# Leetcode 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.

**Solution:**

`class Solution {`

public:

int findMin(vector<int>& nums) {

int start = 0;

int end = nums.size() - 1;

while ( end >= start) {

if (start + 1 == end) {

return nums[start] < nums[end] ? nums[start] : nums[end];

}

int mid = (start + end) / 2;

if (nums[start] <= nums[mid] && nums[mid] <= nums[end]) {

// all are ordered

return nums[start];

} else if (nums[start] <= nums[mid]) {

// left is ordered

start = mid;

} else {

// right is ordered

end = mid;

}

}

return 0;

}

};

Runtime: 9 ms, faster than 24.35% of C++ online submissions for Find Minimum in Rotated Sorted Array.

Memory Usage: 10.1 MB, less than 71.74% of C++ online submissions for Find Minimum in Rotated Sorted Array.