leetcode_gym/_problems/searchInRotatedSortedArray.md

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layout	title	last_solved	category	leetcode_url	status
page	Search in Rotated Sorted Array		binary search	https://leetcode.com/problems/search-in-rotated-sorted-array	Attempted

Problem

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Solution

Do it in two parts:

• first find pivot (similar to find min in rotated sorted array)
• then reset boundaries to do a regular binary search

{% highlight python %}

def search(self, nums: List[int], target: int) -> int:

if not nums: return -1

left, right = 0, len(nums)-1

# get left to the pivot
while left<right:

    mid = (left+right)//2

    if nums[mid]<nums[right]:   # this is the condition that we want to satisfy w.r.t. mid

        right = mid

    else:

        left = mid+1

# left is pivot
start = left
left = 0
right = len(nums)-1

# target is on right side
if target>=nums[start] and target<=nums[right]:
    left = start
else:   # target is on left side
    right = start


# do binary search
while left<=right:
    
    mid = (left+right)//2
    
    if nums[mid]==target:
        return mid
    elif nums[mid]<target:
        left = mid+1
    else:
        right = mid-1

return -1

{% endhighlight %}