# 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, a, a, …, a[n-1]] 1 time results in the array [a[n-1], a, a, a, …, 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
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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
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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
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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

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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];
}
};
``````