I have been trying to figure this assignment but I can't. can this be solved and explained?

A2Driver.java

import java.util.Arrays;

public class A2Driver {

public static void main(String[] args) {

int N = 10;

System.out.println("N = "+N);

String[] ans = A2Work.sum0(N);

System.out.println("Sum0:");

System.out.println((ans[0].equals("f(N) = 5N+4"))?"f(N) Correct (0.5pts)":ans[0]+" : f(N) = 5N+4");

System.out.println((ans[1].equals("O(N) = N"))?"O(N) Correct (0.5pts)":ans[1]+" : O(N) = N");

long count = Long.parseLong(ans[2]);

long correct = 54;

double accuracy = Math.abs(correct-count)/(double)correct*100;//allow up to 5% off

System.out.println((accuracy <= 5)?"opCount Correct (0.5pts)":ans[2]+" : 504");

System.out.println();

System.out.println("Sum0:"+Arrays.toString(A2Work.sum0(N)));

System.out.println("Sum1:"+Arrays.toString(A2Work.sum1(N)));

System.out.println("Sum2:"+Arrays.toString(A2Work.sum2(N)));

System.out.println("Sum3:"+Arrays.toString(A2Work.sum3(N)));

System.out.println("Sum4:"+Arrays.toString(A2Work.sum4(N)));

System.out.println("Sum5:"+Arrays.toString(A2Work.sum5(N)));

System.out.println("Sum6:"+Arrays.toString(A2Work.sum6(N)));

System.out.println();

System.out.println("N = "+N);

System.out.println("Alg 1");

Object[] answer = A2Work.algorithm1(N);

System.out.println("Array Data Check:"+checkArray((int[])answer[0]));

System.out.println(Arrays.toString((String[])answer[1]));

System.out.println("Alg 2");

answer = A2Work.algorithm2(N);

System.out.println("Array Data Check:"+checkArray((int[])answer[0]));

System.out.println(Arrays.toString((String[])answer[1]));

System.out.println("Alg 3");

answer = A2Work.algorithm3(N);

System.out.println("Array Data Check:"+checkArray((int[])answer[0]));

System.out.println(Arrays.toString((String[])answer[1]));

System.out.println();

}

public static boolean checkArray(int[] arr)

{

Arrays.sort(arr);

int size = arr.length;

for(int i = 1; i < size; i++)

{

if(arr[i] == arr[i-1])//all values must be unique

{

return false;

}

if(arr[i] >= size)//no value larger than the size, use zero to N-1

{

return false;

}

}

return true;

}

}

A2Work.java

import java.util.Random;

public class A2Work {

/*

* NOTES:

* Include notes about f(N) for partial credit, similar to what I put in sum0()

* your f(N) does not need to produce the exact answer as your op count for all sums, but it should be close

* use estimates based on rules we discussed to determine f(N)

* For algorithm 1-3, assume random number generator never produces the same value twice in a given loop, but that it does produce the worst possible order of values

*

*/

/*

* Grading:

* THIS IS AN EXAMPLE

*/

public static String[] sum0(int N)

{

//DO NOT COUNT IN opCount

long opCount = 0;

String fn = "f(N) = 5N+4";//sum assignment, loop, and return combined

String On = "O(N) = N";//Growth rate of function as value of N changes - Linear

//BEGIN opCounts

long sum = 0;

opCount++;//assignment of sum

opCount++;//assignment of i - happens as loop starts

opCount++;//comparison - happens as loop starts

for(int i = 0; i < N; i++)//5N+2 - +2 for components that happen as loop starts

{

//all work inside of loop is 5 units

//loop runs N times

//internal work of loop is 5N

sum++;

opCount+=2;//sum++ :: sum=sum+1 - addition and assignment

opCount+=2;//i++ :: i=i+1 - addition and assignment at end of each iteration of loop

opCount++;//comparison - happens at end of loop

}

opCount++;//return - don't need to count construction of return array for this assignment

return new String[] {fn, On, opCount+""};

}

/*

* Grading:

* Add operation counts - 0.5pt

* f(N) formula (show your work) - 0.5pt

* O(N) reduction - 0.5pt

*/

public static String[] sum1(int N)

{

//DO NOT COUNT IN opCount

long opCount = 0;

String fn = "f(N) = [answer]";

String On = "O(N) = [answer]";

//BEGIN opCounts

long sum = 0;

for(int i = 0; i < N*N; i++)

{

sum++;

}

return new String[] {fn, On, opCount+""};

}

/*

* Grading:

* Add operation counts - 0.5pt

* f(N) formula (show your work) - 0.5pt

* O(N) reduction - 0.5pt

*/

public static String[] sum2(int N)

{

//DO NOT COUNT IN opCount

long opCount = 0;

String fn = "f(N) = [answer]";

String On = "O(N) = [answer]";

//BEGIN opCounts

long sum = 0;

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

{

for(int j = 0; j < N; j++)

