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:1Explanation:The original array was [1,2,3,4,5] rotated 3 times.

**Example 2:**

Input:nums = [4,5,6,7,0,1,2]Output:0Explanation: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:11Explanation: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.

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

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