# Solving Leetcode Interviews in Seconds with AI: Find Minimum in Rotated Sorted Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "153" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.  

	# Explanation
	Here's the breakdown of the approach, complexity analysis, and the Python code:

*   **Approach:**
    *   Use Binary Search: Exploit the partially sorted nature of the rotated array.
    *   Compare mid element with rightmost element: If `nums[mid] > nums[right]`, the minimum element lies in the right half. Otherwise, it lies in the left half (including the mid).
    *   Iteratively narrow down the search space: Adjust `left` and `right` pointers accordingly until `left == right`, which points to the minimum element.

*   **Complexity:**
    *   Runtime: O(log n), Storage: O(1)

	
	# Code
	```python
	def findMin(nums):
    """
    Finds the minimum element in a rotated sorted array.

    Args:
        nums: A list of integers representing the rotated sorted array.

    Returns:
        The minimum element in the array.
    """
    left, right = 0, len(nums) - 1

    while left < right:
        mid = (left + right) // 2

        if nums[mid] > nums[right]:
            left = mid + 1
        else:
            right = mid

    return nums[left]
	```