ALGORITHM BubbleSort(A[0..n – 1]) I/Sorts a given array by bubble sort //Input: An array A[0..n – 1] of orderable elements //Output: Array A[0..n – 1] sorted in nondecreasing order for i +0 to n – 2 do for j0 to n – 2- i do if A[j+1] < A[j] swap A[j]and A[j+1] a) Demonstrate the steps based on the algorithm above for the following lists: List 1: 3, 7, 8, 9, 10

ALGORITHM BubbleSort(A[0..n – 1]) I/Sorts a given array by bubble sort //Input: An array A[0..n – 1] of orderable elements //Output: Array A[0..n – 1] sorted in nondecreasing order for i +0 to n – 2 do for j0 to n – 2- i do if A[j+1] < A[j] swap A[j]and A[j+1] a) Demonstrate the steps based on the algorithm above for the following lists: List 1: 3, 7, 8, 9, 10

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

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What is time complexity for this Bubble Sort algorithm in seconds/minutes? def bubbleSort(array): length = len(array) for i in range(length-1): for j in range(0, length-i-1): if array[j] > array[j+1] : array[j], array[j+1] = array[j+1], array[j] arr = [56,7,1,9,0,4,34,100,67] print ("Elements of array before sorting:") for i in range(len(arr)): print ("%d" %arr[i]), bubbleSort(arr) print ("Elements of array after sorting:") for i in range(len(arr)): print ("%d" %arr[i])
IN JAVA Exploration: Test the sorting algorithms as follows. For input sizes n = 500, n = 1,000, n = 1,500, …, n = 10,000 [That’s 20 different values of n]. For each n, create an array I of size n. Fill the array with n random integers. Create a copy M of the array I. Pass the array I to your insertion sort and record the number of comparisons. Pass the array M to your mergesort and record the number of comparisons.
Write an efficient algorithm (to the best of your knowledge) for the following problem, briefly describe why it is a correct algorithm, and analyze the time complexity. If you cannot find any polynomial-time algorithm, then give a backtracking algorithm. Problem: Repeated Number Input: An array A[1...3n] of positive integers. Output: If any number in the array appears at least n times, then print such a number. Otherwise, print 'none'. Example: Input: A = [1,3,3,3,1,1]. Here 3n = 6 and n =2. Output: 3. Note that here 1 would also be a valid output. You just need to print one if there are many options. Example: Input: A = [4,3,5,7,9,1]. Here 3n = 6 and n =2. Output: 'none'
Answer the following question for basic sorting algorithms. Here is an array of 7 integers: (1) 8 3 2 0 1 9 7 Draw this array after the FIRST iteration of the large loop in a selection sort (sorting from smallest to largest, and always pick the smallest element in each iteration). Here is an array of 7 integers: (1) 8 3 2 0 1 9 7 Draw this array after the FIRST iteration of the large loop in a bubble sort (sorting from smallest to largest).
The following algorithm adds all the entries in the “upper triangular” part of a square n × n array A. Consider n = 5 and the 5 × 5 array below. The work unit is the addition operation, sum = sum + A[k, j]. How many times is the sum = sum + A[k, j] operation done? sum = 0 for k = 1 to n do for j = k to n do sum = sum + A[k, j] end for end for 1 3 1 0 2 3 1 0 3 3 0 0 1 2 0 0 2 3 1 2 2 2 3 0 1
1. Consider the algorithm for the sorting problem that sorts an array by counting,for each of its elements, the number of smaller elements and then uses thisinformation to put the element in its appropriate position in the sorted array:ALGORITHMComparisonCountingSort(A[0..n − 1])//Sorts an array by comparison counting//Input: Array A[0..n//Output: Array S[0..n− 1] of orderable values− 1] of A’s elements sorted// in nondecreasing orderfor i ← 0 to nCount− 1 do[i]←0for i ← 0 to n − 2 dofor j ← i +1 to n − 1 doif A[i] < A[j ]Count[j ]← Count[j ] + 1else Count[i]← Count[i] + 1for i ←0 to n−1 doS[Count[i]]←A[i]return Sa. Apply this algorithm to sorting the list 60, 35, 81, 98, 14, 47.b. Is this algorithm stable?c. Is it in-place?
Suppose an array sorted in ascending order is rotated at some pivot unknownto you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).Find the minimum element. The complexity must be O(logN)You may assume no duplicate exists in the array."""def find_min_rotate(array): """ Finds the minimum element in a sorted array that has been rotated. """ low = 0 high = len(array) - 1 while low < high: mid = (low + high) // 2 if array[mid] > array[high]: low = mid + 1 else: high = mid return array[low] def find_min_rotate_recur(array, low, high): """ Finds the minimum element in a sorted array that has been rotated. """ mid = (low + high) // 2 if mid == low:.
Have a look at the array below. We are trying to do quicksort on this array to sort it in ascending order. We wrote an algorithm that always takes the middle element as pivot. How the array will look like after first iteration(write in this format: 11,12,13,14,15)Array: 525 76 980 290 38 77 746 266 375 . Show the simulation in handwritten paper. Which sorting algorithm (Insertion, Merge, Quick) perform same in all (best, average and worst case) ?
Given an integer array [A], efficiently find a descending sorted triplet such that: A[i]> A[i]> A[k] and 0 <=i<j <k <n, where n is the array size. For example, Input: A[] = {9, 1, 6, 7, 3, 4, 5) Output: Any one of the following triplets: (9, 7,5) (9, 7, 4) (9, 7, 3) (9, 6, 5) (9, 6, 4) (9, 6, 3) a) Design a brute-force algorithm to solve this problem complexity of your solution b) Design a more efficient algorithm to do the same task with less complexity and analyse the complexity of your solution. c) Develop a python code to implement your efficient algorithm. depend on the correctness of the code, indentation, comments, test-case] d) Prepare a brief report (250 words) comparing the two algorithms
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.
9.). integer array nums and an integer k, return the length of the shortest non-empty subarray of nums with a sum of at least k. If there is no such subarray, return -1. A subarray is a contiguous part of an array. Example 1: Input: nums = [1], k = 1 Output: 1 Example 2: Input: nums = [1,2], k = 4 Output: -1 Example 3: Input: nums = [2,-1,2], k = 3 Output: 3.
: In searching an element in an array, linear search can be used, even though simple to implement, but not efficient, with only O(n) time complexity. Assuming the array is already in sorted order, modify the search function below, using a better algorithm, so the average time complexity for the search function is O(log n). include <iostream> using namespace std; int search(int al), int s, int v) { 1/ Modify below codes. for (int i = 0; i <s; i++) { if (a[i] = v) return i; return -1; int main() { int intArray:10] = { 5, 7, 8, 9, 10, 12, 13, 15, 20, 34); // Search for element '12' in 10-elements integer array. cout << search(intArray, 10, 12); // '5' will be printed out. // Search for element '35' in 10-elements integer array. cout << search(intArray, 10, 35); // '-1' will be printed out. // Index '-l' means that the element is not found. return 0;