In C programming  Mathematically, given a function f, we recursively define fk(n) as follows: if k = 1, f1(n) = f(n). Otherwise, for k > 1, fk(n) = f(fk-1(n)). Assume that there is an existing function f, which takes in a single  integer and returns an integer. Write a recursive function fcomp, which takes in both n and k (k > 0), and  returns fk(n). int f(int n); int fcomp(int n, int k){

Please IN C++ Write the simplest program that will demonstrate iteration vs recursion using the following guidelines -  Write two primary helper functions - one iterative (IsArrayPrimeIter) and one recursive (IsArrayPrimeRecur) - each of which Takes an array of integers and its size as input params and returns a bool such that 'true' ==> all elements of the array are prime, so the array is prime, 'false' ==> at least one element in array is not prime, so array is not prime.  Print out a message "Entering <function_name>" as the first executed statement of each function. Perform the code to test whether every element of the array is a Prime number. Print out a message "Leaving <function_name>" as the last executed statement before returning from the function. Remember - there will be nested loops for the iterative function and there can be no loops at all in the recursive function. For the recursive function - define one other helper function (IsPrimeRecur) which…

In c++ please help me answer this question I will give you a good rating :) implement the three versions of the addupto20() function: iterative, recursive, and divide-and-conquer approach test these functions with a few inputs from your main() Iterative solution * check if there exsits two numbers from vector data that add up to 20 e.g., if data=[2,5,3,15], the function returns true, as data [0] +data [3]==20 e.g., if data=[3,4,0,8], the function return false precondition: vector data has been initialized postcondition: if there are two numbers from list add up to 20, return true; otherwise, return false */ bool AddupTo20 (const vector‹int> & data){ } Come up with a recursive solution to the problem, following the hints given below: * check if there exsits two numbers from vector data[first...right] add up to 20 e.g., if data=[2,5,3,15], first=0, last=3, the function returns true, as data [0] +data [3]==20 e.g., if data=[2,5,3,15], first=2, last=3, the function returns…

Solved in python

write a c++ recursive function to solve the following: str contains a single pair of parenthesis, return a new string // made of only the parenthesis and whatever those parensthesis contain. You can use substr in this problem. // // Pseudocode Example:// findParen("abc(ghj)789") =&gt; "(ghj)" // findParen("(x)7") =&gt; "(x)" // findParen("4agh(y)") =&gt; "(y)" // string findParen(string str) { }

In Kotlin, write a higher-order function with an expression body that takes an int n and a function f from int to int and returns the result of calling f(f(n)). For example if you call the function with n = -5 anda  function that calculates absolute value, your function will return 5, and if you send the value 2 and a function that calculates the sqaure of an int, the function will return 16. in addition to the function, write the syntax to call it with:a. a function name as the function arguementb. some lambda expression as the function argument

‘Write a function nesting(), which takes an arbitrary number of parameters, and returns a list containing the arguments. Write another function unnesting(), which takes a list L as the parameter and retums a 1D list. Notice that, you should be able to use loops to solve thisproblem and should not use recursive functions (which we will not cover in this course). ‘Write assertions to test the two functions.For example nesting(1, nesting(2, 3, nesting(4, 5, [6])), [7, 8], 9)unnesting([1, [2, 3, [4, 5, [6]]], [7. 8], 9])   (1, [2, 3, [4, 5, [6]]], [7, 8], 9][1, 2,3, 4, 5, 6, 7, 8 9]

Given a vector of real numbers r = (r1, r2, ..., rm). We can standardize the vector using the formulation: vi = "im, where m ri-m is the mean of the vector r, and s is the standard deviation of r. The vector v = (v1, v2, ..., Un) will be the scaled vector. Write a Python function scale_vec (r) that takes the vector r as input and returns the scaled vector v. Sample inputs and outputs: ● Input: np.array([1, 3, 5]), output: [-1.22474487 0. 1.22474487] • Input: np. array([3.3, 1.2, -2.7, -0.6]), output: [1.35457092 0.40637128 -1.35457092 -0.40637128] Hint: Use numpy.mean and numpy.std with default parameters. # Write your function here. Let's test your function. [ ] import numpy as np print (scale_vec (np.array([1, 3, 5]))) print (scale_vec (np.array([3.3, 1.2, -2.7, -0.6])))

For input vector x, write a function to check whether there exists at most 5 elements in x whose values are less than 7.

Change two lines of code of:   def solution(A, K):     n = len(A)     for i in range(n - 1):         if (A[i] + 1 < A[i + 1]):             return False     if (A[0] != 1 and A[n - 1] != K):         return False     else:         return True   SO that it solves the question : you are given a implementation of a function def solution (A,K) This function, given a non-empty array A of N integers (sorted in non-decreasing order) and integer K, checks whether A contains numbers 1,2..K (every numbers from 1 to K at least once and no other numbers  (for example given array A, K=3 ; A(0)=1,  A(1)=1,  A(2)=2,  A(3)=3,  A(4)=3 then the function should return True  while for the following array A, K=2; A(0)=1,  A(1)=1,  A(2)=3  then the function should return False.) Assume that: -N and K are integers within the range (1....300,000) -each element of array A is an integer within the range (0...1,000,000,000) -array A is sorted in non-decreasing order