### LeetCode Challenge #153. Find Minimum in Rotated Sorted Array

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.

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.
##### Video Solution
###### Java Solution
```				```
class Solution {
public int findMin(int[] nums) {
if(nums.length==1){
return nums[0];
}else if ( nums[0]<nums[nums.length-1]){
return nums[0];
}

int start = 0 ;
int end = nums.length-1 ;

while(start<=end){
int mid = (start + end )/2;

if(mid!=0 && nums[mid-1]>nums[mid]){
return nums[mid];
}else if( mid!=nums.length-1 && nums[mid]>nums[mid+1] ){
return nums[mid+1];
}else if(nums[start]<=nums[mid]){ // left part is sorted
start = mid+1 ;
}else{ // right part is sorted ;
end = mid-1 ;
}
}
return -1 ;
}
}
```
```