Create a program that contains a function pairs(x) where x is a list of integers. The function should return (not print) another list which contains numbers from x which have a pair. There may be multiple pairs for the same number. The list to be returned must be sorted. Example test cases • pairs([1, 1, 2, 2, 2, 2, 3, 5, 5]) should return [1, 2, 2, 5] • pairs([9, 8, 9, 8, 1, 1, 2]) should return [1, 8, 9]
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- A sml binarySearch function that recursively implements the binary search algorithm to search a sorted integer list for a specified integer and returns true if it is found, false otherwise. For example, binarySearch ([100,200,300,400,500], 200) returns true, whereas binarySearch([100,200,300,400,500], 299) returns false. Hint: Write a helper function mid that returns a tuple (index, value) representing the middle value in a list. For example, mid [10, 2, 40, 8, 22] returns (2,40) because the value 40 at index 2 is the middle value in the list. Similarly, mid [10, 20] would return (1, 20). Use mid in conjunction with slice to implement binarySearch.Please answer it in Python Write a function extrait_pairs that takes as an argument a list l1 and returns the list obtained from l1 keeping only the even integers (and without changing their order). Be careful, l1 must not be modified by the function. Example: extrait_pairs([4, 7, 12, 0, 3]) is [4, 12, 0] and extrait_pairs([21, 17, 3]) is [ ].Write 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.
- Write a function split n l that splits the list l and returns a pairof lists containing the first n elements of l and the rest of l, wheresplit has the type int -> ’a list -> (’a list * ’a list). For ex-ample, split 3 [1,2,3,4,5] returns the pair ([1,2,3], [4,5]) andsplit 3 [1,2,3] returns ([1,2,3], []).Python Write a function revers_all (sequence) that takes an arbitrary nested list sequence with arbitrary elements and returns a new list that has the same nested structure, but where all lists both at the top level and at all nested levels are reserved ("backwards") examplereverse_all ([1, 2, 3, [4, 5, ['x', 7]], 8]) becomes[8, [[7, 'x'], 5, 4], 3, 2, 1] I sent parameters can not be changed. It is thus a completely new list to be returned.It is allowed to solve the task recursively or itrelatively or with a compilation of these methods. Your function may not make any readings or printouts Use this program to test import copy sequence = eval(input()) seq_copy = copy.deepcopy(sequence) result = reverse_all(sequence) print (result) if results is sequence: print("Your function does not create a new list") if sequence != seq_copy: print("Your function changes in that submission list")Consider a (singly) linked list. Describe an algorithm to remove every second element from the list. (HINT: The algorithm is similar to the one that is used when we delete a single node. Here, we need to skip an element after every deletion, and then if we have not reached the end of the list repeat this procedure until the list is exhausted.). (a) Write the steps to removeEverySecondElement from the linked list. (b) Write a function removeEverySecondElement() that implements the task
- Please code this in Java Or C or C++ Given two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and return the sum as linked list: Example : Input : (7-> 1 -> 6) + (5-> 9-> 2). That is 617 + 295 = 912 Output : 2-> 1 -> 9 Expected time complexity : O(m+n), where m and n are no. of nodes in first and second linked list respectively.Suppose we have a singly linked list implemented with the structure below and a function that takes in the head of the list.typedef struct node{int num;struct node* next;} node;int whatDoesItDo (node * head){struct node * current = head;struct node * other, *temp;if (current == NULL)return head;other = current->next;if (other == NULL)return head;other = other->next;temp = current->next;current->next = other->next;current = other->next;if (current == NULL){head->next = temp;return head;}other->next = current->next;current->next = temp;return head;}If we call whatDoesItDo(head) on the following list, show the list after the function has finished.head -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7We have a list. For example [2,4,6,6,8,2,12,54,15]. Without repeating the elements in the list a new list that does not repeat its elements by creating and using the function that writes Create.
- In PYTHON, write a program that reads elements from a list, where each element from the list is a list with three values. Call the function defined in part 1 with the 3 values as arguments in order and print the returning value.Test your code with the list:"numbers=[[3, 5, 4], [7, 4, 3], [2, 3, 1]]" The functions defined in part 1 were: def greatest(a,b,c): if a>b: return c ifnot a>b and a>c: return b ifnot a>b andnot a>c: return a print(greatest(2,3,1))Write a Python Code for the given function and conditions: Given function: def insert(self, newElement) Pre-condition: None. Post-condition: This method inserts newElement at the tail of the list. If an element with the same key as newElement already exists in the list, then it concludes the key already exists and does not insert the key.The function interleave_lists in python takes two parameters, L1 and L2, both lists. Notice that the lists may have different lengths. The function accumulates a new list by appending alternating items from L1 and L2 until one list has been exhausted. The remaining items from the other list are then appended to the end of the new list, and the new list is returned. For example, if L1 = ["hop", "skip", "jump", "rest"] and L2 = ["up", "down"], then the function would return the list: ["hop", "up", "skip", "down", "jump", "rest"]. HINT: Python has a built-in function min() which is helpful here. Initialize accumulator variable newlist to be an empty list Set min_length = min(len(L1), len(L2)), the smaller of the two list lengths Use a for loop to iterate k over range(min_length) to do the first part of this function's work. On each iteration, append to newlist the item from index k in L1, and then append the item from index k in L2 (two appends on each iteration). AFTER the loop…