Binary search if array contains duplicates

Question

Hi,

what is the index of the search key if we search for 24 in the following array using binary search.

array = [10,20,21,24,24,24,24,24,30,40,45]

I have a doubt regarding binary search that how does it works if a array has duplicate values.Can anybody clarify...

Answer 1

The array you proposed has the target value in the middle index, and in the most efficient implementations will return this value before the first level of recursion. This implementation would return '5' (the middle index).

To understand the algorithm, just step through the code in a debugger.

public class BinarySearch {
    public static int binarySearch(int[] array, int value, int left, int right) {
          if (left > right)
                return -1;
          int middle = left + (right-left) / 2;
          if (array[middle] == value)
                return middle;
          else if (array[middle] > value)
                return binarySearch(array, value, left, middle - 1);
          else
                return binarySearch(array, value, middle + 1, right);           
    }

    public static void main(String[] args) {
        int[] data = new int[] {10,20,21,24,24,24,24,24,30,40,45};

        System.out.println(binarySearch(data, 24, 0, data.length - 1));
    }
}

Answer 2

As pointed out by @Pleepleus it will return the index 5 from the first level of recursion itself. However I would like to point out few things about binary search :

1. Instead of using mid = (left + right)/2 , use mid = left + (right-left)/2

2. If you want to search for lower_bound or upper_bound of an element use the following algorithms:

   binLowerBound(a, lo, hi, x)
     if (lo > hi)
       return lo;
   
     mid = lo +  (hi - lo) / 2;
     if (a[mid] == x)
       return binLowerBound(a, lo, mid-1, x);
     else if (a[mid] > x)
       return binLowerBound(a, lo, mid-1, x);
     else
       return binLowerBound(a, mid+1, hi, x);
   
   binHigherBound(a, lo, hi, x)
     if (lo > hi)
       return lo;
     mid = lo + (hi - lo) / 2;
     if (a[mid] == x)
       return binHigherBound(a, mid+1, hi, x);
     else if (a[mid] > x)
       return binHigherBound(a, lo, mid-1, x);
     else
       return binHigherBound(a, mid+1, hi, x);
   

Answer 3

public class a{
    public static int binarySearch(int[] array, int value, int left, int right) {
          if (left > right)
                return -1;
          int middle = (left + right) / 2;
          if (array[middle] == value)
        {
            if(array[middle-1]<array[middle])
                return middle;
                 //return binarySearch(array, value, left, middle - 1);
                 else
                return binarySearch(array, value, left, middle - 1);
        }
          else if (array[middle] > value)
                return binarySearch(array, value, left, middle - 1);
          else
                return binarySearch(array, value, middle + 1, right);           
    }
public static int binarySearch1(int[] array, int value, int left, int right) {
          if (left > right)
                return -1;
          int middle = (left + right) / 2;
          if (array[middle] == value)
        {
            if(array[middle]<array[middle+1])
                return middle; 
                 else

                    return binarySearch1(array, value, middle + 1, right);           
        }
          else if (array[middle] > value)
                return binarySearch1(array, value, left, middle - 1);
          else
                return binarySearch1(array, value, middle + 1, right);           
    }

    public static void main(String[] args) {
        int[] data = new int[] {10,20,21,24,24,24,24,24,30,40,45};


        System.out.println(binarySearch(data, 24, 0, data.length - 1));     //First Index
        System.out.println(binarySearch1(data, 24, 0, data.length - 1));    //Last Index
    }
}