Java BinarySearch

Question

Can I get some help please? I have tried many methods to get this to work i got the array sorted and to print but after that my binary search function doesnt want to run and

Answer 1

Well I know I am posting this answer much later.
But according to me its always better to check boundary condition at first.
That will make your algorithm more efficient.

    public static int binarySearch(int[] array, int element){
    if(array == null || array.length == 0){  // validate array
        return -1;
    }else if(element<array[0] || element > array[array.length-1]){ // validate value our of range that to be search 
        return -1;
    }else if(element == array[0]){  // if element present at very first element of array
        return 0;
    }else if(element == array[array.length-1]){ // if element present at very last element of array
        return array.length-1;
    }

    int start = 0;
    int end = array.length-1;

    while (start<=end){
        int midIndex = start + ((end-start)/2);   // calculate midIndex
        if(element < array[midIndex]){  // focus on left side of midIndex
            end = midIndex-1;
        }else if(element > array[midIndex]){// focus on right side of midIndex
            start = midIndex+1;
        }else {
            return midIndex;   // You are in luck :)
        }
    }
    return -1; // better luck next time :(
}

Answer 2

Here

    if (mid > key) 
        high = mid -1;
    else if (mid < key) 
        low = mid +1;
    else 
        return mid;

You're comparing index to a value (key) in array. You should instead compare it to a[mid]

And,

System.out.println("Found " + key + " at " + binarySearch(double a[], key));

Should be

System.out.println("Found " + key + " at " + binarySearch(a, key));

Answer 3

   /**
     * Find whether 67 is a prime no
     * Domain consists 25 of prime numbers
     * Binary Search
     */

    int primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};

    int min = 0,
            mid,
            max = primes.length,
            key = 67,
            count= 0;

    boolean isFound = false;


    while (!isFound) {
        if (count < 6) {
          mid = (min + max) / 2;
            if (primes[mid] == key) {
                isFound = true;
                System.out.println("Found prime at: " + mid);
            } else if (primes[mid] < key) {
                min = mid + 1; 
                isFound = false;
            } else if (primes[mid] > key) { 
                max = mid - 1; 
                isFound = false;
            }
            count++;

        } else {
            System.out.println("No such number");
            isFound = true;
        }

    }

Answer 4

I somehow find the iterative version not quite easy to read, recursion makes it nice and easy :-)

public class BinarySearch {

    private static int binarySearchMain(int key, int[] arr, int start, int end) {
        int middle = (end-start+1)/2 + start; //get index of the middle element of a particular array portion

        if (arr[middle] == key) {
            return middle;
        }

        if (key < arr[middle] && middle > 0) {
            return binarySearchMain(key, arr, start, middle-1); //recurse lower half
        }

        if (key > arr[middle] && middle < arr.length-1) {
            return binarySearchMain(key, arr, middle+1, end); //recurse higher half
        }

        return Integer.MAX_VALUE; 
    }

    public static int binarySearch(int key, int[] arr) { //entry point here
        return binarySearchMain(key, arr, 0, arr.length-1);
    }

}