Binary_search using recursion

#1
#2

Hi @Megha2468 , your logic is little flawed you are not dividing the intervals in two halfs . let me explain u the algo along with the code
So what we do is basically ignore half of the elements just after one comparison.

Compare x(key) with the middle element.
If x matches with middle element, we return the mid index.
Else If x is greater than the mid element, then x can only lie in right half subarray after the mid element. So we recur for right half.
Else (x is smaller) recur for the left half.

code :-

// A recursive binary search function. It returns 
// location of x in given array arr[l..r] is present, 
// otherwise -1 
int binarySearch(int arr[], int l, int r, int x) 
{ 
    if (r >= l) { 
        int mid = l + (r - l) / 2; 
  
        // If the element is present at the middle 
        // itself 
        if (arr[mid] == x) 
            return mid; 
  
        // If element is smaller than mid, then 
        // it can only be present in left subarray 
        if (arr[mid] > x) 
            return binarySearch(arr, l, mid - 1, x); 
  
        // Else the element can only be present 
        // in right subarray 
        return binarySearch(arr, mid + 1, r, x); 
    } 
  
    // We reach here when element is not 
    // present in array 
    return -1; 
}

Try to implement this yourself and in case of any doubt feel free to ask :slight_smile:
If you got the logic then mark your doubt as resolved