Below is an array of elements that can be traversed via Linear and Binary search. You will be given five elements for each searching algorithms. You need to list down the elements traversed until it reaches to the element we are looking for: 33 54 74 124 148 248 274 203 443 445 452 457 470 512 520 10 11 12 13 14 15 570 500 834 730 783 a03 810 888 887 030 10 17 18 19 20 21 22 23 27 28 29 30 31 A. Linear Search 1. 763 2. 148 3. 476 4. 526 5. 74 B. Binary Search 1. 985 2. 590 3. 476 4. 887 5. 897

Below is an array of elements that can be traversed via Linear and Binary search. You will be given five elements for each searching algorithms. You need to list down the elements traversed until it reaches to the element we are looking for: 33 54 74 124 148 248 274 203 443 445 452 457 470 512 520 10 11 12 13 14 15 570 500 834 730 783 a03 810 888 887 030 10 17 18 19 20 21 22 23 27 28 29 30 31 A. Linear Search 1. 763 2. 148 3. 476 4. 526 5. 74 B. Binary Search 1. 985 2. 590 3. 476 4. 887 5. 897

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

solve this by c programming Write a program that deletes the first element in an array which matches a search key. Andprint the new array after deletion.
Heapsort has heapified an array to: 96 71 57 36 25 and is about to start the second for loop. What is the array after each loop iteration? i= 4: Ex: 86, 75, 30 i = 3: i = 2: i = 1:
A majority element is an element that makes up more than half of the items inan array. Given a positive integers array, find the majority element. If there is no majority element,return -1. Do this in O(N) time and 0(1) space.Input: 1 2 5 9 5 9 5 5 5Output: 5
120191234 2 20211234 Margaret 3 20207654 Mario 4 20193425 Hannah 5 20118888 Owen 6 7 Sue } www Graham Commetti Arzi Money Adelaide SA 5000 wwwwww 123 Station Street 12 High Street Dandenong Victoria 3175 34 Christmas Street Brisbane Queenland 7898 44 Main Road Sydney NSW 2000 344 Bank Drive Berwick Victoria 3806 public static void sortData(ArrayList student ArrayList) 0 I need to sort students from studentID.Thanks 7.9 9.2 6.9 4.3 7.2 PLEASE GIVE CORRECT ANSWER ONLY ASAP WITHIN 60 MINUTES. IT'S URGENT. PLEASE ANSWER CORRECTLY IF YOU DON'T KNOW THE ANSWER DON'T WASTE MY QUESTION. THANK YOU
In c language Consider this array unsigned int data [ ] = { 0xA3A4A5A6, 0x12131415 } write a for loop to print the array values in this order 0x1415 0x1213 0xA5A6 0xA3A4
Need the answer asap
First Pass: ( 5 1 4 2 8 ) \to ( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps them. ( 1 5 4 2 8 ) \to ( 1 4 5 2 8 ), Swap since 5 > 4 ( 1 4 5 2 8 ) \to ( 1 4 2 5 8 ), Swap since 5 > 2 ( 1 4 2 5 8 ) \to ( 1 4 2 5 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them. Second Pass: ( 1 4 2 5 8 ) \to ( 1 4 2 5 8 ) ( 1 4 2 5 8 ) \to ( 1 2 4 5 8 ), Swap since 4 > 2 ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted. Third Pass: ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) Finally, the array is sorted, and the algorithm can terminate. Bubble sort is a simple sorting algorithm. It works by repeatedly stepping through the list to be sorted, comparing each pair of…
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; }
ASAP