{

sum++;

}

}

return new String[] {fn, On, opCount+""};

}

/*

* Grading:

* Add operation counts - 0.5pt

* f(N) formula (show your work) - 0.5pt

* O(N) reduction - 0.5pt

*/

public static String[] sum3(int N)

{

//DO NOT COUNT IN opCount

long opCount = 0;

String fn = "f(N) = [answer]";

String On = "O(N) = [answer]";

//BEGIN opCounts

long sum = 0;

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

{

for(int j = 0; j < N*N; j++)

{

sum++;

}

}

return new String[] {fn, On, opCount+""};

}

/*

* Grading:

* Add operation counts - 0.5pt

* f(N) formula (show your work) - 0.5pt

* O(N) reduction - 0.5pt

*/

public static String[] sum4(int N)

{

//DO NOT COUNT IN opCount

long opCount = 0;

String fn = "f(N) = [answer]";

String On = "O(N) = [answer]";

//BEGIN opCounts

long sum = 0;

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

{

for(int j = 0; j < i; j++)

{

sum++;

}

}

return new String[] {fn, On, opCount+""};

}

/*

* Grading:

* Add operation counts - 0.5pt

* f(N) formula (show your work) - 0.5pt

* O(N) reduction - 0.5pt

*/

public static String[] sum5(int N)

{

//DO NOT COUNT IN opCount

long opCount = 0;

String fn = "f(N) = [answer]";

String On = "O(N) = [answer]";

//BEGIN opCounts

long sum = 0;

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

{

for(int j = 0; j < i*i; j++)

{

for(int k = 0; k < j; k++)

{

sum++;

}

}

}

return new String[] {fn, On, opCount+""};

}

/*

* Grading:

* Add operation counts - 0.5pt

* f(N) formula (show your work) - 0.5pt

* O(N) reduction - 0.5pt

*/

public static String[] sum6(int N)

{

//DO NOT COUNT IN opCount

long opCount = 0;

String fn = "f(N) = [answer]";

String On = "O(N) = [answer]";

//BEGIN opCounts

long sum = 0;

for(int i = 1; i < N; i++)//i starts at 1 to prevent division error in if statement

{

for(int j = 0; j < i*i; j++)

{

if(j%i == 0)

{

for(int k = 0; k < j; k++)

{

sum++;

}

}

}

}

return new String[] {fn, On, opCount+""};

}

/*

* Grading:

* Correctly follow the described algorithm to complete the method - 1.5pt

* Add operation counts - 0.5pt

* f(N) formula (show your work) - 0.5pt

* O(N) reduction - 0.5pt

*/

public static Object[] algorithm1(int N)

{

//DO NOT COUNT IN opCount

long opCount = 0;

String fn = "f(N) = [answer]";

String On = "O(N) = [answer]";

//BEGIN opCounts

int[] arr = new int[N];

Random randGen = new Random(123456789);

/*

* Use the following method to fill the array

* For each position in the array, generate a random number between zero and N

* - If N = 10, random numbers should be 0-9

* Check if that random number is used in any previous position in the array

* - If it is used anywhere else, generate a new number and try again

* - If it is not used anywhere else, place it into the position and move forward

*/

return new Object[] {arr, new String[] {fn, On, opCount+""}};

}

/*

* Grading:

* Correctly follow the described algorithm to complete the method - 1.5pt

* Add operation counts - 0.5pt

* f(N) formula (show your work) - 0.5pt

* O(N) reduction - 0.5pt

*/

public static Object[] algorithm2(int N)

{

//DO NOT COUNT IN opCount

long opCount = 0;

String fn = "f(N) = [answer]";

String On = "O(N) = [answer]";

//BEGIN opCounts

int[] arr = new int[N];

boolean[] used = new boolean[N];

Random randGen = new Random(123456789);

/*

* Use the following method to fill the array

* For each position in the array, generate a random number between zero and N

* - If N = 10, random numbers should be 0-9

* Check if that used[random] is true

* - If it is, generate a new number and try again

* - If it is not, place it into the position, set used[random] = true, and move forward

*/

return new Object[] {arr, new String[] {fn, On, opCount+""}};

}

/*

* Grading:

* Correctly follow the described algorithm to complete the method - 1.5pt

* Add operation counts - 0.5pt

* f(N) formula (show your work) - 0.5pt

* O(N) reduction - 0.5pt

*/

public static Object[] algorithm3(int N)

{

//DO NOT COUNT IN opCount

long opCount = 0;

String fn = "f(N) = [answer]";

String On = "O(N) = [answer]";

//BEGIN opCounts

int[] arr = new int[N];

Random randGen = new Random(123456789);

/*

* Use the following method to fill the array

* Fill the arr with zero to N-1 in order

* Run a loop through each position

* - For each position, swap that position and a randomly chosen position

*/

return new Object[] {arr, new String[] {fn, On, opCount+""}};

}

}