Answer:
Sequence of popped values: h,s,f.
State of stack (from top to bottom): m, d
Explanation:
Assuming that stack is initially empty. Suppose that p contains the popped values. The state of the stack is where the top and bottom are pointing to in the stack. The top of the stack is that end of the stack where the new value is entered and existing values is removed. The sequence works as following:
push(d) -> enters d to the Stack
Stack:
d ->top
push(h) -> enters h to the Stack
Stack:
h ->top
d ->bottom
pop() -> removes h from the Stack:
Stack:
d ->top
p: Suppose p contains popped values so first popped value entered to p is h
p = h
push(f) -> enters f to the Stack
Stack:
f ->top
d ->bottom
push(s) -> enters s to the Stack
Stack:
s ->top
f
d ->bottom
pop() -> removes s from the Stack:
Stack:
f ->top
d -> bottom
p = h, s
pop() -> removes f from the Stack:
Stack:
d ->top
p = h, s, f
push(m) -> enters m to the Stack:
Stack:
m ->top
d ->bottom
So looking at p the sequence of popped values is:
h, s, f
the final state of the stack:
m, d
end that is the top of the stack:
m
Answer:
def SwapMinMax ( myList ):
myList.sort()
myList[0], myList[len(myList)-1] = myList[len(myList)-1], myList[0]
return myList
Explanation:
By sorting the list, you ensure the smallest element will be in the initial position in the list and the largest element will be in the final position of the list.
Using the len method on the list, we can get the length of the list, and we need to subtract 1 to get the maximum element index of the list. Then we simply swap index 0 and the maximum index of the list.
Finally, we return the new sorted list that has swapped the positions of the lowest and highest element values.
Cheers.
