A list of lists of numbers is a list such that all its elements are lists of numbers. Consider the following examples: • [[3,5], [], [1, 1, 1]] is a list of list of numbers; • [ [ [3,5]]], [], [1]] is not a list of list of numbers due to the first element; • [2, [1]] is not a list of list of numbers due to the first element, but note that the tree that encodes this list also encodes [[0,0],[1]] which is a list of list of numbers. Note that an empty list can always be considered a list of numbers and a list of lists of numbers. Answer for EACH of the given trees (a)–(d) the following question: Does it encode a list of numbers? If it does which is the corresponding list of numbers? USE THE ENCODING OF DATATYPES IN D BELOW (a) ((nil.nil).((nil. (nil.nil)).<

def reverse_list (1st: List [Any], start: int, end: int) -> None: """Reverse the order of the items in list , between the indexes and (including the item at the index). If start or end are out of the list bounds, then raise an IndexError. The function must be implemented **recursively**. >>> lst ['2', '2', 'v', 'e'] >>>reverse_list (lst, 0, 3) >>> lst ['e', 'v', 'i', '2'] >>> lst [] >>>reverse_list (lst, 0, 0) >>> lst [0] >>> Ist = [] >>> reverse_list (lst, 0, 1) Traceback (most recent call last): IndexError #1 #1

Python

• find_last(my_list, x): Takes two inputs: the first being a list and the second being any type. Returns the index of the last element of the list which is equal to the second input; if it cannot be found, returns None instead. >> find_last(['a', 'b', 'b', 'a'], 'b') 2 >>> ind = find_last(['a', 'b', 'b', 'a'], 'c') >>> print(ind) None

A-1: Let l = [−1, −2, . . . , −10] be an existing list. Construct a new list from l by dividing each evennumber element in l by 2. Print the new list. Do not use NUMPY library.A-2: Create a new list containing the following elements using list comprehension:[1, −1, 2, −2, 3, −3, ..., 9, −9]A-3: Consider the following coded string. Create a list of all contiguous (connected without space) lettersin the order of their appearance."Vjg dguv rtqitcou ctg ytkvvgp uq vjcv eqorwvkpi ocejkpgu ecp rgthqto vjgoswkemn{0 Cnuq. vjg dguv rtqitcou ctg ytkvvgp uq vjcv jwocp dgkpiu ecp wpfgtuvcpfvjgo engctn{0 C iqqf guuc{kuv cpf c iqqf rtqitcoogt jcxg c nqv kp eqooqp0"A-4: In the list obtained after executing task(s) from Part(A-3), remove newline characters (if any).A-5: For the list obtained after executing task(s) from Part(A-4), for each word in the list do thefollowing:ˆ Breakdown the word into list of letters.ˆ Convert each of the above letters into integer, using ord() function.ˆ Update the…

Word frequencies (dictionaries) Implement the build_dictionary() function to build a word frequency dictionary from a list of words. Ex: If the words list is: ["hey", "hi", "Mark", "hi", "mark"] the dictionary returned from calling build_dictionary(words) is: {'hey': 1, 'hi': 2, 'Mark': 1, 'mark': 1}   half answered: code: def build_dictionary(words):# The frequencies dictionary will be built with your code below.# Each key is a word string and the corresponding value is an integer # indicating that word's frequenc # The following code asks for input, splits the input into a word list, # calls build_dictionary(), and displays the contents sorted by key.if __name__ == '__main__':words = input().split()your_dictionary = build_dictionary(words)sorted_keys = sorted(your_dictionary.keys())for key in sorted_keys:print(key + ': ' + str(your_dictionary[key]))

Create a function that takes two numbers as arguments (num, length) and returns a list of multiples of num until the list length reaches length. Examples list_of_multiples(7, 5) → [7, 14, 21, 28, 35] list_of_multiples(12, 10) → [12, 24, 36, 48, 60, 72, 84, 96, 108, 120] list_of_multiples(17, 6) [17, 34, 51, 68, 85, 102]

def large_matrix(matrix: list[list[int]]) -> int:"""Returns the area of the rectangle in the area of a rectangle is defined by number of 1's that it contains.The matrix will only contain the integers 1 and 0.>>> case = [[1, 0, 1, 0, 0],... [1, 0, 1, 1, 1],... [1, 1, 1, 1, 1],... [1, 0, 0, 1, 0]]>>> largest_in_matrix(case1)6"" You must use this helper code: def large_position(matrix: list[list[int]], row: int, col: int) -> int: a = rowb = colmax = 0temp = 0rows = len(matrix)column = len(matrix[a])while a < rows and matrix[a][col] == 1:temp = 0while b < column and matrix[a][b] == 1:temp = temp + 1b = b + 1column = ba = a + 1if (a != row+1):temp2 = temp * (a - row)else:temp2 = tempif max < temp2:max = temp2b = colreturn max """ remember:  Please do this on python  you should not use any of the following: dictionaries or dictionary methods try-except break and continue statements recursion map / filter import""'

Given an IntNode struct and the operating functions for a linked list, complete the following functions to extend the functionality of the linked list(this is not graded).

def large_rec(m: list[list[int]], r: int, c: int) -> int:"""Do not use any imports,  dictionaries, try-except statements or break and continue  please you must make use of the helper long here as you loop througheach row of the matrix returns the area of the largest rectangle whose top left corner is atposition r, c in m.>>> case1 = [[1, 0, 1, 0, 0],... [1, 0, 1, 1, 1],... [1, 1, 1, 1, 1],... [1, 0, 0, 1, 0]]>>> largest_at_position(case1, 1, 2)6""" Helper code: def long(lst: list[int]) -> int: count = 0i = 0while i < len(lst) and lst[i] != 0:i = i + 1count = count + 1return count