Find Minimum in Rotated Sorted Array

Problem

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.

Solution

Approach: Modified Binary Search

The key insight is that the minimum element is at the “rotation point” where a larger element is followed by a smaller element. We can use binary search by comparing the middle element with the rightmost element to determine which half contains the minimum.

Logic:

If nums[mid] > nums[right], the minimum is in the right half (rotation point is to the right)
If nums[mid] <= nums[right], the minimum is in the left half (including mid)

Implementation

class Solution:
    def findMin(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1

        while left < right:
            mid = (left + right) // 2

            # If mid element is greater than right, minimum is in right half
            if nums[mid] > nums[right]:
                left = mid + 1
            else:
                # Minimum is in left half (including mid)
                right = mid

        return nums[left]

Alternative (Comparing with Left):

class Solution:
    def findMin(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1
        result = nums[0]

        while left <= right:
            # If current subarray is already sorted
            if nums[left] < nums[right]:
                result = min(result, nums[left])
                break

            mid = (left + right) // 2
            result = min(result, nums[mid])

            # Determine which half to search
            if nums[mid] >= nums[left]:
                # Left half is sorted, search right
                left = mid + 1
            else:
                # Right half is sorted, search left
                right = mid - 1

        return result

Complexity Analysis

Time Complexity: O(log n), where n is the length of the array. We halve the search space with each iteration.
Space Complexity: O(1), we only use a constant amount of extra space.

Key Insights

Rotation Point: The minimum element is at the rotation point, where the array “breaks” from its sorted order. Before the rotation point, elements are from the second part of the original array; after it, elements are from the first part.
Binary Search Modification: We compare nums[mid] with nums[right] (not nums[left]) to determine which half is sorted and which contains the rotation point.
Sorted Subarray Check: If the entire current subarray is sorted (nums[left] < nums[right]), then nums[left] is the minimum in that range.
Why Compare with Right: Comparing with the right element gives us a clear decision:
- nums[mid] > nums[right] → rotation point is to the right
- nums[mid] <= nums[right] → rotation point is to the left (or mid is minimum)
No Duplicates: This problem assumes all elements are unique. With duplicates, we’d need to handle the case where nums[mid] == nums[right] by linearly searching.