elements. # TODO 1.5 first_ten_even_elements =
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1.5
Return the top ten entries of the even_list using list indexing. Put the results in first_ten_even_elements.
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- Q4 Suppose, alist is a 2-dimensional list that is defined as follows. alist = [ [1], [2,3], [5,6,7] ] Which is the best option to access the last element of the last element of this list (i.e., 7)? Question 4 options: alist[ len(alist) - 1 ][len(alist[len(alist) - 1]) - 1] alist[ len(alist) - 1 ][len(alist[len(alist) - 1])] alist[3][3] alist[ len(alist) - 1 ][1]Considering all the elements and its indexes at the right side:Write a sequence of List operations that would result to a LIST that contains ONLY the elements with its respective indexes: {index[00]=104, index[01]=165, index[02]=115, index[03]=384, index[04]=248, index[05]=117}.TRUE or FALSE? Answer the following question and state the reason why: The delete operation only involves the removing of the node from the list without breaking the links created by the next node. You need an array to represent each node in a linked list. STL lists are also efficient at adding elements at their back because they have a built-in pointer to the last element in the list. A circular linked list has 2 node pointers. cout<<list.back()<<endl; = The back member function returns a reference to the last element in the list. In a Dynamic Stack, the pointer top stays at the head after a push operation. During a Pop operation in Static Stack, the elements are being moved one step up. In a dynamic implementation of stack, the pointer top has an initial value of null. In a dynamic stack, the node that was popped is deleted. In a dynamic stack, the pointer top stays at the head after push operation. STL function top returns a reference to element at the top of the…
- Question 30 If N represents the number of elements in the list, then the index-based add method of the LBList class is O(N). True False Question 31 A header node does not contain actual list information. True False Question 32 Any class that implements the Comparable interface must provide a compareTo method. True False Question 33 A SortedABList list can only be kept sorted based on the "natural order" of its elements. True False Question 34 O(N) is the order of growth execution time of the add operation when using the SortedArrayCollection class, assuming a collection size of N. True False Question 35 The iterator operation is required by the Iterable interface. Group of answer choices True False Question 36 O(N) is the order of growth execution time of the remove operation when using the SortedArrayCollection class, assuming a collection size of N. True False Question 37 O(N) is the order of growth execution time of the index-based add operation when…Consider the following piece of pseudocode:new DynamicArray dd[1] ← 1d[2] ← 2d[3] ← 3d[2] ← Ød[1] ← ØWhich of the following describes a linked list implementation of this pseudocode?a. Create a new empty linked list, insert a node (with value 1) with next pointer to null, insert a node (with value 2) at the end of list, insert a node (with value 3) at the end of list, remove end node.b. reate a new empty linked list, insert a node (with value 1) with next pointer to null, insert a node (with value 2) at the end of list, insert a node (with value 3) at the end of list, remove middle node, remove first node.c. Create a new empty linked list, insert a node (with value 1) with next pointer to null, insert a node (with value 2) at the beginning of list, insert a node (with value 3) at the end of list, remove end node.Suppose, alist is a 2-dimensional list that is defined as follows. alist = [ [1], [2,3], [5,6,7] ] Which is the best option to access the last element of the last element of this list (i.e., 7)? Question options: a. alist[ len(alist) - 1 ][len(alist[len(alist) - 1]) - 1] b. alist[ len(alist) - 1 ][len(alist[len(alist) - 1])] c. alist[3][3] d. alist[ len(alist) - 1 ][1]
- Add a proper function to the following linked list that will copy all elements with value above average into another list , then apply it to the list created in main function and print your resultant list. Example: l.copy(l2); will copy all elements with value greater than average into l2 #include <iostream> using namespace std; struct node { int data; node *next,*prev; node(int d,node *p=0,node *n=0) { data=d; prev=p; next=n; } }; class list { node *head; public: list(); void print(); void add_end(int el); }; list::list() { head=0; } void list::add_end(int el) { if(head==0) head=new node(el); else { node *t=head; for(;t->next!=0;t=t->next); t->next=new node(el,t); } } void list::print() { for(node *t=head;t!=0;t=t->next) cout<<t->data<<" "; cout<<endl; } void main() { list l; l.add_end(7); l.add_end(3); l.add_end(9);…Please do not copy from chegg. C++ Dividing a linked list into two sublists of almost equal sizes Add the operation divideMid to the class ListData in Chapter17.cpp on Canvas as follows: void divideMid(ListData &sublist); //This operation divides the given list into two sublists //of (almost) equal sizes. //Postcondition: first points to the first node and last // points to the last node of the first sublist. // sublist.first points to the first node // and sublist.last points to the last node // of the second sublist. Consider the following statements: ListData myList; ListData subList; Suppose myList points to the list with elements 34 65 27 89 12 (in this order). The statement: myList.divideMid(subList); divides myList into two sublists: myList points to the list with the elements 34 65 27, and subList points to the sublist with the elements 89 12. Write the definition of the function template to implement the operation divideMid. chapter 17.cpp #include…9.7 Create a CircularList class method named reverse that allows you to reverse the order of any element within the list. This approach ought to be detrimental. 9.8 In order to remove the last element from a single linked list, we must remove each of the n references. Are all n additional preceding references required in a doubly linked list in order to delete the list's tail in constant time? (Hint: What would occur if Nodes and DoublyLinkedNodes were combined?)
- Modify the given code to move the string from leftmost side to the rightmost side of the screen (no trails). Given Code: .model small.stack.data strg db 'Lemon$' row db 12 col db 0.codemain proc mov ax,@data mov ds,ax mov ah,6 mov al,0 mov bh,7 mov ch,0 mov cl,0 mov dh,24 mov dl,79 mov cx,40again:push cx mov ah,2 mov bh,0 mov dh,row ;row mov dl,col ;col int 10h mov ah,9 mov dx,offset strg int 21h inc col mov cx,10 push cx mov cx,0ffffhx: loop x pop cx loop y pop cx loop again mov ah,4ch int 21hmain endpend mainWrite a function with the signature below that returns the sum of the last k elements of a singly linked list that contains integers. int returnSumOfLastKNodes(Node* head, int k) Example: 10 -> 5->8->15->11->9->23 10 represents the head node, returnSumOfLastKNodes(Node* head, 4) will return 58.Consider the below code of circular linked list. Your work is to update the same code, by providing the functionalities with tail pointer only. #include <iostream> using namespace std; class Node{ public: int data; Node *next; }; class List{ public: Node *head; List(){ head = NULL; } void insert(int data){ Node *nn = new Node; nn->data = data; if(head==NULL){ head = nn; nn->next = head; cout<<"Node inserted successfully at head"<<endl; } else{ Node *temp = head; while(temp->next!=head) temp = temp->next; nn->next = head; temp->next=nn; cout<<"Node inserted successfully at end"<<endl; } } void insertInBetween(Node *node, int data){ Node *nn = new Node; nn->data = data; Node *temp = node->next;…