Question

Suppose you are given an array A holding n distinct integers (negative values are allowed) in sorted order; in other words, A[i] < A[i + 1] for each i ∈ [0, n − 2]. We say the ith element is self referential if A[i] = i. Design an O(log n) time algorithm to determine if there is a self referencial element in the array. Your solution must include

a) Statement of your algorithm in plain English. (Pseudo-code is optional.)

b) Short proof of correctness.

c) Time complexity analysis: Derive recurrence and solve by unrolling

Please answer with pseudo code.

Answer 1

a. Algorithm is very simple. Since entire array is sorted in ascending order, we will perform binary search to find an index i such that A[i]=i if it exist. Below is the algorithm.

FIND_INDEX(A, n)

1. Begin = 1, End = n, Mid = (Begin + End)/2

2. While(Begin < End )

3...........If( A[Mid] ==Mid || A[Mid+1] == Mid)

4..................return ( A[Mid] ==Mid ? Mid : Mid+1 ) //Return the index among Mid and Mid+1 whichever is equal with index

5...........else if( A[Mid] > Mid) //Since every index i greater than Mid will have A[ i ] > i, hence search in lower half indices

6...................End = Mid

7............else

8...................Begin = Mid //otherwise search in upper half indices from Mid to End

9. Return FALSE //In case when we are unable to find such index

b. Proof of correctness:- Since entire elements are integers and they are distinct elements and the array is sorted. Hence at any index i , if A[ i ] > i , then for all index j >= i , A[ j ] > j will hold because there will be increment in value of element in array corresponding to every increment in index. Similarly at index i, if A[ i ] < i , then for every index j <= i , A[ j ] < j. Hence our binary search at every iteration decides to search for index i with A[ i ] = i based on whether A[Mid] > Mid or A[Mid] < Mid or A[Mid] = Mid. Hence our algorithm is correct.

c. Since our algorithm, reduces the search space by half in every iteration, hence the time complexity recurrence relation will be

T(n) = T(n/2) + O(1)

Which will be solved as

T(n) = T(n/2) + O(1) = T(n/2²) + 2*O(1) =  T(n/2³) + 3*O(1) = ....=  T(n/2^k) + k*O(1)

If k = log₂ n

Then,

T(n) = T(n/2^k) + k*O(1) = T(1) + (log₂ n)*O(1) = O(log₂ n)

Please comment for any clarification.