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76 lines
1.8 KiB
Python
76 lines
1.8 KiB
Python
"""
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153. Find Minimum in Rotated Sorted Array
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Difficulty: Medium
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https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/
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──────────────────────────────────────────────────
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Suppose an array of length n sorted in ascending order is rotated
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between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7]
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might become:
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• [4,5,6,7,0,1,2] if it was rotated 4 times.
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• [0,1,2,4,5,6,7] if it was rotated 7 times.
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Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time
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results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
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Given the sorted rotated array nums of unique elements, return the
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minimum element of this array.
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You must write an algorithm that runs in O(log n) time.
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Example 1:
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Input: nums = [3,4,5,1,2]
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Output: 1
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Explanation: The original array was [1,2,3,4,5] rotated 3 times.
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Example 2:
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Input: nums = [4,5,6,7,0,1,2]
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Output: 0
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Explanation: The original array was [0,1,2,4,5,6,7] and it was
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rotated 4 times.
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Example 3:
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Input: nums = [11,13,15,17]
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Output: 11
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Explanation: The original array was [11,13,15,17] and it was rotated
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4 times.
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Constraints:
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• n == nums.length
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• 1 <= n <= 5000
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• -5000 <= nums[i] <= 5000
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• All the integers of nums are unique.
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• nums is sorted and rotated between 1 and n times.
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"""
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class Solution:
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def findMin(self, nums: List[int]) -> int:
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l, r = 0, len(nums) - 1
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lowest_index = -1
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while l <= r:
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m = (l + r) // 2
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if nums[m] <= nums[-1]:
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lowest_index = m
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r = m - 1
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else:
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l = m + 1
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return nums[lowest_index]
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