Problem 3: Array Reversal We have come through an old algorithm whose name indicates that it reverses the array elements; its pseudo-code was defined as follows: note n := A.length denotes the size of the array, number of items. Algorithm 2 REVERSE-ARRAY(A) s+ [), e t ) while s >0A e < (n – 1) do swap (A[s], A[e]) s + s – 1 e + e +1 end while a) Implement the algorithm in Python, and test it b) Empirically, show the performance curve of the algorithm using time measurements c) Using the basics of the theoretical analysis, write the complexity of its worst-case time. d) Prove that the algorithm is actually correct

Problem 3: Array Reversal We have come through an old algorithm whose name indicates that it reverses the array elements; its pseudo-code was defined as follows: note n := A.length denotes the size of the array, number of items. Algorithm 2 REVERSE-ARRAY(A) s+ [), e t ) while s >0A e < (n – 1) do swap (A[s], A[e]) s + s – 1 e + e +1 end while a) Implement the algorithm in Python, and test it b) Empirically, show the performance curve of the algorithm using time measurements c) Using the basics of the theoretical analysis, write the complexity of its worst-case time. d) Prove that the algorithm is actually correct

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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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?
Given an array of integers arr, sort the array by performing a series of pancake flips. In one pancake flip we do the following steps: Choose an integer k where 1 <= k <= arr.length. Reverse the sub-array arr[0...k-1] (0-indexed). For example, if arr = [3,2,1,4] and we performed a pancake flip choosing k = 3, we reverse the sub-array [3,2,1], so arr = [1,2,3,4] after the pancake flip at k = 3. Print out the k-values corresponding to a sequence of pancake flips that sort arr. Example 1: Input: arr = [3,2,4,1] Output: 4, 2, 4, 3 Explanation: We perform 4 pancake flips, with k values 4, 2, 4, and 3. Starting state: arr = [3, 2, 4, 1]. After 1st flip (k = 4): arr = [1, 4, 2, 3] After 2nd flip (k = 2): arr = [4, 1, 2, 3] After 3rd flip (k = 4): arr = [3, 2, 1, 4] After 4th flip (k = 3): arr = [1, 2, 3, 4], which is sorted. Another potential solution is: Output = 3, 4, 2, 3, 1, 2, 1, 1 with a similar explanation. All potential solutions that solve the problem are…
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:.
You are given an unsorted array A[1, ..., n].We know that the maximum (or minimum) element in the array can be found in (n-1) comparisons.Write an algorithm that finds both the maximum and the minimum element in the array in ~(3n/2) comparisons, i.e., (3n/2) + THETA(1) comparisons.
The recursive algorithm below takes as input an array A of distinct integers, indexed between s andf, and an integer k. The algorithm returns the index of the integer k in the array A, or ?1 if the integerk is not contained within A. Complete the missing portion of the algorithm in such a way that you makethree recursive calls to subarrays of approximately one third the size of A.• Write and justify a recurrence for the runtime T(n) of the above algorithm.• Use the recursion tree to show that the algorithm runs in time O(n).FindK(A,s,f,k)if s < fif f = s + 1if k = A[s] return sif k = A[f] return felseq1 = b(2s + f)=3cq2 = b(q1 + 1 + f)=2c... to be continued.else... to be continued.
Please Help me with this Problem Given an unsorted array of size n. Array elements are in the range from 1 to n. One number from set {1, 2,..., n} is missing and one number occurs twice in the array. Find these two numbers.
: 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;
Explain the flow of the code. See attached photo for the problem. (This is not graded) bool add(entry* e) { if(s->contains(e->country) == true) { return false; } int i; for (i = 0; i < index; i++) { entry* existing = array + (i * sizeof(entry)); // 0 + 0 * 5 = 0 if(e->compare(existing)) { break; } } for (int j = index; j >= i; j--) { *(array + ( sizeof(entry) * (j) ) ) = *(array + ( sizeof(entry) * (j - 1) ) ); } *(array + ( sizeof(entry) * i ) ) = *e; if(index < 5) { index++; } if(( sizeof(entry) * i ) < ( sizeof(entry) * 5 )) { return true; } else { return false; } } entry* get(int pos) { return array + (sizeof(entry) * (pos-1) ); } bool ban_indefinitely(entry* e) {…
1. Consider a problem of computing the prefix average of a sequence of numbers stored in an array P consisting of P integers. We want to compute an array P such that P is the average of P for P . Consider the two algorithms related to the above problem. Answer the questions given below
“A” array has a members and “B” array has b members which are sorted. I need an algorithm to find the k-th member in a new array which is a mix of “A” and “B”. The runtime must be O(log(a+b))
Let A be an array of numbers. In the maximum sub-array problem, your goal is to determine the sub-array A[x . . . y] of consecutive terms for which the sum of the entries is as large as possible. For example, if A = [−2, −3, 4, −1, −2, 1, 5, −3], the maximum sub-array is [4, −1, 2, 1, 5], and the largest possible sum is S = 4 − 1 − 2 + 1 + 5 = 7. Suppose A = [1, 2, −4, 8, 16, −32, 64, 128, −256, 512, 1024, −2048]. Determine S, the largest possible sum of a sub-array of A.
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])