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.

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.

Idea:

Use Binary Search Template

• Time Complexity: O(logN)
• Space Complexity: O(1)

Solution:

``````/**
* @param {number[]} nums
* @return {number}
*/
var findMin = function(nums) {
// if only have one number
if (nums.length === 1) return nums[0];

let l = 0;
let r = nums.length;
// if this array is sorted
if (nums[0] < nums[r - 1]) return nums[0];

while (l < r) {
let mid = l + Math.floor((r - l) / 2);

// if previous element is greater than next element,
// next element is the smallest
if (nums[mid] > nums[mid + 1])
return nums[mid + 1];
if (nums[mid - 1] > nums[mid])
return nums[mid];

// first half is sorted: min is on right side
if (nums[mid] > nums[0])
l = mid;
else
r = mid + 1;
}
return -1;
};``````

Solution 2:

``````var findMin = function(nums) {
return searchMin(nums, 0, nums.length - 1);
};

function searchMin(nums, l, r) {
// exit recursion:
// only one or two elements in the array
if (l + 1 >= r)
return Math.min(nums[l], nums[r]);
// this array is sorted
if (nums[l] < nums[r])
return nums[l];

let mid = l + Math.floor((r - l) / 2);
return Math.min(searchMin(nums, l , mid), searchMin(nums, mid + 1, r));
}``````