What will be the list after the following operations: a = [4, 2, 5] a.append(3)
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- def items_in_sets(items: List) -> int: """Given a list of numbers that represent distinct items, how many ways are there to select a single item from the union of all sets? E.g., a list of [1, 2] would mean there are 2 sets of objects, containing 1 and 2 items each respectively." items_in_sets = None # YOUR CODE HERE #raise NotImplementedError() return items_in_sets# Create a list of tuplestuples = [("p", 1), ("q", 2), ("r", 3)]# Use enumerate() to iterate over the tuples and print the index and valuefor index, value in enumerate(tuples):print(index, value)This code will print the following output:0 p1 q2 r Could you try to modify your code to get the output mentioned by you? Also, the output for the above function should be0 ('p', 1)1 ('q', 2)2 ('r', 3) Could you please confirm?Programming language : Java Write the program the should determine if there exists a 3 partition of the given list, P1, P2, and P3 such that the sum of the elements in P1 minus the sum of the elements in P2 are exactly k and the sum of the elements in P2 minus the sum of the elements in P3 are also exactly k. k is a positive integer, or zero. P1, P2, and P3 are the partitions such that |sum(P1) - sum(P2)| = k and |sum(P2) - sum(P3)| = k. For example if list is given below: 1. List = [3 1 3 2 1] k = 4 Output: True The 3 partition form above list which satifies the condition is: P1 = {2}, P2 = {3, 3}, P3 = {1, 1} Sum(P1) = 2 Sum(P2) = 6 Sum(P3) = 1 + 1 = 2 |Sum(P1) - Sum(P2)| = 4 |Sum(P2) - Sum(P3)| = 4 2. List = [49 49 36 44 43 49 48 36 32 26 38] k = 88 Output: True A 3 partition that works is…
- Programming language : Java Write the program the should determine if there exists a 3 partition of the given list, P1, P2, and P3 such that the sum of the elements in P1 minus the sum of the elements in P2 are exactly k and the sum of the elements in P2 minus the sum of the elements in P3 are also exactly k. k is a positive integer, or zero. P1, P2, and P3 are the partitions such that |sum(P1) - sum(P2)| = k and |sum(P2) - sum(P3)| = k. For example if list is given below: 1. List = [3 1 3 2 1] k = 4 Output: True The 3 partition form above list which satifies the condition is: P1 = {2}, P2 = {3, 3}, P3 = {1, 1} Sum(P1) = 2 Sum(P2) = 6 Sum(P3) = 1 + 1 = 2 |Sum(P1) - Sum(P2)| = 4 |Sum(P2) - Sum(P3)| = 4 2. List = [49 49 36 44 43 49 48 36 32 26 38] k = 88 Output: True A 3 partition that works is…Write Code to Split Even and Odd Numbers in Separate List using this list : [22,8,7,0,13,99,38,63]) (Num = Solve this question in any programming language you prefer (It is preferable to use Dart) || You can upload an image or code file 1 Add file *Write a code that contains a tuple of first name and another set contains a list of the last name, then concatenate the index of tuple A with the index of list B. Example: x= ("Ibrahim", "Foad") y = ["Adeshola", "Fortune"] Output: Ibrahim Adeshola Foad Fortune
- /** removeDuplicates returns a new array containing the unique values in the * array. There should not be any extra space in the array --- there should * be exactly one space for each unique element (Hint: numUnique tells you * how big the array should be). You may assume that the list is sorted, as * you did for numUnique. * * * Your solution may call numUnique, but should not call any other * functions. After the call to numUnique, you must go through the array * exactly one more time. Here are some examples (using "==" informally): * * * * * * * * * } new double[] { } == removeDuplicates (new double[] { }) new