Find Minimum in Rotated Sorted Array

Problem Description: Given an array of distinct integers nums sorted in ascending order, the array is rotated at some unknown pivot point. Find the minimum element in the array.

Example:

python

# Example usage of the function
nums = [4,5,6,7,0,1,2]
print(findMin(nums))  # Output: 0

1. Clarify the problem: Before starting, let's ask some clarifying questions to ensure we fully understand the problem requirements:

• Can the array contain duplicate elements?
• Can the array be empty?
• Are there any constraints on the size of the array?

2. Analyze the problem: Let's break down the problem to better understand its components:

• Input: An array nums of distinct integers sorted in ascending order.
• Output: The minimum element in the array.
• Constraints: The array may be rotated at some unknown pivot point.

3. Design an algorithm: To solve this problem, we can use a modified binary search algorithm to find the minimum element in the rotated sorted array. By comparing the mid element with the start and end elements, we can determine the direction to move our search.

Here's an outline of the algorithm:

1. Initialize pointers left and right to the start and end indices of the array.
2. While left is less than right:
   • Calculate the middle index mid.
   • If nums[mid] is greater than nums[right], the minimum element lies to the right of mid. Set left = mid + 1.
   • If nums[mid] is less than nums[right], the minimum element lies to the left of mid or is equal to nums[mid]. Set right = mid.
   • If nums[mid] is equal to nums[right], decrement right by 1.
3. Return nums[left], which contains the minimum element.

4. Explain your approach: We will use a modified binary search algorithm to find the minimum element in the rotated sorted array. By comparing the middle element with the start and end elements, we can determine the direction to move our search.

5. Write clean and readable code:

Here's an implementation of the algorithm in Python:

python

def findMin(nums):
    left = 0
    right = len(nums) - 1

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

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

    return nums[left]

6. Test your code: Now, let's test the code with the example usage and some additional test cases:

python

# Example test case
nums = [4, 5, 6, 7, 0, 1, 2]
print(findMin(nums))  # Output: 0

# Additional test cases
nums = [3, 4, 5, 1, 2]
print(findMin(nums))  # Output: 1

nums = [11, 13, 15, 17]
print(findMin(nums))  # Output: 11

nums = [1]
print(findMin(nums))  # Output: 1

7. Optimize if necessary: The given solution has a time complexity of O(log n) and a space complexity of O(1), which is already optimal for this problem. There is no further optimization required.

8. Handle error cases: The given code assumes that the input array nums is not empty and contains distinct integers. It doesn't handle cases where the array is empty or contains duplicate elements. Error handling strategies can be discussed with the interviewer.

9. Discuss complexity analysis:

• Time complexity: The algorithm has a time complexity of O(log n), where n is the length of the input array. This is because the algorithm uses binary search to find the minimum element.
• Space complexity: The space complexity is O(1) since the algorithm uses a constant amount of extra space to store the pointers left and right.