Complete the recursive function remove_last which creates a new list identical to the input list s but with the last element in the sequence that is equal to x removed. Hint: Remember that you can use negative indexing on lists! For example 1st [-1] refers to the last element in a list 1st, lst [-2] refers to the second to last element... def remove_last(x, s): """Create a new list that is identical to s but with the last element from the list that is equal to x removed. >>> remove_last(1, []) [] remove_last(1, [1]) >>> remove_last(1, [1,1]) [1] >>> remove_last(1, [2,1]) [2] >>> [] >>> remove_last(1, [3,1,2]) [3, 2] >>> remove_last(1, [3,1,2,1]) [3, 1, 2] >>> remove_last (5, [3, 5, 2, 5, 11]) [3, 5, 2, 11] HIGH "*** YOUR CODE HERE ***"

Complete the recursive function remove_last which creates a new list identical to the input list s but with the last element in the sequence that is equal to x removed. Hint: Remember that you can use negative indexing on lists! For example 1st [-1] refers to the last element in a list 1st, lst [-2] refers to the second to last element... def remove_last(x, s): """Create a new list that is identical to s but with the last element from the list that is equal to x removed. >>> remove_last(1, []) [] remove_last(1, [1]) >>> remove_last(1, [1,1]) [1] >>> remove_last(1, [2,1]) [2] >>> [] >>> remove_last(1, [3,1,2]) [3, 2] >>> remove_last(1, [3,1,2,1]) [3, 1, 2] >>> remove_last (5, [3, 5, 2, 5, 11]) [3, 5, 2, 11] HIGH "* YOUR CODE HERE *"

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

Complete the recursive function remove_last which creates a new list identical to the input list s but
with the last element in the sequence that is equal to x removed.
Hint: Remember that you can use negative indexing on lists! For example 1st [-1] refers to the last
element in a list 1st, lst [-2] refers to the second to last element...
def remove_last(x, s):
"""Create a new list that is identical to s but with the last
element from the list that is equal to x removed.
>>> remove_last(1, [])
[]
>>> remove_last(1, [1])
[]
>>>
[1]
>>>
[2]
remove_last(1, [1,1])
remove_last(1, [2,1])
>>> remove_last(1, [3,1,2])
[3, 2]
>>> remove_last(1, [3,1,2,1])
[3, 1, 2]
>>> remove_last (5, [3, 5, 2, 5, 11])
[3, 5, 2, 11]
IIIII
"*** YOUR CODE HERE ***"
Illustrated here is a more complete doctest that shows good testing methodology. It is a little cumbersome
as documentation, but you'll want to think about it for your projects. Test every condition that might come
up. Then you won't be surprised when it does.

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A sequential search of a sorted list can halt when the target is less than a given element in the list. Modify the program to stop when the target becomes less than the current value being compared. In a sorted list, this would indicate that the target is not in the list and searching the remaining values is unnecessary. ----------------------------------------------------------------------------------- """ File: search.py Project 11.1 """ def sequentialSearch(target, lyst): """Returns the position of the target item if found, or -1 otherwise. The lyst is assumed to be sorted in ascending order.""" position = 0 while position < len(lyst): if target == lyst[position]: return position position += 1 return -1 def main(): """Tests with three lists.""" print(sequentialSearch(3, [0, 1, 2, 3, 4])) print(sequentialSearch(3, [0, 1, 2])) # Should stop at second position. print(sequentialSearch(3, [0, 4, 5, 6])) if…