Question

Does this look correct?

Add operation counts - 0.5pt
   * f(N) formula (show your work) - 0.5pt
   * O(N) reduction - 0.5pt

public static long sum1(int N)//f(N) = ; O(N) =
   {
       long opCount = 0;
       long sum = 0; //1
       opCount++;
       opCount++;
       opCount++;
       for(int i = 0; i < N; i++) //2 + 2N
       {
           sum++; //N
           opCount+=2;
       }
       opCount++;
       System.out.println("f(N) = [f(N)= 3N + 4]");
       System.out.println("O(N) = [O(N)=O(N)]");
       System.out.println("OpCount : "+opCount);
       return sum; // 1
   }

Answer 1

//do comment if any problem arises

//highlighted code is your function

//there are some bugs in given code according to formula f(N)= 3N + 4 which has been fixed

import java.util.Scanner;

class Sum {

public static long sum1(int N)// f(N) = ; O(N) =

{

long opCount = 0;

long sum = 0; // 1

for (int i = 0; i < N; i++) // 2 + 2N

{

sum++; // N

opCount += 3;

}

opCount++;

opCount++;

opCount++;

opCount++;

System.out.println("f(N) = [f(N)= 3N + 4]");

System.out.println("O(N) = [O(N)=O(N)]");

System.out.println("OpCount : " + opCount);

return sum; // 1

}

public static void main(String[] args) {

Sum object = new Sum();

Scanner sc = new Scanner(System.in);

System.out.print("Enter value of N: ");

int N = sc.nextInt();

System.out.println("O(N) = " + object.sum1(N) + "\n");

}

}

Output: