Write a pseudocode to solve the following problem: Odd index element problem input: a non-empty sequence S of n elements output: a list of all the elements in the odd index, or None if S is empty

import numpy as np%matplotlib inlinefrom matplotlib import pyplot as pltimport mathfrom math import exp np.random.seed(9999)def is_good_peak(mu, min_dist=0.8):    if mu is None:        return False    smu = np.sort(mu)    if smu[0] < 0.5:        return False    if smu[-1] > 2.5:        return False    for p, n in zip(smu, smu[1:]):        #print(abs(p-n))        if abs(p-n) < min_dist:            return False    return True maxx = 3ndata = 500nset = 10l = []answers = []for iset in range(1, nset):        npeak = np.random.randint(2,4)    xs = np.linspace(0,maxx,ndata)    ys = np.zeros(ndata)    mu = None        while not is_good_peak(mu):        mu = np.random.random(npeak)*maxx    for ipeak in range(npeak):        m = mu[ipeak]        sigma = np.random.random()*0.3 + 0.2        height = np.random.random()*0.5 + 1        ys += height*np.exp(-(xs-m)**2/sigma**2)        ys += np.random.randn(ndata)*0.07    l.append(ys)    answers.append(mu) p6_ys = lp6_xs =…

Write a version of the binary search algorithm that can be used to search a list of strings. (Use the selection sort that you designed to sort the list.)

In Java, Apply replacement selection sort to the following list assuming array size M = 3; 27 47 35 7 67 21 32 18 24 20 12 8 Show the content of each sorted file.

Create three problem instances of size n is around 10, representing the best-case, worst-case and average-case for Radix Sort respectively. Show how the Radix Sort works on the best-case, and explain why each of them represents the best-case, worst-case and average-case respectively.   psudocode for Radixsort attached

First Pass: ( 5 1 4 2 8 ) \to ( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps them. ( 1 5 4 2 8 ) \to ( 1 4 5 2 8 ), Swap since 5 > 4 ( 1 4 5 2 8 ) \to ( 1 4 2 5 8 ), Swap since 5 > 2 ( 1 4 2 5 8 ) \to ( 1 4 2 5 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them. Second Pass: ( 1 4 2 5 8 ) \to ( 1 4 2 5 8 ) ( 1 4 2 5 8 ) \to ( 1 2 4 5 8 ), Swap since 4 > 2 ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted. Third Pass: ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) Finally, the array is sorted, and the algorithm can terminate. Bubble sort is a simple sorting algorithm. It works by repeatedly stepping through the list to be sorted, comparing each pair of…

Heapsort has heapified an array to: 98 71 63 38 16 and is about to start the second for loop. What is the array after each loop iteration? i = 4: Ex: 86, 75, 30 i = 3: i = 2: i = 1: