Question

1. Shown below is the code for the insertion sort consisting of two recursive methods that
replace the two nested loops that would be used in its iterative counterpart:
void insertionSort(int array[])
{
insert(array, 1);
}
void insert(int[] array, int i)
{
if (i < array.length)
{
int value = array[i];
int j = shift(array, value, i);
array[j] = value;
insert(array, i + 1);
}
}
int shift(int[] array, int value, int i)
{
int insert = i;
if (i > 0 && array[i - 1] > value)
{
array[i] = array[i - 1];
insert = shift(array, value, i - 1);
}
return insert;
}
Draw the recursion tree for insertionSort when it is called for an array of length 5 with
data that represents the worst case. Show the activations of insertionSort, insert and
shift in the tree. Explain how the recursion tree would be different in the best case.

2. Refer back to the recursion tree you provided in the previous problem. Determine a
formula that counts the numbers of nodes in that tree. What is Big- for execution time?
Determine a formula that expresses the height of the tree. What is the Big- for memory?

**I only need help on number 2.**

Answer 1

2)
Formula for that counts the number of nodes in recursion tree: C(n) = n*(n-1)/2 //n is number of elements //derived from C(n) = C(n-1)+n
execution time : T(n) = n*(n-1)/2 = O(n^2)
formula for height: 2 //it constant, the max height possible is 2 only //because we consider recursive calls as parallel nodes
bigO for memory:
since we are using recursion, it needs stack space so: O(n^2) //for each recursive call
if we are not considering stack space then
space is :O(1)