Find Minimum in Rotated Sorted Array

Problem

https://leetcode.com/problems/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.

Pattern

Array, Binary Search

Approaches

Code

def findMin(nums: list[int]) -> int:
    left = 0
    right = len(nums) - 1
    while left < right:
        mid = (left + right) // 2
        if nums[mid] < nums[right]:
            right = mid
        else:
            left = mid + 1
    return nums[left]

Test

>>> from find_minimum_in_rotated_sorted_array__binary_search import findMin
>>> findMin([3, 4, 5, 1, 2])
1
>>> findMin([4, 5, 6, 7, 0, 1, 2])
0
>>> findMin([11, 13, 15, 17])
11
find_minimum_in_rotated_sorted_array__binary_search.findMin(nums: list[int]) → int