## Given problem

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. 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.

## Using Binary Search algorithm

1. Using Template #1

`````` class Solution {
public int findMin(int[] arr) {
int len = arr.length;
int left = 0;
int right = len - 1;

while (left <= right) {
if (arr[left] <= arr[right]) {  // Case 1: sorted array
return arr[left];
}

int mid = left + (right - left) / 2;
int next = (mid + 1) % len;
int prev = (mid + len - 1) % len;
if (arr[mid] <= arr[next] && arr[mid] <= arr[prev]) {   // Case 2: mid index points to the minimum element
return arr[mid];
} else if (arr[mid] <= arr[right]) {    // Case 4
right = mid - 1;
} else if (arr[mid] >= arr[left]) {     // Case 3
left = mid + 1;
}
}

return -1;
}
}
``````
2. Using Template #2

`````` class Solution {
public int findMin(int[] nums) {
if (nums == null) {
return -1;
}

int left = 0;
int right = nums.length - 1;

while (left < right) {
int mid = left + (right - left) / 2;

if (nums[mid] > nums[right]) {
left = mid + 1;
} else {
right = mid;
}
}

return nums[left];
}
}
``````
3. Using Template #3

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

while (low + 1 < high) {
int mid = low + (high - low) / 2;

if (nums[mid] > nums[high]) {
low = mid;
} else {
high = mid;
}
}

return nums[low] < nums[high] ? nums[low] : nums[high];
}
}
``````

The complexity of this solution:

• Time complexity: O(logn)
• Space complexity: O(1)

## Wrapping up

• Understanding how to use some templates of Binary Search algorithm.

Refer:

153. Find Minimum in Rotated Sorted Array