Question

Write Java program to compare time consumed by linear search and binary search to search for non-exit element in arrays of length 1000, 100000,500000,1000000. Hints: - Use Random class, method nextInt(a.length) to generate random numbers. - Select an element that is greater than a.length. - Use Arrays.sort(a) to sort the array before using binarySearch.

Answer 1

PROGRAM:


elapsed = endTime-startTime; this was missing many times
Updated Code
import java.util.Arrays;
import java.util.Random;
public class Temp {
public static void main(String[] args) {
  
int a1[] = new int[1000];
int a2[] = new int [100000];
int a3[] = new int [500000];
int a4[] = new int[1000000];
long startTime = 0;
long endTime = 0;
long elapsed;
  
  
Random r = new Random();
  
for(int i = 0; i<a1.length; i++)
a1[i]=r.nextInt(1000);
  
for(int i = 0; i<a2.length; i++)
a2[i]=r.nextInt(100000);
  
for(int i = 0; i<a3.length; i++)
a3[i]=r.nextInt(500000);
  
for(int i = 0; i<a4.length; i++)
a4[i]=r.nextInt(10000000);
  
// testing linear search
  
System.out.println("Array with length 1000");
  
startTime = System.nanoTime();
linearSearch(a1, 1100);
endTime = System.nanoTime();
elapsed = endTime-startTime;
System.out.print("Linear search time: ");
System.out.println(elapsed);
  
  
Arrays.sort(a1);
startTime = System.nanoTime();
binarySearch(a1,1100);
endTime = System.nanoTime();
elapsed = endTime-startTime;
System.out.print("Binary search time: ");
System.out.println(elapsed);
  
////////////////////////////////////////////
  
System.out.println(" ");
System.out.println("Array with size 100000");
  
startTime = System.nanoTime();
linearSearch(a2, 110000);
endTime = System.nanoTime();
System.out.print("Linear search time: ");
elapsed = endTime-startTime;
System.out.println(elapsed);
  
Arrays.sort(a2);
startTime = System.nanoTime();
binarySearch(a2,110000);
endTime = System.nanoTime();
elapsed = endTime-startTime;
System.out.print("Binary search time: ");
System.out.println(elapsed);
  
System.out.println(" ");
System.out.println("Array with size 500000");
  
startTime = System.nanoTime();
linearSearch(a3, 550000);
endTime = System.nanoTime();
elapsed = endTime-startTime;
System.out.print("Linear search time: ");
System.out.println(elapsed);
  
Arrays.sort(a3);
startTime = System.nanoTime();
binarySearch(a3,550000);
endTime = System.nanoTime();
elapsed = endTime-startTime;
System.out.print("Binary search time: ");
System.out.println(elapsed);
  
System.out.println(" ");
System.out.println("Array with size 1000000");
  
startTime = System.nanoTime();
linearSearch(a4, 1100000);
endTime = System.nanoTime();
elapsed = endTime-startTime;
System.out.print("Linear search time: ");
System.out.println(elapsed);
  
Arrays.sort(a4);
startTime = System.nanoTime();
binarySearch(a4,1100000);
endTime = System.nanoTime();
elapsed = endTime-startTime;
System.out.print("Binary search time: ");
System.out.println(elapsed);
  
}
  
///// linear search method
public static int linearSearch(int a[], int x) {
  
for(int i = 0; i< a.length; i++) {
if(a[i] == x)
return i;
}
return -1;  
}
  
///// binary search method
  
public static int binarySearch(int a[], int x) {
int lo = 0;
int hi = a.length-1;
int mid;
  
while (lo<=hi) {
mid = (lo+hi)/2;
if(a[mid] == x)
return mid;
else {
if (a[mid] < x)
lo = mid + 1; // search in the second half
else
hi = mid - 1;
}
}
return -1;
}
}

OUTPUT:

Array with length 1000 Linear search time: 66000 Binary search time: 10800 Array with size 100000 Linear search time: 4581800 Binary search time: 9600 Array with size 589999 Linear search time: 911500 Binary search time: 7200 Array with size 1000000 Linear search time: 538900 Binary search time: 6800