How do you solve Find Minimum in Rotated Sorted Array II on LeetCode?

Find Minimum in Rotated Sorted Array II (LeetCode #154) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The presence of duplicates can complicate the binary search approach, requiring careful handling to avoid infinite loops.

What is the brute force solution for Find Minimum in Rotated Sorted Array II?

The brute force approach for Find Minimum in Rotated Sorted Array II is Brute Force, with O(n) time complexity and O(1) space complexity. The simplest way to find the minimum element is to iterate through the entire array and keep track of the smallest number found. This approach is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Minimum in Rotated Sorted Array II?

Find Minimum in Rotated Sorted Array II uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Two Pointers.

What are the interview tips for Find Minimum in Rotated Sorted Array II?

Always clarify the constraints and edge cases with the interviewer before diving into the solution. Explain your thought process as you work through the problem, especially when transitioning from brute force to optimal solutions. Practice implementing both brute force and optimal solutions to build a strong foundation.

What are common mistakes when solving Find Minimum in Rotated Sorted Array II?

Failing to handle duplicates correctly, which can lead to infinite loops in the binary search. Not considering edge cases, such as arrays with all identical elements.

#154

Find Minimum in Rotated Sorted Array II

Hard

ArrayBinary SearchBinary SearchTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Using a modified binary search allows us to efficiently find the minimum element in the rotated sorted array, even with duplicates. This method reduces the number of comparisons needed.

⚙️

Algorithm

6 steps

1Step 1: Initialize left and right pointers to the start and end of the array.
2Step 2: While left is less than or equal to right, calculate the mid index.
3Step 3: If the middle element is greater than the right element, the minimum must be in the right half, so move left to mid + 1.
4Step 4: If the middle element is less than the right element, the minimum is in the left half or could be mid, so move right to mid.
5Step 5: If the middle element is equal to the right element, decrement right to skip duplicates.
6Step 6: When left equals right, return the element at that index.

solution.py17 lines

1# Full working Python code
2
3def findMin(nums):
4    left, right = 0, len(nums) - 1
5    while left < right:
6        mid = (left + right) // 2
7        if nums[mid] > nums[right]:
8            left = mid + 1
9        elif nums[mid] < nums[right]:
10            right = mid
11        else:
12            right -= 1
13    return nums[left]
14
15# Example usage
16print(findMin([1, 3, 5]))  # Output: 1
17print(findMin([2, 2, 2, 0, 1]))  # Output: 0

ℹ

Complexity note: The time complexity is O(n) in the worst case due to the presence of duplicates, which can lead to a linear scan in some scenarios. However, the average case is O(log n) when duplicates are not present.

1The presence of duplicates can complicate the binary search approach, requiring careful handling to avoid infinite loops.
2Understanding the properties of rotated sorted arrays is crucial for applying binary search effectively.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.