Students Marks Sum || Hackerrank Solution

 Students Marks Sum

You are given an array of integers, , denoting the marks scored by students in a class.

• The alternating elements , ,  and so on denote the marks of boys.
• Similarly, , ,  and so on denote the marks of girls.

The array name, , works as a pointer which stores the base address of that array. In other words,  contains the address where  is stored in the memory.

For example, let  and  stores 0x7fff9575c05f. Then, 0x7fff9575c05f is the memory address of .

Complete the function, marks_summation(int* marks, char gender, int number_of_students) which returns the total sum of:

• marks of boys if 
• marks of girls if 

The locked code stub reads the elements of  along with . Then, it calls the function marks_summation(marks, gender, number_of_students) to get the sum of alternate elements as explained above and then prints the sum.

Input Format

• The first line contains , denoting the number of students in the class, hence the number of elements in .
• Each of the  subsequent lines contains .
• The next line contains .

Constraints

•  (where )
•  =  or 

Output Format

The output should contain the sum of all the aternate elements in  as explained above.

Sample Input 0

3
3
2
5
b

Sample Output 0

8

Explanation 0

 = [3, 2, 5] and  = .

So, .

Sample Input 1

5
1
2
3
4
5
g

Sample Output 1

6

Explanation 1

 = [1, 2, 3, 4, 5] and  = 

So,  =  = .

Sample Input 2

1
5
g

Sample Output 2

0

Explanation 2

 = [5] and  = 

Here,  does not exist. So,  = .

• Solution:

//Complete the following function.

int marks_summation(int* marks, int number_of_students, char gender) {

  int sum = 0;

    

    if(gender == 'b'){

        for(int i = 0; i < number_of_students; i += 2)

            sum += marks[i];

    }

    else{

        for(int i = 1; i < number_of_students; i += 2)

            sum += marks[i];

    }

    

    return sum;

}

🙏THANKS FOR VISIT🙏