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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<bits/stdc++.h>
using namespace std;
 
// Returns count of passing cars
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])
                 result++;
       }
    }
    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




// 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




<?php
// Simple PHP program to count passing cars
 
// Returns count of passing cars
function getPassingCars($A, $n)
{
    $result = 0;
    for ($i = 0; $i < $n - 1; $i++)
    {
        if ($A[$i] == 0)
        {
            for ($j = $i + 1; $j < $n; $j++)
                if ($A[$j])
                    $result++;
        }
    }
    return $result;
}
 
    // Driver Code
    $A = array(0, 1, 0, 1, 1);
    $n = sizeof($A);
    echo getPassingCars($A, $n);
 
// This code is contributed by m_kit
?>

Javascript




<script>
 
    // Simple Javascript program
    // to count passing cars
     
    // Returns count of passing cars
    function getPassingCars(A, n)
    {
        let result = 0;
        for (let i = 0; i < n - 1; i++)
        {
          if (A[i] == 0)
          {
              for (let j = i + 1; j < n; j++)
                  if (A[j] == 1)
                      result++;
          }
        }
        return result;
    }
     
    let A = [0, 1, 0, 1, 1];
    let n = A.length;
    document.write(getPassingCars(A, n));
     
</script>

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<bits/stdc++.h>
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




<?php
// Efficient PHP program to count passing cars
 
// Returns count of passing cars
function getPassingCars($A, $n)
{
    // Initialize count of 1s from right
    // and result
    $countOne = 0; $result = 0;
    while ($n >= 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




<script>
    // Efficient Javascript program to count passing cars
     
    // Returns count of passing cars
    function getPassingCars(A, n)
    {
        // Initialize count of 1s from right
        // and result
        let countOne = 0, result = 0;
        while (n >= 1)
        {
            if (A[n - 1] == 1)
                countOne++;
            else
                result += countOne;
            n--;
        }
 
        return result;
    }
     
    let A = [0, 1, 0, 1, 1];
    let n = A.length;
    document.write(getPassingCars(A, n));
     
</script>

Output: 

5

Time Complexity: O(n) 
Auxiliary Space: O(1)
Reference : 
http://stackoverflow.com/questions/23774985/codility-passing-car
This article is contributed by Rakesh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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