Day 20: Search in Rotated Sorted Array: Binary Search - leetcode - Python3

 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

SOLUTION

class Solution:

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

        l, r = 0, len(nums)-1

        while(l<=r):

            mid: int = (l+r)//2

            if target == nums[mid]:

                return mid

            elif nums[l] <= nums[mid]:

                if target > nums[mid] or nums[l] > target:

                    l = mid + 1

                else:

                    r = mid - 1

            else:

                if target < nums[mid] or nums[r] < target:

                    r = mid - 1

                else:

                    l = mid + 1

                   

        return -1

Time Complexity: O(log n)

Space Complexity:O(1)

How it works:

1. The l and r variables are initialized to the beginning and end of the array, respectively.
2. The while loop iterates until l is greater than or equal to r.
3. In each iteration, the mid variable is calculated as the middle element of the array.
4. The target value is compared to the mid value.
5. If the target value is equal to the mid value, the algorithm returns the mid value.
6. If the target value is less than the mid value, the algorithm sets the r variable to the mid value minus 1.
7. If the target value is greater than the mid value, the algorithm sets the l variable to the mid value plus 1.
8. If the target value is not found in the array, the algorithm returns -1.