(d) The pseudocode below outlines a bucket sorting algorithm where all elements of the unsorted input A[1...n] are in the interval [0,1). BUCKETSORT(A) 1 n = A. length 2 buckets B[0...n-1] 3 4 for i = 1 to n insert A[i] into bucket B[[nA[i]]] 5 sort each bucket 6 concatenate sorted buckets Write down the contents of each bucket immediately before execution of line 5 for input A = [0.2,0.1,0.7].
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- Language: Python 3 Autocomplete Ready O 1 v import ast 3. Hybrid Sort input() lst %3D 3 lst = ast.literal_eval(lst) 4 Insertion sort is a simple sorting algorithm that builds the final sorted array one item at a time. In each iteration, insertion sort inserts an element into an already sorted list (on left). The position where the item will be inserted is found through linear search. You decided to improve insertion sort by using binary search to find the position p where the new insertion should take place. 6 print(BinaryInsertionSort(lst)) Algorithm BinarylnsertionSort Input/output: takes an integer array a = {a[0], ..., a[n – 1]} of size n begin BinarylnsertionSort for i =1 to n val = a[i] p = BinarySearch(a, val, 0, i – 1) for j = i-1 to p a[j + 1]= a[i] j= j-1 end for a[p] = val i i+1 end for end BinarylnsertionSort Here, val = a[i] is the current value to be inserted at each step i into the already sorted part a[0], ..., ați – 1] of the array a. The binary search along that part…Rank SortWrite an efficient parallel program that reads n integers from any input file and implements rank sort using p processors (p << n). Assume n is divisible by p.Given n integers stored in an array, rank sort computes rank of every a[j] as follows: Rank a[j] = no. of array elements that are ≤ a[j] for k = 0, 1, 2 … j-1 + no. of array elements that are < a[j] for k = j+1, j+2, j+3,… n – 1 Note: Every a[j] is compared with all array elements.1r. ""Implementation of the Misra-Gries algorithm.Given a list of items and a value k, it returns the every item in the listthat appears at least n/k times, where n is the length of the array By default, k is set to 2, solving the majority problem. For the majority problem, this algorithm only guarantees that if there isan element that appears more than n/2 times, it will be outputed. If thereis no such element, any arbitrary element is returned by the algorithm.Therefore, we need to iterate through again at the end. But since we have filtredout the suspects, the memory complexity is significantly lower thanit would be to create counter for every element in the list. For example:Input misras_gries([1,4,4,4,5,4,4])Output {'4':5}Input misras_gries([0,0,0,1,1,1,1])Output {'1':4}Input misras_gries([0,0,0,0,1,1,1,2,2],3)Output {'0':4,'1':3}Input misras_gries([0,0,0,1,1,1]Output None""".
- Kindly solve the below function in cpp Write a program / Algorithm to find intersection of two sets in O(n). void setA_Intersection_setB (int setA[], int counterA, int setB[], int counterB)MASM IRVINE 32 library PROGRAM. How do you take a SORTED array of elements and count the amount of each integer in that list and then display it. So if i had a list of [1,2,4,1,4,2,1,4,1,2,4,5,2]. I need to make a program that can iterate through that and find the amount of 1s, 2s, 3s, 4s etc and display it. MUST BE IN MASM IRVINE 32 LIBRARY.Language: Python 3 • Autocomplete Ready O 1 v import ast lst = input(O lst = ast.literal_eval(lst) def binarysearch(lst,x,low,high): if low - high x: 10 11 Algorithm BinarylnsertionSort 12 Input/output: takes an integer array a = {a[0], ..., a[n – 1]} of size n 13 begin BinarylnsertionSort return binarysearch (lst, x, mid, high) 14 for i =1 to n val = a[i] p = BinarySearch(a, val, 0, i – 1) for j = i-1 to p alj + 1]= a[j] j= j-1 end for 15 else: 16 return mid 17 18 def BinaryInsertionSort(lst): 19 print (BinaryInsertionSort(lst)) 20 a[p] = val j=i+1 end for end BinarylnsertionSort Here, val = a[i] is the current value to be inserted at each step i into the already sorted part a[0], ..., ați – 1] of the array a. The binary search along that part returns the position p where the val will be inserted. After finding p, the data values in the subsequent positions j = i- 1, ..., p are sequentially moved one position up to i, ..., p+1 so that the value val can be inserted into the proper…
