153. Find Minimum in Rotated Sorted Array
Medium LeetCodeSuppose 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.
How to solve the problem
LinearSearch with T = O(n)
Binary Search: Compare the middle element and the right boundary element instead of
mid + 1
element. Because this is an ascending order array.
Code
- Approach 1: Linear Search
class Solution:
def findMin(self, nums: List[int]) -> int:
# Liner Search with T(n) = O(n)
ans = nums[0]
for i in range(len(nums)):
ans = min(ans, nums[i])
return ans
- Approach 2: Binary Search
class Solution:
def findMin(self, nums: List[int]) -> int:
# Binary Search
left = 0
right = len(nums) - 1
while left < right:
mid = (left + right) // 2
if nums[mid] > nums[right]: # Compare the middle element and the right boundary element instead of `mid + 1` element
left = mid + 1 # Make sure `left` doesn't get into a dead loop
else:
right = mid
return nums[right]
Complexity
Time complexity: O(log n)
Space complexity: O(1)
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