Identify the unit count for each portion of the code given below to identify the

Question

Identify the unit count for each portion of the code given below to identify the overall Big Oh speed. Use the example below as a template for your answer, or some classes may use summations. The table answers below are NOT correct!! But the overall setup is. Partial credit will be given if some portion of the work shown is correct.

Code

Cost

# of Times

Total

int sum1(int N)           

{

     int s = 0;

     for(int i = 1; i <= N; i++)

     {

           s = s + i;

     }

    

     return s;

}

-

-

1

1

1

-

-

1

1

1

1 (int i) + 1 (<=) + 1 (++)

1 + 2(n + 1)

2n + 3

-

-

2

n

2n

-

-

-

-

1

1

1

-

-

Final

4n + 6

O(n)

MAKE THE TABLE FOR THIS EXAMPLE

#4

sum = 0;

for ( i = 0; i < n; i++ )

for (j = 0; j < n; j++ )

++sum;

Code

Cost

# of Times

Total

int sum1(int N)           

{

     int s = 0;

     for(int i = 1; i <= N; i++)

     {

           s = s + i;

     }

    

     return s;

}

-

-

1

1

1

-

-

1

1

1

1 (int i) + 1 (<=) + 1 (++)

1 + 2(n + 1)

2n + 3

-

-

2

n

2n

-

-

-

-

1

1

1

-

-

Final

4n + 6

O(n)

Explanation / Answer

first of all I explained your questions in brief...

The running time for the operation sum++ is a constant. The outer loop runs n times, The nested loop also runs n times, hence the complexity would be O( n 2 ) .

n2 + n +1

O(n2)

final answers is O(n2).

Code Cost Of times Total Sum=0 - - - for(i=0;i<n;i++) o(i)+o(<)+o(++) o+(n) n for(j=0;j<n;j++) o(j)+o(<)+o(++) o+n(n) o(n2) ++sum 1 1 1 Final

n2 + n +1

O(n2)

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