Leetcode - 153. Find Minimum in Rotated Sorted Array

Medium

Dec 22, 2025

#binary-search

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.

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.

Constraints:

Notes

Intuition

In a rotated sorted array, the minimum element is the pivot point where the rotation occurred. It's the only element where both neighbors are larger (or it's at a boundary).

Implementation

Use binary search. The stop condition is when the element before and after mid are both greater (or mid is at a boundary). To decide which half to search, compare nums[mid] with nums[r]—if mid is larger, the minimum is in the right half; otherwise, it's in the left half.

Edge-cases

Boundary checks are essential. When mid is 0 or n-1, we can't compare with neighbors, so include these cases in the stop condition to avoid index out of bounds.

Time: O(log n) — binary search halves the range each iteration
Space: O(1) — only tracking pointers

Complexity

Solution

from typing import List

class Solution:
    def findMin(self, nums: List[int]) -> int:
        n = len(nums)
        l, r = 0, n - 1

        while l <= r:
            m = (l + r) // 2
            if (m == 0 or nums[m - 1] > nums[m]) and (m == n - 1 or nums[m + 1] > nums[m]):
                return nums[m]
            elif nums[r] < nums[m]:
                # move towards smaller value
                l = m + 1
            else:
                # move towards smaller value
                r = m - 1
        return -1 # shouldn't get here
        



if __name__ == "__main__":
    # Include one-off tests here or debugging logic that can be run by running this file
    # e.g. print(solution.two_sum([1, 2, 3, 4], 3))
    solution = Solution()