search-in-rotated-sorted-array_33.py

# 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


# ---------------------------------------Runtime 60 ms Beats 22.48% Memory 16.6 MB Beats 42.58%---------------------------------------

# My solution
# Time complexity O(n)


class Solution:
    def search(self, nums: list[int], target: int) -> int:
        start, end = 0, len(nums) - 1
        while start <= end:
            if nums[start] == target:
                return start
            if nums[end] == target:
                return end
            start += 1
            end -= 1
        return -1


# Solution @neetcode
# Time complexity O(log n)


class Solution:
    def search(self, nums: list[int], target: int) -> int:
        l, r = 0, len(nums) - 1

        while l <= r:
            mid = (l + r) // 2
            if target == nums[mid]:
                return mid

            # left sorted portion
            if nums[l] <= nums[mid]:
                if target > nums[mid] or target < nums[l]:
                    l = mid + 1
                else:
                    r = mid - 1
            # right sorted portion
            else:
                if target < nums[mid] or target > nums[r]:
                    r = mid - 1
                else:
                    l = mid + 1
        return -1