Find the Fine

Problem

Given an array of penalties fine[], an array of car numbers car[], and also the date. The task is to find the total fine which will be collected on the given date. Fine is collected from odd-numbered cars on even dates and vice versa.

Examples:

Example 1:

Input:
N = 4, date = 12
car[] = {2375, 7682, 2325, 2352}
fine[] = {250, 500, 350, 200}
Output:
600
Explantion:
The date is 12 (even), so we collect the fine from odd numbered cars. The odd numbered cars and the fines associated with them are as follows:
2375 -> 250
2325 -> 350
The sum of the fines is 250+350 = 600

Example 2:

Input:
N = 3, date = 8
car[] = {2222, 2223, 2224}
fine[] = {200, 300, 400}
Output:
300

Your task:

You don't need to read input or print anything. Your task is to complete the function totalFine() which takes the array car[] and fine[] its size N and an integer date as inputs and returns the total fine.

• Expected Time Complexity: $O(N)$
• Expected Auxiliary Space: $O(1)$

Constraints:

• $1<=N<=10^5$
• $1 <= Car_i <= 10^5$
• $1 <= Date <= 10^5$
• $1 <= Fine_i <= 10^5$

Solution

Python

def totalFine(self, n, date, car, fine):
    totalFine = 0
    for i in range(n):
        if date % 2 == 0:  
            if car[i] % 2 != 0:  
                totalFine += fine[i]
        else:  
            if car[i] % 2 == 0:  
                totalFine += fine[i]
    return totalFine

Java

public long totalFine( long n, long date, long car[], long fine[]) {
    long totalFine = 0;
    for (int i = 0; i < n; i++) {
        if (date % 2 == 0) {  /
            if (car[i] % 2 != 0) {  
                totalFine += fine[i];
            }
        } 
        else {  
            if (car[i] % 2 == 0) {  
                totalFine += fine[i];
            }
        }
    }
    return totalFine;  
}

C++

long long int totalFine(int n, int date, int car[], int fine[]) {
    long long int totalFine = 0;
    for (int i = 0; i < n; i++) {
        if (date % 2 == 0) {  
            if (car[i] % 2 != 0) {  
                totalFine += fine[i];
            }
        } 
        else {  
            if (car[i] % 2 == 0) {  
                totalFine += fine[i];
            }
        }
    }
    return totalFine;
}

C

long long int totalFine(int n, int date, int car[], int fine[]) {
    long long int totalFine = 0;
    for (int i = 0; i < n; i++) {
        if (date % 2 == 0) {
            if (car[i] % 2 != 0) {
                totalFine += fine[i];
            }
        } 
        else {
            if (car[i] % 2 == 0) {
                totalFine += fine[i];
            }
        }
    }
    return totalFine;
}

• Time Complexity: $O(N)$
• Auxiliary Space: $O(1)$