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author | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
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committer | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
commit | d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch) | |
tree | 96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /find_min_rotated_arr.py | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
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Diffstat (limited to 'find_min_rotated_arr.py')
-rw-r--r-- | find_min_rotated_arr.py | 34 |
1 files changed, 34 insertions, 0 deletions
diff --git a/find_min_rotated_arr.py b/find_min_rotated_arr.py new file mode 100644 index 0000000..18f5390 --- /dev/null +++ b/find_min_rotated_arr.py @@ -0,0 +1,34 @@ +# 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' mine become a +different array. return the sorted array unique elements minimum times. +algorithm must be o log n time. +""" + + +class Solution(object): + def findMin(self, nums): + """ + :type nums: List[int] + :rtype: int + """ + if len(nums) == 1 or nums[0] < nums[-1]: + return nums[0] + left, right = 0, len(nums) - 1 + while left <= right: + mid = left + (right - left) // 2 + if mid > 0 and nums[mid - 1] > nums[mid]: + return nums[mid] + if nums[mid] > nums[right]: + left = mid + 1 + else: + right = mid - 1 + + +if __name__ == "__main__": + obj = Solution() + print(obj.findMin([3, 4, 5, 1, 2])) + print(obj.findMin([4, 5, 6, 7, 0, 1, 2])) + print(obj.findMin([11, 13, 15, 17])) |