One approach that might come up is to run the binary search to find the pivot and then search the element.
The interesting property of a sorted and rotated array is that when you divide it into two halves, atleast one of the two halves will always be sorted. We can use this property and code it to search in the sorted half or non-sorted half. Refer to the program below that uses O(log n) time complexity and O(1) memory.
You've successfully subscribed to Coding Today
Great! Next, complete checkout for full access to Coding Today
Welcome back! You've successfully signed in.
Unable to sign you in. Please try again.
Success! Your account is fully activated, you now have access to all content.