double[] {11} removeDuplicates (new double[] {11}) == removeDuplicates (new double[] { 11, 11, 11, 11 }) new double[] { 11, 22, 33, 44, 55, 66, 77, 88 } == removeDuplicates (new double[] { 11, 11, 11, 11, 22, 33, 44, 44, 44, 44, 44, 55, 55, 66, 77, 88, 88 }) removeDuplicates (new double[] { 11, 22, 33, 44, 44, 44, 44, 44, 55, 55, 66, 77, 88 }) == == * */ public static double…Code NOT working. What to do? Here is code: appleList = [["Apple",52,14,0,0],["McIntosh red",80,18,0,0.5],["Gala (Apple)",52,11.4,0.2,0.3],["Fuji Apple",52,11.4,0.2,0.3],["Honey Crisp Apples",90,21,0,0],["Granny Smith Apples",52,11.4,0.2,0.3],["Red Delicious Apples",80,17,0,0],["Braeburn Apples",71.7,16,0.5,0.1],["Golden Delicious Apples",130,29,1,0],["Jonagold",130.7,34,1,0],["Cripps Pink Apple",80,18,0,0],["Empire Apples",80,17,0,0],["Produce Cortland Apples",70,6,1,5],["Jazz Apples",53.8,12,0.6,0.5],["Cameo Apples", 80,22,0,0]["Winesap Apples",80,22,0,0],["Rome Apples",80,22,0,0],["Ambrosia Apples",90,17,0.4,0],] print(":Type of Apple : Calories : Carbs (in GRAMS) : Protein (in GRAMS) : Fat (in GRAMS) :") for item in appleList:print(":",item[0]," "*(9-len(item[0])),":",item[1]," "*(13-len(item[1])),":",item[2]," "*(4-len(str(item[2]))),":")3. Write the remove_evens() function that receives a list of integers as a parameter and returns a new list of integers containing only the odd numbers from the original list. Ex1) If n = [1, 2, 3, 4, 5, 6], remove_evens(n) returns [1, 3, 5]. Ex2) If n = [2, 4, 8], remov move_evens(n) returns [].
- def remove_occurences[T](xs:List[T], elem: T, n:Int) : List[A] = {} Using the function above, can you show me how to make a polymorphic function in scala where the program removes elem , n number of times from the list. If n > elem then remove all elem from the list For example: remove_occurences(List(4,5,6,7,4),4,1) -> List(5,6,7,4)remove_occurences(List(4,5,6,7,4),4,2) -> List(5,6,7)remove_occurences(List(4,5,6,7,4),2,2) -> List(4,5,6,7,4)def trip_planning(city_list): """ Question 2 You are planning the perfect trip in Europe! To do so ... - First, remove the duplicate cities in the list provided - Then, remove the countries that are not in Europe - Finally, order the cities alphabetically - Don't forget to return the resulting list of cities! Args: city_list (list) Returns: list >>> trip_planning(["Amsterdam, Europe", "Beijing, Asia", "Madrid, Europe", "Bucharest, Europe", "Lima, America"]) ["Amsterdam, Europe", "Bucharest, Europe", "Madrid, Europe"] >>> trip_planning(["London, Europe", "London, Europe", "Cairo, Africa", "Rome, Europe", "Sacramento, America"]) ["London, Europe", "Rome, Europe"] # print(trip_planning(["Amsterdam, Europe", "Beijing, Asia", "Madrid, Europe", "Bucharest, Europe", "Lima, America"])) # print(trip_planning(["London, Europe", "London, Europe", "Cairo, Africa", "Rome, Europe", "Sacramento, America"]))Having trouble with creating the InsertAtEnd function in the ItemNode.h file below. " // TODO: Define InsertAtEnd() function that inserts a node // to the end of the linked list" Given main(), define an InsertAtEnd() member function in the ItemNode class that adds an element to the end of a linked list. DO NOT print the dummy head node. Ex. if the input is: 4 Kale Lettuce Carrots Peanuts ------------------------------------------------------- main.cpp -------------------------------------------------------- #include "ItemNode.h" int main() { ItemNode *headNode; // Create intNode objects ItemNode *currNode; ItemNode *lastNode; string item; int i; int input; // Front of nodes list headNode = new ItemNode(); lastNode = headNode; cin >> input; for (i = 0; i < input; i++) { cin >> item;…