Based off the question below, I have created a method and I have determined that regardless of data in random, ordered or reversed data, this has a complexity of O(n^2).

I was wondering if you could explain to me how to find the number of comparisons necessary to identify larges and smallest integers with the set of l elements, as would be determined by my algorithm. I am having troubel conceptualizing.

Lets sort using a method not discussed in class. Suppose you have n data values in in array A. Declare an array called Count. Look at the value in A[i]. Count the number of items in A that are smaller than the value in A[i]. Assign that result to count[i]. Declare an output array Output. Assign Output[count[i]] = A[i]. Think about what the size of Output needs to be. Is it n or something else? Write a method to sort based on this strategy.

Public class Array {

//main method

Public static void main (String args[] {

int j = 0;

int l = 16;

int counter;

//create array to hold values

int[] countNumbers = new int[n];

int[A] = {15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0};

int[] OutputArray = new int[n];

//create nested for loop that has counter count up to biggest number held in array.

for (int i=0; i

counter =0;

for (int k = 0; k

j++;

if (A[k]>A[i])

counter++;

countNumber[i] = counter;

output[countNumber[i]] = A[i];

For (int i = 0; i

System.out.println(OutputArray[i]);

j++;