**source:**https://leetcode.com/problems/find-minimum-in-rotated-sorted-array**C/C++**

**Solution to LeetCode**problem

**153**.

**Find Minimum in Rotated Sorted Array**.

## 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[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.

## Examples

**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: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.

## Solution

The logic is pretty similar to the problem #33:

- We identify the sorted part of the array:
- If the value at the lower index
`<`

current min value:- Update minimum value.
- Next search will be in this part of the array.

- If not, then the minimum value is on the unsorted part.

- If the value at the lower index

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class Solution {
public:
int findMin(vector<int>& nums) {
int min = nums[0];
int l = 0;
int h = nums.size() - 1;
int m;
while (l <= h) {
m = (l + h) / 2;
if (nums[m] < min)
min = nums[m];
if (nums[l] <= nums[m]) {
if (nums[l] < min)
h = m - 1;
else
l = m + 1;
} else {
if (nums[l] < min)
l = m + 1;
else
h = m - 1;
}
}
return min;
}
};