Question

Problem 2. Consider sorting n numbers stored in array A by first finding the smallest element of A and exchanging it with the element in A[1]. Then find the second smallest element of A, and exchange it with A[2]. Continue in this manner for the first n − 1 elements of A.

a. Write pseudocode for this algorithm, which is known as selection sort.
b. Why does it need to run for only the first n−1 elements, rather than for all n elements?

c. Give the best-case and worst-case running times of selection sort in Θ-notation.

Answer 1

a)

for i=0 to n-1:

small= i;

for j=i+1 to n:

if(a[small]<a[j]) small=j;

end if;

end for:

if(small!=i) swap(a[small],a[i])

end if;

end for;

B)

We are running the loop only till n-1. Because in selection sort we are checking for the any small elements presents than the element after it. If we use n instead of n there are no elements in n+1 as we sets j =i+1 and if i=n; So it avoid array out of bound exception.

C) How ever way the array is sorted it the outer loop runs n-1 and inner loope runs [1+2+3+4+........+n-1] times.

So worst case= O(n² ) =(n²)

best case=O(n² )=(n²)