- Function PrintArray(integer array(?) dataList) returns nothing integer i for i = 0; i < dataList.size; i = i + 1 dataList[i] = Get next input Put dataList to output Put "_" to output // Your solution goes here. Modify as needed i = 0 Complete the PrintArray function to iterate over each element in dataList. Each iteration should put the element to output. Then, put "_" to output. Ex: If dataList's elements are 2 4 7, then output is: 2_4_7_ Function Main() returns nothing integer array(3) userNums integer i for i = 0; i < userNums.size; i = i + 1 userNums[i] = Get next input PrintArray(userNums)1. You, Alice and Bob are working on recursive search algorithms and have been studying avariant of binary search called trinary search. Alice has created the following pseudocodefor this algorithm:TSearch(A[a...b], t)If a > b return -1Let p1 = a + Floor((b - a)/3)If A[p1] = t return p1If A[p1] > t return TSearch(A[a...p1-1],t)Let p2 = a + Ceiling(2(b - a)/3)If A[p2] = t return p2If A[p2] > t return TSearch(A[p1+1...p2-1],t)Return TSearch(A[p2+1...b],t)EndTSearcha) State a recurrence relation that expresses the number of operations carried out bythis recursive algorithm when called on an input array of size n.b) Bob has heard that trinary search is no more efficient than binary search whenconsidering asymptotic growth. Help prove him correct by using induction to showthat your recurrence relation is in Θ(log2 n) as well.i. Split the tight bound into and upper (big-O) and lower (big-Ω).ii. For each bound select a function from Θ(log2 n) to use in your proof, likea log2 n or a…Select the appropriate code that performs selection sort. a) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length-1; k++) { if(arr[k] < arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; } b) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length; k++) { if(arr[k] < arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; } c) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length-1; k++) { if(arr[k] > arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; } d) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length; k++) { if(arr[k] > arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; }
- Quicksort is a powerful divide-and-conquer sorting algorithm that can be described in just four lines ofpseudocode. The key to Quicksort is the PARTITION(A, p, r) procedure, which inputs elementsptorof array A,and chooses the final element x = A[r] as the pivot element. The output is an array where all elementsto the left ofxare less thanx, and all elements to the right of x are greater than x. In this question, we will use the Lomuto Partition Method from class and assume that the pivot isalwaysthe last (right-most) element of the input array. Question: Let A be an array withn= 2k−1 elements, where k is some positive integer. Determine a formula (in terms of n) for the minimum possible number of total comparisons required by Quicksort, as well as a formula for the maximum possible number of total comparisons required by Quicksort. Use your formulas to show that the running time of Quicksort is O(nlogn) in the best case and O(n2) in the worst case.#include <bits/stdc++.h>using namespace std;int getMedian(int ar1[], int ar2[], int n){int j = 0;int i = n - 1;while (ar1[i] > ar2[j] && j < n && i > -1)swap(ar1[i--], ar2[j++]);sort(ar1, ar1 + n);sort(ar2, ar2 + n);return (ar1[n - 1] + ar2[0]) / 2;}// Driver Codeint main(){int ar1[] = { 1, 12, 15, 26, 38 };int ar2[] = { 2, 13, 17, 30, 45 };int n1 = sizeof(ar1) / sizeof(ar1[0]);int n2 = sizeof(ar2) / sizeof(ar2[0]);if (n1 == n2)cout << "Median is " << getMedian(ar1, ar2, n1);elsecout << "Doesn't work for arrays"<< " of unequal size";getchar();return 0;} PLEASE SOLVE THIS USING 'class' . Thanks a lot in advance:)Python Numpy Function to complete: def w14(v): Inputs: v: A numpy array of shape (N, 1) Returns: The L2 norm of v: norm = (sum_i^N v[i]^2)^(1/2) You MAY NOT use np.linalg.norm