C function takes a simply-linked list as an input argument. It modifies the list by moving the last element to the front of the list and returns the modified list. Some part of the code is left blank. Choose the correct alternative to replace the blank line. typedef struct node { int value; struct node *next; }Node;   Node *move_to_front(Node *head) { Node *p, *q; if ((head == NULL: || (head->next == NULL)) return head; q = NULL; p = head; while (p-> next !=NULL) { q = p; p = p->next; } _______________________________ return head; } a) q = NULL; p->next = head; head = p; b) q->next = NULL; head = p; p->next = head; c) head = p; p->next = q;

Hi I need help please I need to make a function that converts an infix expression to a post fix expression such as converting a + b * (c-d) To a b c d - * +

[] [] partite_sets In the cell below, you are to write a function "partite_sets (graph)" that takes in a BIPARTITE graph as its input, and then returns a single list whose two entries are the partite sets of the graph as lists (the order of the sets outputed does not matter). After compiling the above cell, you should be able to compile the following cell and obtain the desired outputs. print (partite_sets({"A" : ["B", "C"], "B" : partite_sets({"A" : ["B", "C"], "B" : ["A"], "C" : ["A"]}), ["A", "D"], "C" : ["A", "D"], "D" : ["B", "C"]})) This should return [["B","C"], ["A"]] [["A","D"], ["B","C"]] (the order in which the entries appear does not matter) Python Python

please go in to the detail and explain the purpose and function to each line & function of code listed below.  #include <stdio.h>    //Represent a node of the singly linked list  struct node{      int data;      struct node *next;  };         //Represent the head and tail of the singly linked list  struct node *head, *tail = NULL;     //addNode() will add a new node to the list  void addNode(int data) {      //Create a new node      struct node *newNode = (struct node*)malloc(sizeof(struct node));      newNode->data = data;      newNode->next = NULL;            //Checks if the list is empty      if(head == NULL) {          //If list is empty, both head and tail will point to new node          head = newNode;          tail = newNode;      }      else {          //newNode will be added after tail such that tail's next will point to newNode          tail->next = newNode;          //newNode will become new tail of the list          tail = newNode;      }  }…