Trace counting sort on the following list L of unordered keys which lie within the range [10, 18]. L[1:10]: 16 11 18 13 11 12 15 15 18 16
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Trace counting sort on the following list L of unordered keys which lie within
the range [10, 18].
L[1:10]: 16 11 18 13 11 12 15 15 18 16
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- Write a function that takes in n ADT Lists in an array. The function is going to be called Intersection, and it will return a new list with the elements common to all the n lists. Determine the time complexityRecamán's sequencedef recaman(n):Compute and return the first n terms of the Recamán's sequence, starting from the term a1 = 1. n Expected result 10 [1, 3, 6, 2, 7, 13, 20, 12, 21, 11] 1000000 (a list of million elements whose last five elements are [2057162,1057165, 2057163, 1057164, 2057164] )The program written for binary search, calculates the midpoint of the span as mid: (Low + High)/2. The program works well if the number of elements in the list is small (about 32,000) but it behaves abnormally when the number of elements is large. This can be avoided by performing the calculation as: =
- Apply Selection Sort on the following list of elements: 16, 23, 19, 6, 20, 10, 34, 54Recamán's sequence def recaman_item(n):Compute and return the n:th term of the Recamán's sequence. See the linked definition for this sequence on Wolfram Mathworld. Note how the definition for the current element in this sequence depends on whether some particular number can be found in the previously generated prefix of the sequence. To allow your function to execute in a sure and speedy fashion even for millions of elements, you should use a set to keep track of which elements are already part of the prefix of the sequence generated so far. This way you can generate each element in constant time, instead of having to iterate through the entire previously generated list the way some "Shlemiel" might approach this problem. n Expected result 1 1 8 12 13 23 57 87 395743 258330 The previous version of this problem, simply called recaman in the very first versions of this problem collection,…Language: Python 3 Autocomplete Ready O 1 v import ast 3. Hybrid Sort input() lst %3D 3 lst = ast.literal_eval(lst) 4 Insertion sort is a simple sorting algorithm that builds the final sorted array one item at a time. In each iteration, insertion sort inserts an element into an already sorted list (on left). The position where the item will be inserted is found through linear search. You decided to improve insertion sort by using binary search to find the position p where the new insertion should take place. 6 print(BinaryInsertionSort(lst)) Algorithm BinarylnsertionSort Input/output: takes an integer array a = {a[0], ..., a[n – 1]} of size n begin BinarylnsertionSort for i =1 to n val = a[i] p = BinarySearch(a, val, 0, i – 1) for j = i-1 to p a[j + 1]= a[i] j= j-1 end for a[p] = val i i+1 end for end BinarylnsertionSort Here, val = a[i] is the current value to be inserted at each step i into the already sorted part a[0], ..., ați – 1] of the array a. The binary search along that part…
- 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…Sort-by-Length Write a function sort_by_length(words:List[str]) -> List[str] that returns a list of words, sorted by the length of each word and if some words have the same length, sort them alphabetically. For example, sort_by_length(['this', 'is', 'a', 'test', 'for', 'sorting', 'by', 'length']) == ['a', 'by You can see that the given list is sorted by the length of each word. The shortest word 'a' is in the first position. The by and is have the same length but b comes before i. How do you implement this function? Here are some steps you can follow: 1. Start with an empty list length_word_tuples 2. For each word, create a tuple where its first element is the length of the word and the second argument is the word itself e.g. (4, 'this'). Append the tuple into length_word_tuples. 3. Sort the list length_word_tuples using sorted() function. What is the type of each element of length_word_tuples? How is this different from calling sorted (words)? 4. Create another empty list called ans.…Define the function (siftNum lst). This function should resolve to all the numbers in the list, lst. For example: (siftNum '(a 8 2 1 m l 90 p 2)) resolves to '(8 2 1 1 90 2).
- Given the following linked lists: Trace the following codeon these two linked lists and show what will be printed.Write a Python Code for the given function and conditions: Given function: def insert(self, newElement) Pre-condition: None. Post-condition: This method inserts newElement at the tail of the list. If an element with the same key as newElement already exists in the list, then it concludes the key already exists and does not insert the key.Quick Sort is used for most default sorting functions. Why is QuickSort the preferred algorithm when something like MergeSort has better/more predicatable run time?
