Search A [i] = i in a sorted array with duplicates

Question

Given a sorted array of integers with possibly duplicates , how do you find an index isuch asA[i]=i

This is a problem in one of my books on programming (Cracking code interview). The solution is as follows:

 public static int magicFast(int[] array, int start, int end) {

    if (end < start || start < 0 || end >= array.length) {
     return -1;
     }

     int midlndex = (start + end) / 2;
     int midValue = array[midlndex];
     if (midValue == midlndex) {
       return midlndex;
     }

     /* Search left */
     int leftlndex = Math.min(midlndex - 1, midValue);
     int left = magicFast(array, start, leftlndex);
     if (left >= 0) {
     return left;
     }

     /* Search right */
     int rightlndex = Math.max(midlndex + i, midValue);
     int right = magicFast(array, rightlndex, end);
     return right;    
}

The author does not comment on the complexity of time. However, this is apparently a solution to O (n), since we need to look at both sides of the “midpoint”, in contrast to the problem in which the elements of the array are different. Recurrence ratio: T (n) = 2T (n / 2) + c (c is a constant time to check if the middle element is the answer)

How is it better than a simple linear scan? It is too complicated, just to achieve linear time efficiency. Did I miss something?

Answer 1

, . , , , .

. a,

a(i) = i + 1 for i ≠ j
a(j) = j

j. , , n - 1.

CTCI, , - a(i) - i , .