Question

Java

For each of the algorithms unique1 and unique2, which solve the element uniqueness problem, perform an experimental analysis to determine the largest value of n such that the given algorithm runs in one minute or less. Hint: Do a type of “binary search” to determine the maximum effective value of n for each algorithm.

public static boolean unique1(int[] data) {
int n = data.length;
for (int j=0; j < n-1; j++)
for (int k=j+1; k < n; k++)
if (data[j] == data[k])
return false; // found duplicate pair
return true; // if we reach this, elements are unique
}

/** Returns true if there are no duplicate elements in the array. */
public static boolean unique2(int[] data) {
int n = data.length;
int[] temp = Arrays.copyOf(data, n); // make copy of data
Arrays.sort(temp); // and sort the copy
for (int j=0; j < n-1; j++)
if (temp[j] == temp[j+1]) // check neighboring entries
return false; // found duplicate pair
return true; // if we reach this, elements are unique
}

Answer 1

Java code:

import java.util.concurrent.TimeUnit;
import java.lang.Math;
import java.util.Arrays;

public class timeSearch
{
public static boolean unique1(int [] data,int n)
{
   //int n=data.length;
   for(int j=0;j<n-1;j++)
   {
       for(int k=j+1;k<n;k++)
       {
           if(data[j]==data[k])
               return(false);
       }
   }
   return(true);
}

public static boolean unique2(int [] data,int n)
{
   int []temp=Arrays.copyOf(data,n);
   Arrays.sort(temp);
   for(int j=0;j<n-1;j++)
   {
       if(temp[j]==temp[j+1])
           return(false);
   }
   return(true);
}

public static void main(String args[])
{
   int sz=(int)Math.pow(2,20);
   int data[]= new int[sz];
   int i;

   for(i=0;i<sz;i++)   //fill array
   {
       data[i]=sz-i+1;
   }

   long startTime,endTime,totalTime;

   i=1;
   totalTime=0;
   while(totalTime/Math.pow(10,6)<1)
   {
       startTime = System.nanoTime();
       unique1(data,i);
       endTime = System.nanoTime();
       totalTime = endTime - startTime;
       i++;
   }

   System.out.println("Value of n for which unique1 takes 10^(-6)second(1 microsecond):"+i);

   totalTime=0;
   i=1;
   while(totalTime/Math.pow(10,6)<1)
   {
       startTime = System.nanoTime();
       unique2(data,i);
       endTime = System.nanoTime();
       totalTime = endTime - startTime;
   i++;
   }
   System.out.println("Value of n for which unique1 takes 10^(-6)second(1 microsecond):"+i);
}
}

output: