // Driver code public static void main(String[] args) {     int A[] = {0, 1, 0, 1, 1};     int n = A.length;     System.out.println(getPassingCars(A, n)); } }    // This code is contributed by Mukul Singh.

Python3

 # Efficient Python3 program to  # count passing cars     # Returns count of passing cars  def getPassingCars(A, n):        # Initialize count of 1s      # from right and result      countOne = 0; result = 0     while n >= 1:         if A[n - 1] == 1:             countOne += 1         else:             result += countOne         n -= 1     return result    # Driver code A = [0, 1, 0, 1, 1] n = len(A) print(getPassingCars(A, n))    # This code is contributed # by Shrikant13

C#

 // Efficient C# program to // count passing cars using System;    class GFG {        // Returns count of passing cars static int getPassingCars(int []A, int n) {     // Initialize count of 1s from right     // and result     int countOne = 0, result = 0;     while (n >= 1)     {         if (A[n - 1] == 1)             countOne++;         else             result += countOne;         n--;     }        return result; }    // Driver code public static void Main() {     int []A = {0, 1, 0, 1, 1};     int n = A.Length;     Console.Write(getPassingCars(A, n)); } }    // This code is contributed // by Akanksha Rai

PHP

 = 1)     {         if (\$A[\$n-1] == 1)             \$countOne++;         else             \$result += \$countOne;         \$n--;     }        return \$result; }    // Driver Code \$A = array(0, 1, 0, 1, 1); \$n = sizeof(\$A); echo getPassingCars(\$A, \$n);    // This code is contributed  // by Akanksha Rai ?>

Output:

5

Time Complexity : O(n)
Auxiliary Space : O(1)

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