Question

Suppose you have a sorted array of positive and negative integers and would like to determine if there exist some value of x such that both x and -x are in the array. Consider the following three algorithms: Algorithm #1: For each element in the array, do a sequential search to see if its negative is also in the array. Algorithm #2:For each element in the array, do a binary search to see if its negative is also in the array. Algorithm #3: Maintain two indices i and j, initialized to the first and last element in the array, respectively. If the two elements being indexed sum to 0, then x has been found. Otherwise, if the sum is smaller then 0, advance i; if the sum is larger then 0 retreat j, and repeatedly test the sum until either x is found, or i and j meet. Determine the running times of each algorithm, and implement all three obtaining actual timing data for various values of N. Confirm your analysis with the timing data. (Language in Java)

Answer 1

Run time of first algorithm is O(n²). Outer loop will run for n times worst case and inner loop will also run n times.

Run time of second algorithm is O(n log(n)). Outer loop will run for n times worst case in binarySearch1 method and inner loop of binary search1 will run log(n) times.

Run time of third algorithm is O(n). There's only one loop here which will run over the array once if element is not found.

Code for the three algorithms with 3 different functions are calculated for given array,calling one function each time and commenting others to jot down the timings of runtime.

public class SortCheck {

   public static int linerSearch(int[] arr){
      
       int size = arr.length;
       for(int j=0;j<size;j++)
       {
           int key =-arr[j];
           for(int i=0;i<size;i++){
               if(arr[i] == key){
                   return i;
               }
           }
       }
       return -1;
   }
   public static int binarySearch(int[] arr){
      
       int size = arr.length;
       int value = -1;
       for(int j=0;j<size;j++)
       {
           int key =-arr[j];
           value = binarySearch1(arr,0,size-1,key);
           if(value>-1)
               return value;
       }
       return value;
      
   }
   public static int binarySearch1(int arr[], int l, int r, int x)
{
if (r >= l) {
int mid = l + (r - l) / 2;
if (arr[mid] == x)
return mid;

if (arr[mid] > x)
return binarySearch1(arr, l, mid - 1, x);
  
return binarySearch1(arr, mid + 1, r, x);
}
  
// when element is not present control reaches here
return -1;
}
  
   public static int thirdAlgo(int[] arr){
  
   int size = arr.length,i=0,j=size-1;
  
   while(i!=j)
   {
       int sum = arr[i]+arr[j];
       if(sum ==0)
           return i;
       else if(sum >0)
           j--;
       else
           i++;
   }
   return -1;
   }
  
  
   public static void main(String a[]){
       final long startTime = System.nanoTime();
       int[] arr1= {-23,-17,-4, -3,-2,1,4,5,7,8,14,25};
       //System.out.println("Element found at "+linerSearch(arr1));
       //System.out.println("Element found at "+binarySearch(arr1));
       System.out.println("Element found at "+thirdAlgo(arr1));
       final long duration = System.nanoTime() - startTime;
       System.out.println(duration);
   }
}

Sample output for 3rd algorithm

Running time tabulated for given array above

Algorithm	Running time(ns)
Linear Search	583661
Binary Search	488943
Third algo	369055