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---
title: '153. Find Minimum in Rotated Sorted Array'
description: '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:'
sidebar:
label: 'Find Minimum in Rotated Sorted Array'
badge: 'Medium'
---
<Badge variant="accent">Array</Badge>
::::warning
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]].
::::
### 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.
## Approach
```mermaid
flowchart TD
S(["findMin(nums)"]) --> I["l = 0, r = n-1, lowest_index = -1"]
I --> W{"l <= r?"}
W -- no --> E(["return nums[lowest_index]"])
W -- yes --> M["m = (l + r) // 2"]
M --> C{"nums[m] <= nums[-1]?"}
C -- yes --> A["in the right sorted run: record lowest_index = m, r = m - 1"]
A --> W
C -- no --> B["still in the left sorted run: l = m + 1"]
B --> W
```
## Solution
```py
class Solution:
def findMin(self, nums: List[int]) -> int:
l, r = 0, len(nums) - 1
lowest_index = -1
while l <= r:
m = (l + r) // 2
if nums[m] <= nums[-1]:
lowest_index = m
r = m - 1
else:
l = m + 1
return nums[lowest_index]
```
## Explanation
<YouTube id="nIVW4P8b1VA" title="Find Minimum in Rotated Sorted Array - Binary Search - Leetcode 153 - Python" />