The problem is asking to find the minimum element in a rotated sorted array. The array was initially sorted in ascending order, and then rotated. The rotation is to the right, and it can be done between 1 and n times.

Here is a Python solution using Binary Search, which ensures the time complexity is O(log n):

```python
def findMin(nums):
if len(nums) == 1:
return nums[0]
left = 0
right = len(nums) - 1
if nums[right] nums[0]:
return nums[0]
while right = left:
mid = left + (right - left) // 2
if nums[mid] nums[mid + 1]:
return nums[mid + 1]
if nums[mid - 1] nums[mid]:
return nums[mid]
if nums[mid] nums[0]:
left = mid + 1
else:
right = mid - 1
```

This function works by first checking if the last element of the array is greater than the first element. If it is, that means the array is not rotated, so we return the first element.

Then, we enter a while loop where we calculate the middle index. If the element at the middle index is greater than the element next to it, that means we found the point of rotation, so we return the next element.

If the element before the middle element is greater than the middle element, we found the point of rotation, so we return the middle element.

If the middle element is greater than the first element, that means the smallest element is on the right side of the middle element, so we move the left pointer to mid + 1.

Otherwise, the smallest element is on the left side, so we move the right pointer to mid - 1.

We keep doing this until we find the smallest element.

Question

The problem is asking to find the minimum element in a rotated sorted array. The array was initially sorted in ascending order, and then rotated. The rotation is to the right, and it can be done between 1 and n times.

Here is a Python solution using Binary Search, which ensures the time complexity is O(log n):

```python
def findMin(nums):
    if len(nums) == 1:
        return nums[0]
    left = 0
    right = len(nums) - 1
    if nums[right] > nums[0]:
        return nums[0]
    while right >= left:
        mid = left + (right - left) // 2
        if nums[mid] > nums[mid + 1]:
            return nums[mid + 1]
        if nums[mid - 1] > nums[mid]:
            return nums[mid]
        if nums[mid] > nums[0]:
            left = mid + 1
        else:
            right = mid - 1
```

This function works by first checking if the last element of the array is greater than the first element. If it is, that means the array is not rotated, so we return the first element.

Then, we enter a while loop where we calculate the middle index. If the element at the middle index is greater than the element next to it, that means we found the point of rotation, so we return the next element.

If the element before the middle element is greater than the middle element, we found the point of rotation, so we return the middle element.

If the middle element is greater than the first element, that means the smallest element is on the right side of the middle element, so we move the left pointer to mid + 1.

Otherwise, the smallest element is on the left side, so we move the right pointer to mid - 1.

We keep doing this until we find the smallest element.

Knowee AI · Accepted Answer