Day 23: Find Minimum in Rotated Sorted Array: Binary Search - leetcode - Python3

 Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

• [4,5,6,7,0,1,2] if it was rotated 4 times.
• [0,1,2,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Example 1:

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 

Constraints:

• n == nums.length
• 1 <= n <= 5000
• -5000 <= nums[i] <= 5000
• All the integers of nums are unique.
• nums is sorted and rotated between 1 and n times.

                              SOLUTION 1

class Solution:

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

        result: int=nums[0]

        l, r = 0, len(nums)-1

        while l < r:

            mid = (l+r)//2

            result = min(result, nums[mid])

            if nums[r] < nums[mid]:

                l = mid +1

            else:

                r = mid - 1

       

        return min(result, nums[l])      

Time Complexity: O(log n)

Space Complexity:O(1)

How it works:

1. The function first initializes the result variable to the value of the first element in the list.
2. The l and r variables are initialized to 0 and the length of the list minus 1, respectively.
3. The while loop iterates while l is less than r.
4. In each iteration of the loop, the mid variable is calculated as the average of l and r.
5. The result variable is updated to the minimum of result and nums[mid].
6. The if statement checks if the element at index r is less than the element at index mid.
7. If the element at index r is less than the element at index mid, then the l variable is updated to mid + 1.
8. Otherwise, the r variable is updated to mid - 1.
9. The return statement returns the minimum of result and nums[l].

                                     SOLUTION 2

class Solution:

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

        h = min(nums)

        return h

Time Complexity: O(n)

Space Complexity:O(1)