an array 19,8,2 copy om last tO first to a second 5-DWORD-element array Note 1: Use the loop as well as push and pop instructions Note 2: Use the LENGTHOF operator to get the number of elements of the array Note 3: Your algorithm should work for any array irrespective of its size with minimal mod- ifications

an array 19,8,2 copy om last tO first to a second 5-DWORD-element array Note 1: Use the loop as well as push and pop instructions Note 2: Use the LENGTHOF operator to get the number of elements of the array Note 3: Your algorithm should work for any array irrespective of its size with minimal mod- ifications

Microsoft Visual C#

7th Edition

ISBN:9781337102100

Author:Joyce, Farrell.

Publisher:Joyce, Farrell.

Chapter6: Using Arrays

Section: Chapter Questions

Problem 19RQ

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The Sieve of Eratosthenes is an algorithm that finds all prime numbers up to a given limit, n. It works by creating an array of Booleans, flag, of size n+1, initializing the array to true (assuming every number is a prime number in the beginning), and iteratively traversing the numbers from 2 to n, setting the values of indices that are multiples of other numbers to false. Write a program using the algorithm above, and display all the prime numbers up to 50.
In the "Common Array Algorithms" section, you learned how to find the position of an element in an array. Suppose you want to find the position of the second match. You could remove the first match and start over, but that would be wasteful. Instead, adapt the algorithm from that section so that it starts the search at a given position, not zero. Complete the following code.
Given an array, find the next greater element for each element in the array, ifavailable. If not available, print the element itself. The next greater element y for anelement x in the array is the first element that is greater than x and occurs on its rightside. The next greater element of the right most element in an array is the elementitself.Example: Given A = [ 6 8 4 3 9] the next greater element listB = [8 9 9 9 9]
// function to generate all subsequences of given arrayfunction generate_subseq(arr): // length of arr n = len(arr) // total subsequences for array of length n m = 2**n // to store all subsequences seqs = [] // running a loop for m times for i in range(1, m): // creating an array of zeros of length n a = [0]*n num = i // to use as an index for 'a' j = n-1 // run this loop till num > 0 while num > 0: if num is odd: a[j] = 1 // divide num by 2 num = num/2 // subtract 1 from 'j' j -= 1 // to store current subsequence seq = [] // iterating for n times for i in range(n): // add ith index value to 'seq' if a[i] == 1: seq.append(arr[i]) // add 'seq' to 'seqs' seqs.append(seq) // return seqs return seqs // given listsS1 = ['B','C','D','A','A','C','D']S2 = ['A','C','D','B','A','C']…
given an array of integer values , return true if 6 appears as either the first or last element in the array. firstLast6([1,2,36]) true firstLast([6,1,2,3]) true firstLast([13,6,1,2,3]) false
Given an ArrayList of integers, return an ArrayList where all the odds are before the evens in the order that they occur. public ArrayList<Integer> solution(ArrayList<Integer> nums) {ArrayList<Integer> finalList = new ArrayList<>();int index = 0;for (int i = 0; i < nums.size(); i++) {if (nums.get(i) % 2 == 1) {finalList.add(index, nums.get(i));index++;} elsefinalList.add(nums.get(i));}return finalList;} This is my failing code. for [1.-3.3] where [1,3,-3] is being returned
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