Question

Suppose that we are given a sorted array of distinct integers A[1, ......,  n] and we want to decide whether there is an index i for which A[i] = i. Describe an efficient divide-and-conquer algorithm that solves this problem and explain the time complexity.

1. Describe the steps of your algorithm in plain English.
2. Write a recurrence equation for the runtime complexity.
3. Solve the equation by the master theorem.

Answer 1

We call the element fixed point if it is equal to index

Algorithm->

First check whether middle element is Fixed Point or not.

If it is, then return it;

otherwise check whether index of middle element is greater than value at the index.

If index is greater, then Fixed Point lies on the right side of the middle point .

Else the Fixed Point lies on left side.

Recurrence-> T(n) = T(n/2)+O(1)

By masters theorem,

a=1, b=2, f(n) = n^0 => c=0 (n^c)

since logb(a) = 0 = c

by case 2 of masters theorem, T(n) = O(n^0 * log(n)) = O(logn)