0153: Find Minimum in Rotated Sorted Array

Problem Statement

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 Solution

import bisect

class Solution:
    def findMin(self, nums):
        def bsearch(arr, lx, rx, target):
            if lx <= rx:
                mx = lx + (rx - lx) // 2
                if int(arr[mx] < nums[0]) < target:
                    return bsearch(arr, mx + 1, rx, target)
                else:
                    temp = bsearch(arr, lx, mx - 1, target)
                    if temp == -1:
                        return mx
                    return temp
            return -1
        ix = bsearch(nums, 0, len(nums) - 1, 0.5)
        if ix == -1:
            return nums[0]
        return nums[ix % len(nums)]