Let A₁ = - [14]. Find Answer: nAi Hint: Write you answer in an interval notation such as [a,b], (a,b], [a,b), (a,b) or set notation such as {a,b,... } Write: {}for empty set (no solution). Write: inf instead of ...
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- Considering a singly linked list of integers implementation, assuming that elements of the list are always in ascending order, write a function (find_median) that finds and returns median value. The function cannot scan the list more than ONCE. class SinglyList: class _Node: def __init__(self, e, next): self._element = e self._next = next def __init__(self): self._head = self._tail = None def find_median(self):def smallest_score(lst: list[int]) -> int:'''Valid sublists have at least 2 elements.Return the smallest score of any valid sublist of lst.lst is guaranteed to have at least 2 elements.'''in_order = lst[:]in_order.sort()res = in_order[1] - in_order[0]for i in range(len(in_order) - 1):res = min(res, in_order[i+1] - in_order[i])return res"""Unfortunately, your code is not correct.Below, write a Pytest test case that fails,but that should pass if your smallest_score function were correct.Write ONLY the Pytest test case; don't try to fix the busted smallest_score code."""# TODO Your pytest test function heredef smallest_score(lst: list[int]) -> int: ''' Valid sublists have at least 2 elements. Return the smallest score of any valid sublist of lst. lst is guaranteed to have at least 2 elements. ''' in_order = lst[:] in_order.sort() res = in_order[1] - in_order[0] for i in range(len(in_order) - 1): res = min(res, in_order[i+1] - in_order[i]) return res"""Unfortunately, your code is not correct.Below, write a Pytest test case that fails,but that should pass if your smallest_score function were correct.Write ONLY the Pytest test case; don't try to fix the busted smallest_score code.""" # write pytest test function here
- How can I create a list by asking for input (x) and create a list from 1 to x using recursion? Example: Input: 12 Output: [1,2,3,4,5,6,7,8,9,10,11,12] def newlist(n): #Base Case/s #Add conditions here for your base case/s #if <condition> : #return <value> #Recursive Case/s #Add conditions here for your recursive case/s #else: #return <operation and recursive call> function return []Correct answer will be upvoted else downvoted. Computer science. You are given two integers n and k. You are approached to pick greatest number of particular integers from 1 to n so that there is no subset of picked numbers with aggregate equivalent to k. A subset of a set is a set that can be gotten from starting one by eliminating a few (potentially all or none) components of it. Input The main line contains the number of experiments T (1≤T≤100). Every one of the following T lines contains two integers n and k (1≤k≤n≤1000) — the depiction of experiments. Output For each experiment output two lines. In the main line output a solitary integer m — the number of picked integers. In the subsequent line output m particular integers from 1 to n — the picked numbers. In case there are various replies, print any. You can print the numbers in any requestImplement Flatten Arrays.Given an array that may contain nested arrays,produce a single resultant array."""from collections.abc import Iterable # return listdef flatten(input_arr, output_arr=None): if output_arr is None: output_arr = [] for ele in input_arr: if not isinstance(ele, str) and isinstance(ele, Iterable): flatten(ele, output_arr) #tail-recursion else: output_arr.append(ele) #produce the result return output_arr # returns iteratordef flatten_iter(iterable): """ Takes as input multi dimensional iterable and returns generator which produces one dimensional output. """ for element in iterable: if not isinstance(element, str) and.
- code in scala def unfold[A, S](z: S)(f: S => Option[(A, S)]): LazyList[A] = f(z) match { case Some((h, s)) => h #:: unfold(s)(f) case None => LazyList() } A positive integer is perfect if it equals the sum of all of its factors, excluding the number itself. For example, 6, 28, 496, and so on. Define a LazyList to generate an infinite list of perfect numbers. Use higher-order functions whenever possible. [Hint: One way to do this is by (1) having one function construct a list of factors (use unfold and filter for this), (2) having another function which folds the list of factors into their sum, and (3) creating a LazyList using unfold, filter and the previously constructed functions.] e number itself. For example, 6, 28, 496, and so on. Define a LazyList to generate an infinite list of perfectarr = list(map(int, input("Enter array values: ").split())) n = int(input("\nEnter number to search: ")) def linear_search(arr,n): for i in range(len(arr)): if arr[i]==n: return i return -1 #testop = linear_search(arr, n) if op>=0: print("\nNumber found at index: {}".format(op))else: print("\nNot found: {}".format(op)) Check this code and using time measurements, test your algorithm and plot a graph that shows its growth rate when variating the size.Implement MSD string sorting using queues, as follows: Keep onequeue for each bin. On a first pass through the items to be sorted, insert each item intothe appropriate queue, according to its leading character value. Then, sort the sublistsand stitch together all the queues to make a sorted whole. Note that this method doesnot involve keeping the count[] arrays within the recursive method.
- Please answer it in Python Write a function pairsbypairslist that would return pair by pair from the smallest value to the biggest value For example [12, 13, 11, 15], # return : (11, 12) and (13, 15) [9, 15, 4, 24, 19, 2] # return : (2,4) (9,15) (19,24) [1, 5, 9, 15, 21, 9, 12, 9], # return : (1,5) (9,9) (9,12) (15,21)(In Python) Write a generator function that will take a number n and generate all of the combinations using the sequence of numbers, ex. N = 3, (0, 1, 2) and create all combination (0,0) (0,1) (0,2) (1,1) (1,2) (2,2) N! = 6 and show its operation in using it in a list and print its generation.Let L={x1,x2,…,xn} be a list of n elements. Let us search for a key K in the list L. If the key is presented in the list L at index(or position) j then partition the list L into disjoint lists L1 and L2 such that L1={x[i]:x[i]εL such that i≤j} and L2={x[i]:x[i]εL such that i>j}. If the key is not present in the list output is “no”. Write an algorithm (using single linked list) and subsequent C program for your algorithm to compute lists L1 and L2 for the given list L and key K. Note: Don’t use any inbuilt functions in your program. Example1: If L={16, 15, 1, 27, 19, 100, 200,3} and key k= 27 then L1={16, 15,1,27} and L2={19, 100,200, 3}. Example 2: If L={16, 15, 1, 27, 19, 100, 200,3} and key k= 127 then no.