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.