# Find Minimum in Rotated Sorted Array

[Question]: 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, 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` of unique elements, return the minimum element of this array.
You must write an algorithm that runs in `O(log n) time.`

Example 1:

```Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.```
``````// TC: O(log(N))
// SC: O(1)
func findMin(_ nums: [Int]) ->Int {

var low = 0
var high = nums.count - 1
var ans = Int.max
while low <= high {
let mid = (low + high) / 2
//then arr[low] will always be
//the minimum in that search space:
if nums[low] <= nums[high] {
ans = min(ans, nums[low])
break
}

// If left part is sorted:
if nums[low] <= nums[mid] {
// Keep the minimum:
ans = min(ans, nums[low])

// Eliminate left half:
low = mid + 1
} else { // If right part is sorted:
// Keep the minimum:
ans = min(ans, nums[mid])

// Eliminate right half:
high = mid - 1
}
}
return ans
}

let rotArrayInput = [4, 5, 6, 7, 0, 1, 2, 3]
let rotArrayOutput = findMin(rotArrayInput)
print("The minimum element is: ", rotArrayOutput)// 0``````