What is the run time of the following algorithm?Counting-Sort(A, B, k)let c[0..k] be a new arrayfor i = 0 to kC[i] = 0for j = 1 to A.lengthC[A[j]] = C[A[j]]+1for i=1 to kC[i] = [[i] + [i-1]for j = A.length downto 1B[C[A[j]]] = A[j]C[A[j]] = C[A[j]] - 1О Ө (п 1g n)O O (lg n)O 0 (n)O e (n²)Question 5When the counting-sort algorithm in the previous question completes, B will contain the same elements as A,but in sorted order. Prove the correctness of this algorithm.

Option: C Time Complexity: O(n+k) where n is the number of elements in input array and k is the…

Counting-Sort(A, B, k) let [[0..k] be a new array for i = 0 to k [[i] = 0 for j = 1 to A.length C[A[j]] = C[A[j]]+1 %3D for i=1 to k C[i] = C[i] + C[i-1] for j = A.length downto 1 B[C[A[j]]] = A[j] C[A[j]] = C[A[j]] - 1 O O (n lg n) O O (lg n) O 0 (n) O e (n²) Question 5 When the counting-sort algorithm in the previous question completes, B will contain the same elements as A, but in sorted order. Prove the correctness of this algorithm.

Counting-Sort(A, B, k) let [[0..k] be a new array for i = 0 to k [[i] = 0 for j = 1 to A.length C[A[j]] = C[A[j]]+1 %3D for i=1 to k C[i] = C[i] + C[i-1] for j = A.length downto 1 B[C[A[j]]] = A[j] C[A[j]] = C[A[j]] - 1 O O (n lg n) O O (lg n) O 0 (n) O e (n²) Question 5 When the counting-sort algorithm in the previous question completes, B will contain the same elements as A, but in sorted order. Prove the correctness of this algorithm.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

Create a function that takes two "sorted" arrays of numbers and returns an array of numbers which are common to both the input arrays. Examples common Elements([-1, 3, 4, 6, 7, 9], [1, 3]) → [3] -> [1, 3, 4, 7] common Elements([1, 3, 4, 6, 7, 9], [1, 2, 3, 4, 7, 10]) common Elements([1, 2, 2, 2, 3, 4, 5], [1, 2, 4, 5]) → [1, 2, 4, 5] common Elements([1, 2, 3, 4, 5], [10, 12, 13, 15]) -> [] Notes • Arrays are sorted. • Try doing this problem with O(n + m) time complexity.
Create a function that can nest a flat array to represent an incremental depth level sequence. Examples incrementalDepth([1, 2]) → incrementalDepth ([1, 2, 3, 4, 5]) - incrementalDepth ([1, 3, 2, 6]) → [1, [3, [2, [6]]]] [1, [2]] [1, [2, [3, [4, [5]]]]] IN PHP
In a non-empty array of integers, the priority of a number is determined by the frequency of its occurrence. Elements are added as they come but the most frequent element is deleted first. If two elements have the same priority, then the element which came first will be deleted first. For example: nums = [1,2,3,1,3,1,3,2,3]. After the first deletion now the array will be num=[2,3,3,3,2,3]. Implement a priority queue for the aforementioned scenario. give the answer in brief and the correct one.
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
Consider the following list, which is sorted by insertion sort for array-based lists: 28, 18, 21, 10, 25, 30, 12, 71, 32, 58, 15 Show the resulting list after six passes of the sorting phase—that is, after six iterations of the for-loop.
Consider the Insertion Sort Function below:public static void insertionSort(int array[]){ int n = array.length; for (int j = 1; j < n; j++) { int key = array[j]; int i = j-1; while ( (i > -1) && ( array [i] > key ) ) { array [i+1] = array [i]; i--; } array[i+1] = key;} }Discuss when is its best case, and worst case performance.
ALGORITHM Quicksort(A[l..r])//Sorts a subarray by quicksort//Input: Subarray of array A[0..n − 1], defined by its left and right// indices l and r//Output: Subarray A[l..r] sorted in nondecreasing order if l < r s = Partition(A[l..r]) //s is a split position Quicksort(A[l..s − 1]) Quicksort(A[s + 1..r]) The above code is the quicksort algorithm.Perform a time complexity analysis for quicksort. You need to explain 1) what makes a worst case, what makes a best case, 2) the time complexity for both the worst case and the best case, 3) how to avoid the worst case in practice.
15. Consider the problem of finding the first position in which an array b occurs as a subsequence of an array a. Write two nested loops: let result undefined for (let i = 0; i < a.length - b.length; i++) { for (let j = 0; j < b.length; j++) { if (a[i+j] = b[j]) . . . } } Complete with labeled break and continue statements.
Consider an array of length 5, arr[5] = {9,7,4,2,1}. What are the steps of insertions done while running insertion sort on the array? a) 7 9 4 2 1 4 7 9 2 1 2 4 7 9 1 1 2 4 7 9 b) 9 7 4 1 2 9 7 1 2 4 9 1 2 4 7 1 2 4 7 9 c) 7 4 2 1 9 4 2 1 9 7 2 1 9 7 4 1 9 7 4 2 d) 7 9 4 2 1 2 4 7 9 1 4 7 9 2 1 1 2 4 7 9
Reverse the order of a n-element array in C What am I doing incorrectly and how can I fix? My code: #include <stdio.h> sort(int *p, int m) { //Write your code here for(int i = 0; i<n/2; i++){ temp = arr[i]; arr[i] = arr[n-i-1]; arr[n-i-1] = temp; } } main(){ int i, n; int *p, num[20]; printf("input n:"); scanf("%d", &n); printf("please input these numbers:\n"); for (i = 0; i < n; i++) scanf("%d", &num[i]); p = &num[0]; sort(p, n); printf("Now, the sequence is:\n"); for (i = 0; i < n; i++) printf("%d ", num[i]); }
IN C Programing Let us define the weighted sum of an integer array a[0], a[1], a[2], …, a[n-1] be a[0]*1 + a[1]*2 + a[2]*3 + …+a[n-1]*n. For example, the weighted sum of the array [5,2,6] would be 5*1+2*2+6*3 = 27. Write a recursive function that takes in an array numbers and its length n, and returns its weighter sum. You can assume n is non-negative integer. int weightedSum(int numbers[], int n) {
Let M(n) be the minimum number of comparisons needed to sort an array A with exactly n ele- ments. For example, M(1) = 0, M(2) = 1, and M(4) = 4. If n is an even number, clearly explain why M(n) = 2M(n/2) + n/2.