1+ # Suppose an array of length n sorted in ascending order is rotated between 1 and n times.
2+ # For example, the array nums = [0,1,4,4,5,6,7] might become:
3+ 4+ # [4,5,6,7,0,1,4] if it was rotated 4 times.
5+ # [0,1,4,4,5,6,7] if it was rotated 7 times.
6+ # Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array
7+ # [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
8+ 9+ # Given the sorted rotated array nums that may contain duplicates, return the minimum element
10+ # of this array.
11+ 12+ # You must decrease the overall operation steps as much as possible.
13+ 14+ 15+ # Example 1:
16+ 17+ # Input: nums = [1,3,5]
18+ # Output: 1
19+ # Example 2:
20+ 21+ # Input: nums = [2,2,2,0,1]
22+ # Output: 0
23+ 24+ 25+ # Constraints:
26+ 27+ # n == nums.length
28+ # 1 <= n <= 5000
29+ # -5000 <= nums[i] <= 5000
30+ 31+ 32+ from typing import List
33+ class Solution :
34+ def findMin (self , nums : List [int ]) -> int :
35+ l = 0
36+ r = len (nums ) - 1
37+ 38+ while l < r :
39+ m = (l + r )// 2
40+ if nums [m ] == nums [r ]:
41+ r -= 1
42+ elif nums [m ] < nums [r ]:
43+ r = m
44+ else :
45+ l += 1
46+ return nums [l ]
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