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

Search in Rotated Sorted Array II (LeetCode #81) 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: Understanding how rotation affects the sorted order is crucial.

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

The brute force approach for Search in Rotated Sorted Array II is Brute Force, with O(n) time complexity and O(1) space complexity. The simplest way to solve this problem is to iterate through the entire array and check if the target exists. 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 Search in Rotated Sorted Array II?

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

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

Always clarify if duplicates are allowed in the array. Explain your thought process while coding to show your understanding. Practice identifying patterns in problems to improve your problem-solving skills.

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

Not handling duplicates correctly, leading to infinite loops. Assuming the array is strictly sorted without considering duplicates.

#81

Search in Rotated Sorted Array II

Medium

ArrayBinary SearchBinary SearchArray

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 locate the target in a rotated sorted array, even with duplicates. This approach reduces the number of comparisons significantly compared to brute force.

⚙️

Algorithm

6 steps

1Step 1: Initialize two pointers, left and right, 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: Check if nums[mid] equals the target; if so, return true.
4Step 4: If nums[left] is equal to nums[mid], increment left to skip duplicates.
5Step 5: Determine which side is sorted. If the left side is sorted and target lies within that range, move right; otherwise, move left.
6Step 6: If the right side is sorted and target lies within that range, move left; otherwise, move right.

solution.py21 lines

1# Full working Python code
2
3def search(nums, target):
4    left, right = 0, len(nums) - 1
5    while left <= right:
6        mid = (left + right) // 2
7        if nums[mid] == target:
8            return True
9        if nums[left] == nums[mid]:
10            left += 1
11        elif nums[left] <= nums[mid]:
12            if nums[left] <= target < nums[mid]:
13                right = mid - 1
14            else:
15                left = mid + 1
16        else:
17            if nums[mid] < target <= nums[right]:
18                left = mid + 1
19            else:
20                right = mid - 1
21    return False

ℹ

Complexity note: This complexity is due to the worst-case scenario where we might have to check every element in the array, especially when duplicates are present.

1Understanding how rotation affects the sorted order is crucial.
2Identifying duplicates can complicate the search process.

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