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# Count passing car pairs

• Difficulty Level : Basic
• Last Updated : 27 Apr, 2021

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[0], A[1]), (A[0], A[3]), (A[0], A[4]),
(A[2], A[3]) and (A[2], A[4]). 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```

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

 ``

## Javascript

 ``

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 the right. Whenever we see a 0, we increment the result by a 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[0]);``    ``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``?>`

## Javascript

 ``

Output:

`5`

Time Complexity: O(n)
Auxiliary Space: O(1)
Reference :
http://stackoverflow.com/questions/23774985/codility-passing-car
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