Java: Consider the following algorithm for searching in an unsorted array. If the size of the array is 1, then check if it contains the element to be searched. Otherwise, divide the array into two halves, and  recursively search both halves. Which of (a)–(c) is false? The running time of this algorithm is O(N) The actual running time of this algorithm is likely to be better than sequential search. This is an example of a divide‐and‐conquer algorithm all of the above are true none of the above is true

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
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Java: Consider the following algorithm for searching in an unsorted array. If the size of the array is 1, then check if it contains the element to be searched. Otherwise, divide the array into two halves, and  recursively search both halves. Which of (a)–(c) is false?
The running time of this algorithm is O(N)
The actual running time of this algorithm is likely to be better than sequential search.
This is an example of a divide‐and‐conquer algorithm
all of the above are true
none of the above is true
Expert Solution
Step 1

This Algorithm is based on Merge sort. 

Pseudocode for Merge sort:

We shall now see the pseudocodes for merge sort functions. As our algorithms point out two main functions − divide & merge.

Merge sort works with recursion and we shall see our implementation in the same way.

procedure mergesort( var a as array )
   if ( n == 1 ) return a

   var l1 as array = a[0] ... a[n/2]
   var l2 as array = a[n/2+1] ... a[n]

   l1 = mergesort( l1 )
   l2 = mergesort( l2 )

   return merge( l1, l2 )
end procedure

procedure merge( var a as array, var b as array )

   var c as array
   while ( a and b have elements )
      if ( a[0] > b[0] )
         add b[0] to the end of c
         remove b[0] from b
      else
         add a[0] to the end of c
         remove a[0] from a
      end if
   end while
   
   while ( a has elements )
      add a[0] to the end of c
      remove a[0] from a
   end while
   
   while ( b has elements )
      add b[0] to the end of c
      remove b[0] from b
   end while
   
   return c
	
end procedure

 

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