Calculate the total fine to be collected
Last Updated :
01 Sep, 2022
Given a date and an array of integer containing the numbers of the cars traveling on that date(an integer), the task is to calculate the total fine collected based on the following rules:
- Odd numbered cars can travel on only odd dates.
- Even numbered cars on only even dates.
- Otherwise a car would be fined 250 Rs.
Examples:
Input: car_num[] = {3, 4, 1, 2}, date = 15
Output: 500
Car with numbers '4' and '2' will be fined
250 each.
Input: car_num[] = {1, 2, 3} , date = 16
Output: 500
Car with numbers '1' and '3' will be fined
250 each.
Approach:
- Start traversing the given array.
- Check if the current car number and date don’t match i.e. one is even and other is odd or vice-versa.
- If not matched charge the fine on that car number. Else, not.
- Print the total fine.
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int totFine( int car_num[], int n, int date, int fine)
{
int tot_fine = 0;
for ( int i = 0; i < n; i++)
if (((car_num[i] ^ date) & 1) == 1)
tot_fine += fine;
return tot_fine;
}
int main()
{
int car_num[] = { 3, 4, 1, 2 };
int n = sizeof (car_num) / sizeof (car_num[0]);
int date = 15, fine = 250;
cout << totFine(car_num, n, date, fine);
return 0;
}
|
Java
class GFG
{
static int totFine( int car_num[], int n,
int date, int fine)
{
int tot_fine = 0 ;
for ( int i = 0 ; i < n; i++)
if (((car_num[i] ^ date) & 1 ) == 1 )
tot_fine += fine;
return tot_fine;
}
public static void main(String[] args)
{
int car_num[] = { 3 , 4 , 1 , 2 };
int n = car_num.length;
int date = 15 , fine = 250 ;
System.out.println(totFine(car_num, n,
date, fine));
}
}
|
Python 3
def totFine(car_num, n, date, fine) :
tot_fine = 0
for i in range (n) :
if (((car_num[i] ^ date) & 1 ) = = 1 ):
tot_fine + = fine
return tot_fine
if __name__ = = "__main__" :
car_num = [ 3 , 4 , 1 , 2 ]
n = len (car_num)
date, fine = 15 , 250
print (totFine(car_num, n, date, fine))
|
C#
using System;
class GFG
{
static int totFine( int [] car_num, int n,
int date, int fine)
{
int tot_fine = 0;
for ( int i = 0; i < n; i++)
if (((car_num[i] ^ date) & 1) == 1)
tot_fine += fine;
return tot_fine;
}
public static void Main()
{
int [] car_num = { 3, 4, 1, 2 };
int n = car_num.Length;
int date = 15, fine = 250;
Console.Write(totFine(car_num, n,
date, fine));
}
}
|
PHP
<?php
function totFine(& $car_num , $n ,
$date , $fine )
{
$tot_fine = 0;
for ( $i = 0; $i < $n ; $i ++)
if ((( $car_num [ $i ] ^
$date ) & 1) == 1)
$tot_fine += $fine ;
return $tot_fine ;
}
$car_num = array (3, 4, 1, 2 );
$n = sizeof( $car_num );
$date = 15;
$fine = 250;
echo totFine( $car_num , $n ,
$date , $fine );
?>
|
Javascript
<script>
function totFine(car_num, n, date, fine)
{
let tot_fine = 0;
for (let i = 0; i < n; i++)
if (((car_num[i] ^ date) & 1) == 1)
tot_fine += fine;
return tot_fine;
}
let car_num = [ 3, 4, 1, 2 ];
let n = car_num.length;
let date = 15, fine = 250;
document.write(totFine(car_num, n, date, fine));
</script>
|
Complexity Analysis:
- Time complexity: O(n)
- Auxiliary Space: O(1)
Source:https://www.geeksforgeeks.org/microsoft-interview-experience-for-internship/
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