# Find Minimum in Rotated Sorted Array

Updated: Mar 25, 2021

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, return *the minimum element of this array*.

**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.

**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:**

**Approach1 O(n):**

```
class Solution {
public int findMin(int[] nums) {
if(nums.length==1) return nums[0];
for(int i=0;i<nums.length;i++)
{
if(nums[i]>nums[(i+1)%nums.length])
{
return nums[(i+1)%nums.length];
}
}
return -1;
}
}
```

**Approach 2 O(log n) Binary Search:**

```
class Solution {
public int findMin(int[] nums) {
if(nums.length==1) return nums[0];
int start = 0, end=nums.length-1;
if(nums[0]<nums[end]) return nums[0];
while(start<=end)
{
int mid = start+(end-start)/2;
if(nums[mid+1]<nums[mid]) return nums[mid+1];
if(nums[mid-1]>nums[mid]) return nums[mid];
if(nums[mid]>nums[0])
{
start = mid+1;
}
else end = mid-1;
}
return -1;
}
}
```