### 154. Find Minimum in Rotated Sorted Array II

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

- [4,5,6,7,0,1,4] if it was rotated 4 times.
- [0,1,4,4,5,6,7] 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 that may contain duplicates, return the minimum element of this array.

You must decrease the overall operation steps as much as possible.

##### Example 1:

##### Example 2:

##### Constraints:

- n == nums.length
- 1 <= n <= 5000
- -5000 <= nums[i] <= 5000
- nums is sorted and rotated between 1 and n times.

From: LeetCode

Link: 154. Find Minimum in Rotated Sorted Array II

#### Solution:

###### Ideas:

**1. Initialization:** Set two pointers, left and right, at the beginning and end of the array, respectively.

**2. While Loop:** Continue searching as long as left is less than right.

**3. Middle Element:** Calculate the middle position mid between left and right.

**4. Decision Tree:**

- If nums[mid] is greater than nums[right], the minimum is in the right half (excluding mid), so move left to mid + 1.
- If nums[mid] is less than nums[right], the minimum could be mid or to the left of mid, so move right to mid.
- If nums[mid] equals nums[right], reduce right by one to gradually eliminate duplicates without skipping the minimum.

**5. Conclusion:** Once left equals right, the minimum element is found, as the search space is narrowed down to a single element.

###### Code:

```
int findMin(int* nums, int numsSize) {
int left = 0, right = numsSize - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] > nums[right]) {
left = mid + 1;
} else if (nums[mid] < nums[right]) {
right = mid;
} else { // nums[mid] == nums[right]
right--;
}
}
return nums[left];
}
```