Objective: Given a sorted array of distinct integers, Find the Magic index or Fixed point in the array.
Magic Index or Fixed Point: Magic index or a Fixed point in an array is an index i in the array such that A[i] = i.
Example :
int[] A = { -1, 0, 1, 2, 4, 10 };
Magic index or fixed point is : 4
Approach:
Naive approach is to do the linear scan and find the magic index in O(n).
Better solution would Modify the Binary Search – Time Complexity O(logN).
- Check mid = (start+end)/2, with A[mid], check if A[mid] = mid. if yes then return mid.
- If mid>A[mid] means fixed point might be on the right half of the array, make a recursive call to search(A, mid + 1, end).
- If mid<A[mid] means fixed point might be on the left half of the array, make a recursive call to search(A, start, mid – 1).
Code:
Output:
Magic index 4
in the example you have given, the array is not sorted
Thanks kamal for pointing it out, corrected it.
this algorithm is wrong
Here the counter example :
array={1,2,3,3,4,5,6,7,8,9,10}
will return -1 not 3
Please check it out
The program is works for distinct integers, My bad the objective of the problem was confusing, i have corrected it. Thanks for finding it out.
Is there a similar binary search method for arrays with repeated elements or is linear scan the only way?
Thanks for post! Did you look into the case when elements are not distinct?