Use racket, plait language (define ( func n x) ...) returns a list containing n copies of x
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- 3. Write a LISP function, call it "count", of two arguments, an atom x and a list L of sub-lists, which returns the number of sub-lists in L that contain the atom x. Use recursion. example: > (count 'b ( (ab c) (b c d) (c a) (d ab c)) ). Returns: 3 > (count 'd '((a b) (b) (c a))) Returns: 0Given a singly linked list, print reverse of it using a recursive function printLinkedList( node *first ) where first is the pointer pointing to the first data node. For example, if the given linked list is 1->2->3->4, then output should be: 4 3 2 1 (note the whitespace in between each data value)Write a LISP function f10 that finds the intersection of two lists. The intersection meansthe common elements of the two lists. Example: (f10 ‘(a (a b) c d) ‘(b(a b) d)) returns ((a b) d) CAN ONLY USE CAR AND CDR AND RECURSION SETQ AND DO LIST CANT BE USED
- Subtract 10 from all list entries without any looping in the haskell programming language. Use of map and fmap Furthermore, let us suppose that there are two distinct lists.In this question, you are to implement two functions for reversing linked lists. One function, reverse_iter(..), must use an iterative approach (i.e., loops), whereas the other function, reverse_rec(..), must use recursion instead. We have already implemented the linked list as well as all its helper-functions in the Ilist-module.Implements clone which duplicates a list. Pay attention, because if there are sublists, they must be duplicated as well. Understand the implementation of the following function, which recursively displays the ids of each node in a list Develop your solution as follows: First copy the nodes of the current list (self) Create a new list with the copied nodes Loop through the nodes of the new list checking the value field If this field is also a list (use isinstance as in the show_ids function) then it calls clone on that list and substitutes the value. Complete the code: def L4(*args,**kwargs): class L4_class(L): def clone(self): def clone_node(node): return <... YOUR CODE HERE ...> r = <... YOUR CODE HERE...> return r return L4_class(*args,**kwargs)
- This is assignment used C++. Develop a linked-list processing function, IsolateTargetSoloAsTail, to process a linked list as follows. ● If a target cannot be found on the given list, a new node containing the target is created and added to the list's end (made the new tail node). ► This includes the case where the given list is empty, in which the new tail node added is also the new head node. (This is so because the only node in a 1-node list is the list's head and tail node.) ● If the target appears only once on the given list, the target-matching node is moved to the list's end (made the new tail node). ► Nothing needs to be done if the target-matching node is already the tail node (of the given list). ● If the target appears multiple times on the given list, the first target-matching node is moved to the list's end (made the new tail node), and all other target-matching nodes are to be deleted from the list. ► Note that although…Please do this in JAVA PROGRAMMING. Given: List L of pairs of charactersString SOutput: TRUE if S is a valid string, FALSE otherwise. A string is considered valid if each character in the string can be paired with another character in the string, where the pair belongs to the input list L. Furthermore, two pairs cannot cross each other. In other words, a pair most completely enclose another, or be completely separate. Design and implement an efficient dynamic programming solution to this problem. Examples: Input L: (a b) (b c) (c d) (a a) Input S: aaba Output: True (pairs shown color-coded: aaba, "aa" fully encloses "ab") Input L: (a b) (b c) (c d) (a a) Input S: abcaad Output: True (pairs shown color-coded: abcaad, "ab" is separate from the other pairs) Input L: (a b) (b c) (c d) (a a) Input S: acbd Output: False (acbd is not valid because the pairs cross each other.) Input L: (a b) (b c) (c d) (a a) Input S: aaac Output: FalseWrite a lisp function f9 that removes duplicates from a list. Example: (f9 ‘(c (a b) c d (a b))) returns (c (a b) d) should mainly just use cond car and cdr
- Let xHead, yHead and zHead be the head pointers of 3 linked lists of integers (called X-list, Y-list and Z-list, respectively, from here). X-list and Y-list are each sorted (in non-decreasing order) in itself and each may be empty. Z-list is initially empty (i.e., zHead initially contains the null pointer). Develop and test a recursive C++ function called Merge2AscListsRecur that combines the nodes in X-list and Y-list into Z-list such that, after calling the function, Z-list is a sorted list (in non-decreasing order) containing all the nodes initially contained in X-list and Y-list – X-list and Y-list should both be empty after the call. Other specifications/requirements: ● Each node of the list has the following structure: struct Node{ int data; Node *link;}; ● The function should: ► Have only three parameters (each a pointer-to-Node) and no return value (be a void function). ► Not use any global variables or static local…Let xHead, yHead and zHead be the head pointers of 3 linked lists of integers (called X-list, Y-list and Z-list, respectively, from here). X-list and Y-list are each sorted (in non-decreasing order) in itself and each may be empty. Z-list is initially empty (i.e., zHead initially contains the null pointer). Develop and test a recursive C++ function called SortedMergeRecur that combines the nodes in X-list and Y-list into Z-list such that, after calling the function, Z-list is a sorted list (in non-decreasing order) containing all the nodes initially contained in X-list and Y-list – X-list and Y-list should both be empty after the call. Other specifications/requirements: ● Each node of the list has the following structure: struct Node{ int data; Node *link;}; ● The function should: ► Have only three parameters (each a pointer-to-Node) and no return value (be a void function). ► Not use any global variables or static local…a. Write the recursive method for adding a node in a linked list. b. Consider the infix expression 15 + 2 – 10 / 2 ∗ 2 and convert this into postfix expression using stack. Next Token Action Effect on operatorStack Effect on postfix 15 + 2 - 10 / 2 * 2