Question

(7 marks) Consider the array A = 17, 9, 2, 5). Trace through a stack-based implemen- tation of selection sort on this array. Assume that during each step, you scan through the entire array to find the minimum value, then replace that index of the array with null. After finding the minimum value in each step (write down what this number is), draw 3 stacks: (1) the sorted stack with the new number in it, (2) the temp stack holding whatever numbers are necessary to have drawn the previous sorted stack, and (3) the sorted stack with everything from temp in it.

Answer 1

Given, A = { 7, 9, 2 , 5 }

• sorted - after the min element is added in each step
• temp - after min element is popped and pushed to sorted (since we are asked to draw temp holding whatever numbers are required to draw the sorted stack)
• unsorted (this is the initial stack, which holds the numbers to sort. It is this stack to which we push numbers back from the temp stack) - after numbers are pushed from the temp stack.

Step #1:

• Iterate throgh the stack A, find the minimum element. For this, we need to pop all elements from A and push to the temp stack, min = 2
• pop all elements from temp upto min (excluding), and push to A
• pop min from temp and push to sorted stack.
• push NULL to A
• pop all elements from temp, and push to A

min = 2   

sorted	temp	unsorted A
2	5	7 9 NULL 5

Step #2:

min = 5

sorted	temp	unsorted A
2 5		7 9 NULL NULL

Step #3:

min = 7

sorted	temp	unsorted A
2 5 7	NULL NULL 9	NULL 9 NULL NULL

Step #4:

min = 9

sorted	temp	unsorted A
2 5 7 9	NULL NULL	NULL NULL NULL NULL

I hope it's clear. For the stack based implementation of selection sort, we iterate through the given stack, popping elements from it and pushing to temp and finding minimum element in each step. Once we have the minimum element, we have to pop out all elements from temp upto min and push to the initial stack. And pop min from temp andpush to sorted stack.