# Leetcode 153 - Find Min in rotated array

`Note:`

- Use
`binary search`

so we can finish it with`O(logn)`

. - Use
`nums[mid] > nums[right]`

to check which part the mid is in. - If yes, it means we have to move
`left`

pointer to`mid + 1`

. - If not, in case
`nums[mid]`

is the answer, we need to do`right = mid`

.- For example, in
`[4,5,1,2,3]`

, nums[mid] = 1 < 3. If we do`right = mid - 1`

, we’d miss the ans.

- For example, in

`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[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] |

`Code:`

1 | /** |