Find an element in a sorted rotated array

Problem:
Given a sorted integer array which is rotated any number of times and an integer num, find the index of num in the array.
If not found, return -1.

Solution:
1: Find index of pivot element (minimum element).
2: If num lies between start element and element at pivot-1 position, then find num in array[start…pivot-1] using binary search.
3: Else if num lies between pivot and last element, then find num in array[pivot…end] using binary search.

[IDeserve]

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[DhananjayKumarThakur]

Very Good Explaination !!! Liked it!!!

[Flower_withanshi]

does it same for ascending and descending?

[akshaygoyal]

Instead of checking arr[mid+1] as pivot can we go for checking arr[mid] as pivot?

[priyaranjansingh]

@IDeserve we can also find pivot by applying min heap which is logN complexity

[devkashyap]

Hi Saurabh,
Nice video . Great Job :)

[boloyeung]

suppose we have [6, 13, 71, 72, 73, 73] -> [73, 6, 13, 71, 72, 73]. Now array[0]<=array[len-1] is True but it is rotated... I mean if you have any duplicates (not necessary at the end) your algorithm for a pivot will not work. There was said nothing in the description that all elements in the array were unique.

[JimitRupani]

How do you find a pivot and as you said sorted array 10 is not in a sorted form in an array.