## Algorithm

Problem Name: 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.
```

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.

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
#include
#include

int findMin(int *nums, int numsSize) {
int l = 0;
int r = numsSize - 1;
while (l < r) {
int m = l + (r - l) / 2;
if (nums[l] > nums[m]) { /* right side is sorted */
r = m;
}
else if (nums[r] < nums[m]) { /* left side is sorted */
l = m + 1;
}
else { /* the sub-array is not rotated */
r = m;
}
}
return nums[l];
}

int main() {
int nums[] = { 3, 1, 2 };

assert(findMin(nums, sizeof(nums) / sizeof(nums[0])) == 1);

printf("all tests passed!\n");

return 0;
}
``````
Copy The Code &

Input

cmd
nums = [3,4,5,1,2]

Output

cmd
1

### #2 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int findMin(int[] nums) {
if (nums.length == 1) {
return nums[0];
}
int left = 0;
int right = nums.length - 1;
// Sorted but not rotated
if (nums[right] > nums[left]) {
return nums[left];
}
while (left <= right) {
int mid = (left + right) / 2;
// mid + 1 is point of rotation
if (nums[mid] > nums[mid + 1]) {
return nums[mid + 1];
}
// mid is point of rotation
if (nums[mid - 1] > nums[mid]) {
return nums[mid];
}
if (nums[mid] > nums[0]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return Integer.MAX_VALUE;
}
}
``````
Copy The Code &

Input

cmd
nums = [3,4,5,1,2]

Output

cmd
1

### #3 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const findMin = function (nums) {
let low = 0,
high = nums.length - 1

while (low < high) {
const mid = low + ((high - low) >> 1)
if (nums[mid] <= nums[high]) high = mid
else if (nums[mid] > nums[high]) low = mid + 1
}

return nums[low]
}
``````
Copy The Code &

Input

cmd
nums = [4,5,6,7,0,1,2]

### #4 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
findMin = min
``````
Copy The Code &

Input

cmd
nums = [4,5,6,7,0,1,2]

### #5 Code Example with C# Programming

```Code - C# Programming```

``````
namespace LeetCode
{
public class _0153_FindMinimumInRotatedSortedArray
{
public int FindMin(int[] nums)
{
if (nums.Length == 1) return nums[0];
if (nums[nums.Length - 1] > nums[0]) return nums[0];

int left = 0, right = nums.Length - 1;
while (left < right)
{
var mid = left + (right - left) / 2;

if (nums[mid] < nums[right]) right = mid;
else left = mid + 1;
}

return nums[right];
}
}
}
``````
Copy The Code &

Input

cmd
nums = [11,13,15,17]

Output

cmd
11