An integer sorted array rotated to left by unknown no of times, write an efficient algorithm to search for an element.
Example: 4 5 6 7 8 9 1 2 3 4
I am thinking every time I find a mid in Binary Search I compare the element with extreme end element and take a decision on which half to pick to repeat the process. Is it wrong? Or is there any efficient algo?
Your example array contains duplicates. When there are duplicates there is no efficient algorithm - you must always do O(n) work in the worst case.
To prove this, consider arrays of this form:
000000000000000010000000
It is a rotation of a sorted array, but in the worst cast you must iterate over every element to see where the 1 is.