# Find Minimum In Rotated Sorted Array Problem

## Description

LeetCode Problem 153.

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 inO(log n) time.

Example 1:

1
2
3

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

Example 2:

1
2
3

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:

1
2
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.

## Sample C++ Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29

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