Search in Rotated Sorted Array

33. Search in Rotated Sorted Array

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

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

Example 3:

Input: nums = [1], target = 0
Output: -1

Constraints:

• 1 <= nums.length <= 5000
• -104 <= nums[i] <= 104
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -104 <= target <= 104

Analysis

We can divide the binary search by the relationship between mid , end and target. For example, there are 3 scenarios that we should consider

1. [4, 5, 6, 1, 2] with mid = 6, end = 2, target = 4

   In this case, if we are in current index of 2 with value 6, and the target value is 4, we need to move our right pointer to mid pos 2.

   The relationship is mid > end and target > end and mid > target

2. [6, 1, 2, 3, 4, 5] with mid =2, end = 5

• if the target is 6. we have the relationship mid <= end and target > end, so we need to move our right pointer to mid pos 2
• if the target is 1, we have the relationsip mid <= end and mid > target, so we need to move our right pointer to mid pos 2

Solution

class Solution:
    def search(self, nums: List[int], t: int) -> int:

        def is_blue(i): # should we move the right pointer to the mid
            x = nums[i]
            e = nums[-1]
            # e < t < x
            # 4 5 6 1 2 [t=4,x=6]
            if x > e:
                return t > e and x >= t
            else:
                # x < e and t > e
                # 4 5 0 1 2 3 [t=4,x=0,e=3] x <= e
                # x < e and x >= t
                # 4 0 1 2 3 [t=0,x=1,e=3]  x <= e
                return t > e or x >= t 

        l = -1
        r = len(nums)

        while l+1 < r:
            mid = (l+r)//2
            if is_blue(mid): # move the right pointer to the mid
                r = mid
            else:
                l = mid
        if r == len(nums) or nums[r] != t:
            return -1
        return r