# Find Minimum in Rotated Sorted Array II LeetCode Solution Share:

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, a, a, ..., a[n-1]]` 1 time results in the array `[a[n-1], a, a, a, ..., 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:

```Input: nums = [1,3,5]
Output: 1
```

Example 2:

```Input: nums = [2,2,2,0,1]
Output: 0
```

Constraints:

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

### Find Minimum in Rotated Sorted Array II Solutions

Time: O(logn)→O(n)
Space: O(1)

### C++

``````class Solution {
public:
int findMin(vector<int>& nums) {
int l = 0;
int r = nums.size() - 1;

while (l < r) {
const int m = l + (r - l) / 2;
if (nums[m] == nums[r])
--r;
else if (nums[m] < nums[r])
r = m;
else
l = m + 1;
}

return nums[l];
}
};
``````

### Java

`````` class Solution {
public int findMin(int[] nums) {
int l = 0;
int r = nums.length - 1;

while (l < r) {
final int m = l + (r - l) / 2;
if (nums[m] == nums[r])
--r;
else if (nums[m] < nums[r])
r = m;
else
l = m + 1;
}

return nums[l];
}
}
``````

### Python

``````
class Solution:
def findMin(self, nums: List[int]) -> int:
l = 0
r = len(nums) - 1

while l < r:
m = (l + r) // 2
if nums[m] == nums[r]:
r -= 1
elif nums[m] < nums[r]:
r = m
else:
l = m + 1

return nums[l]
``````

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