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finlep [7]
3 years ago
13

A member of your team wants to add some funny pictures that he took during the company party to lighten the mood at the beginnin

g of the presentation. Does this violate any copyright laws?
Computers and Technology
2 answers:
san4es73 [151]3 years ago
7 0
No, because they are personal photos taken of an event that was not copyrighted. If the event was copyrighted then they would but if not then no.
adoni [48]3 years ago
7 0
I don't see how it can, seeing as the member took the pictures them self. 
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1 year ago
Digital learning can help students who enjoy
mr Goodwill [35]

Answer:

flexibility and independence

Explanation:

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2 years ago
You are given 4-bit ripple carry adders and logic gates you have learnt about in the class (AND, OR, NOT, XOR, XNOR, NAND, NOR).
katrin2010 [14]

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5 0
3 years ago
What are the major differences between searching and sorting in Java? What are some of the differences in the techniques used in
Studentka2010 [4]

Answer:  

  • Searching is a technique to look for an item or target value in a data structure like searching for a phone number in a directory.Data structure can be an array,List etc. Searching algorithms are used for searching. Most common examples are Linear Search and Binary Search. Lets take the example Linear Search in order to explain it using JAVA. Its the simplest searching algorithm. To search for a specific element, look at each element in the data structure sequentially and check if it matches with the element being searched for.
  • Sorting is a technique of arranging the elements in a specific order e.g. numerical sorting, ordering students according to their exam score. This order can be ascending or descending or alphabetical order. Contrary to search it returns the data structure e.g. an array in which the elements of array are sorted in a particular order. Sorting algorithms are used to sort elements in a data structure. Some common examples of sorting algorithms are Bubble Sort, Insertion Sort, Selection Sort, Merge Sort, Quick Sort, Heap Sort etc. JAVA uses Array.Sort() built-in function for sorting an array. By default it sorts the input array in ascending order.
  • Selection Sort: It is a sorting technique which divides an array into two subarrays. One subarray in the left is sorted and the other one at right is unsorted. This is an in-place algorithm. It is not a good option for large data. Initially the sorted part is empty and all elements are placed in unsorted array. First the element which is the smallest in the unsorted array is selected and swapped with the leftmost array element and becomes part of the sorted array. In each iteration the smallest element from the unsorted array is selected and moved to sorted part of the array.The worst case time complexity of this algorithm is O(n)^2 as we have to find the smallest for every element in the array.
  • Merge Sort: It is a comparison based algorithm. It works on divide and conquer technique. It uses recursion approach for sorting. This means it breaks the problem(lets say array list to be sorted) into sub problems (smaller parts) and then solves (in this case sorts) each sub problem in a recursive manner. At the end it merges the solutions (hence the merged sorted array). Although selection sort works faster when data set is small merge sort outperforms it for larger data sets. Merge sort is a stable algorithm and works best for linked lists. Its not an in place algorithm. Time complexity of merge sort is O(n*log n) for best, average and worst cases because it always divides the array in two parts and takes linear time to merge these part. O(n(logn)) time complexity makes it better,more efficient and faster to sort large data sets.
  •  Big Oh O notation is an asymptotic annotation written as O(n) which is a mathematical way to represent the upper bound of the running time of   algorithm (sorting algorithm in this case). It computes the worst case time complexity. Worst case time complexity means that the longest amount of time or maximum number of operations that will be required for a sorting algorithm to complete. The time complexity mostly gets affected as the size of the input varies.
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  • Searching algorithms are used when there is a need to find a specific data item from bulk of data item. Searching algorithms make this hectic process easier. For example you want to find phone number of person from directory. without searching algorithm looking for each phone number in the directory manually can be very time consuming. For example you have to find address of a customer number 254 from database to deliver a product. Instead  of manually looking for customer numbers you can simply use  linear search algorithm that will start from customer 1 and sequentially searches for specific customer 254 number and provides the address in a shorter time.
  • Sorting reduces complexity of problems e.g reducing the searching complexity. It is easier to locate data elements in a sorted list than unsorted. For example comparing two large data sets containing millions of records. If both the data sets are ordered, the comparison gets easier. Moreover every sorting algorithm has certain usage. Like merge sort is useful for linked lists,heap sort is good with arrays and uses less memory. If data is small with large values, selections sort is better for this. It doesn’t require any additional space. Databases use merge sort to arrange data that is too large to be loaded completely into memory.  Heap sort is used in reading bar codes on plastic cards. Quick sort is used to maintain sports score on the basis of win-loss ratio.
5 0
3 years ago
Let's implement a classic algorithm: binary search on an array. Implement a class named BinarySearcher that provides one static
yKpoI14uk [10]

Answer:

Hope this helped you, and if it did , do consider giving brainliest.

