153. Find Minimum in Rotated Sorted Array

Medium LeetCode

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.

How to solve the problem

1. LinearSearch with T = O(n)

2. 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)