The function interleave_lists in python takes two parameters, L1 and L2, both lists. Notice that the lists may have different lengths. The function accumulates a new list by appending alternating items from L1 and L2 until one list has been exhausted. The remaining items from the other list are then appended to the end of the new list, and the new list is returned. For example, if L1 = ["hop", "skip", "jump", "rest"] and L2 = ["up", "down"], then the function would return the list: ["hop", "up", "skip", "down", "jump", "rest"]. HINT: Python has a built-in function min() which is helpful here. Initialize accumulator variable newlist to be an empty list Set min_length = min(len(L1), len(L2)), the smaller of the two list lengths Use a for loop to iterate k over range(min_length) to do the first part of this function's work. On each iteration, append to newlist the item from index k in L1, and then append the item from index k in L2 (two appends on each iteration). AFTER the loop finishes, append the remaining items (if any) to newlist; use a decision statement to determine which list has remaining items -- then it just takes one line of code to append those items to newlist (using a slice of L1 or L2) return newlist For example: Test Result L1 = ["hop", "skip", "jump"] L2 = ["A", "B", "C", "D", "E"] print(interleave_lists(L1, L2)) ['hop', 'A', 'skip', 'B', 'jump', 'C', 'D', 'E'] L1 = ["dive"] L2 = ["hop", "skip", "jump"] print(interleave_lists(L1, L2)) ['dive', 'hop', 'skip', 'jump'] cats = ["Ziggy", "Troll Boy"] dogs = ["Ruthie", "Ame"] print(interleave_lists(cats, dogs)) ['Ziggy', 'Ruthie', 'Troll Boy', 'Ame

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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The function interleave_lists in python takes two parameters, L1 and L2, both lists. Notice that the lists may have different lengths.

The function accumulates a new list by appending alternating items from L1 and L2 until one list has been exhausted. The remaining items from the other list are then appended to the end of the new list, and the new list is returned.

For example, if L1 = ["hop", "skip", "jump", "rest"] and L2 = ["up", "down"], then the function would return the list: ["hop", "up", "skip", "down", "jump", "rest"].

HINT: Python has a built-in function min() which is helpful here.

Initialize accumulator variable newlist to be an empty list
Set min_length = min(len(L1), len(L2)), the smaller of the two list lengths
Use a for loop to iterate k over range(min_length) to do the first part of this function's work. On each iteration, append to newlist the item from index k in L1, and then append the item from index k in L2 (two appends on each iteration).
AFTER the loop finishes, append the remaining items (if any) to newlist; use a decision statement to determine which list has remaining items -- then it just takes one line of code to append those items to newlist (using a slice of L1 or L2)
return newlist

For example:

Test	Result
L1 = ["hop", "skip", "jump"] L2 = ["A", "B", "C", "D", "E"] print(interleave_lists(L1, L2))	['hop', 'A', 'skip', 'B', 'jump', 'C', 'D', 'E']
L1 = ["dive"] L2 = ["hop", "skip", "jump"] print(interleave_lists(L1, L2))	['dive', 'hop', 'skip', 'jump']
cats = ["Ziggy", "Troll Boy"] dogs = ["Ruthie", "Ame"] print(interleave_lists(cats, dogs))	['Ziggy', 'Ruthie', 'Troll Boy', 'Ame']

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