Explanation:

import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

 

//   main method

  public static void main(String[] args) {

     

//   create a list of Comparable type

     

      List<Comparable> list = new ArrayList<>();

     

//       add elements

     

      list.add(1);

      list.add(2);

      list.add(3);

      list.add(4);

      list.add(5);

      list.add(6);

      list.add(7);

     

//       print list

     

      System.out.println("\nList : "+list);

     

//       test search method

     

      Comparable a = 7;

      System.out.println("\nSearch for 7 : "+search(list,a));

     

      Comparable b = 3;

      System.out.println("\nSearch for 3 : "+search(list,b));

     

      Comparable c = 9;

      System.out.println("\nSearch for 9 : "+search(list,c));

     

      Comparable d = 1;

      System.out.println("\nSearch for 1 : "+search(list,d));

     

      Comparable e = 12;

      System.out.println("\nSearch for 12 : "+search(list,e));

     

      Comparable f = 0;

      System.out.println("\nSearch for 0 : "+search(list,f));

     

  }

 

 

 

//   static method named search takes arguments Comparable list and Comparable parameter

  public static boolean search(List<Comparable> list, Comparable par) {

     

//       if list is empty or parameter is null the throw IllegalArgumentException

     

      if(list.isEmpty() || par == null ) {

         

          throw new IllegalArgumentException();

         

      }

     

//       binary search

     

//       declare variables

     

      int start=0;

     

      int end =list.size()-1;

     

//       using while loop

     

      while(start<=end) {

         

//           mid element

         

          int mid =(start+end)/2;

         

//           if par equal to mid element then return

         

          if(list.get(mid).equals(par) )

          {

              return true ;

             

          }  

         

//           if mid is less than parameter

         

          else if (list.get(mid).compareTo(par) < 0 ) {

                 

              start=mid+1;

          }

         

//           if mid is greater than parameter

         

          else {

              end=mid-1;

          }

      }

     

//       if not found then retuen false

     

      return false;

     

  }

 

 

}import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

 

//   main method

  public static void main(String[] args) {

     

//   create a list of Comparable type

     

      List<Comparable> list = new ArrayList<>();

     

//       add elements

     

      list.add(1);

      list.add(2);

      list.add(3);

      list.add(4);

      list.add(5);

      list.add(6);

      list.add(7);

     

//       print list

     

      System.out.println("\nList : "+list);

     

//       test search method

     

      Comparable a = 7;

      System.out.println("\nSearch for 7 : "+search(list,a));

     

      Comparable b = 3;

      System.out.println("\nSearch for 3 : "+search(list,b));

     

      Comparable c = 9;

      System.out.println("\nSearch for 9 : "+search(list,c));

     

      Comparable d = 1;

      System.out.println("\nSearch for 1 : "+search(list,d));

     

      Comparable e = 12;

      System.out.println("\nSearch for 12 : "+search(list,e));

     

      Comparable f = 0;

      System.out.println("\nSearch for 0 : "+search(list,f));

     

  }

 

 

 

//   static method named search takes arguments Comparable list and Comparable parameter

  public static boolean search(List<Comparable> list, Comparable par) {

     

//       if list is empty or parameter is null the throw IllegalArgumentException

     

      if(list.isEmpty() || par == null ) {

         

          throw new IllegalArgumentException();

         

      }

     

//       binary search

     

//       declare variables

     

      int start=0;

     

      int end =list.size()-1;

     

//       using while loop

     

      while(start<=end) {

         

//           mid element

         

          int mid =(start+end)/2;

         

//           if par equal to mid element then return

         

          if(list.get(mid).equals(par) )

          {

              return true ;

             

          }  

         

//           if mid is less than parameter

         

          else if (list.get(mid).compareTo(par) < 0 ) {

                 

              start=mid+1;

          }

         

//           if mid is greater than parameter

         

          else {

              end=mid-1;

          }

      }

     

//       if not found then retuen false

     

      return false;

     

  }

 

 

}

7 0
2 years ago
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