Search in Rotated Sorted Array LeetCode Solution

In this blog post, we will tackle the Search In Rotated Sorted Array LeetCode Problem.

We’ll walk you through the problem, discuss the approach to solving it efficiently, and provide a Python code solution.

This problem is a great example of a binary search type problem, and we’ll break it down step by step to make it easy to understand, especially for beginners.

Problem Overview

The problem presents a scenario where you are given an integer array originally sorted in ascending order with distinct values.

However, this array has been rotated at an unknown pivot index (k), resulting in a partially sorted array.

Your task is to find a target integer in this rotated array and return its index.

If the target is not in the array, return -1.

The challenge here is to devise an algorithm with a runtime complexity of O(log n).

Example 1:

Let’s consider an example.

We are given the rotated array: [4, 5, 6, 7, 0, 1, 2], and the target is 0.

The expected output is 4, as the target is found at index 4 in the array.

Understanding the Search in Rotated Sorted Array LeetCode Problem Constraints

Before we delve into the solution, let’s understand the constraints of this problem:

The length of the input array, nums, is between 1 and 5000.
The values in nums range from -10,000 to 10,000.
All values in nums are unique.
The array nums is ascending but possibly rotated.
The target to find can also be in the range from -10,000 to 10,000.

Now that we’ve grasped the problem’s requirements, let’s explore the approach to solving it efficiently.

Efficient Solution to Search in Rotated Sorted Array

The most efficient way to approach this problem is by using binary search.

Binary search is known for its O(log n) runtime complexity and is a perfect fit for this problem.

We will break down the algorithm into several cases and iterate through the array, narrowing down the search range until we find the target element.

Here’s the Python code solution for this problem:

View on GitHub ➹

def search(self, nums: List[int], target: int) -> int:
    left, right = 0, len(nums) - 1

    while left &lt;&equals; right:
        mid = (left + right) // 2

        if nums[mid] == target:
            return mid

        if nums[left] &lt;&equals; nums[mid]:
            if nums[left] &lt;&equals; target &lt; nums[mid]:
                right = mid - 1
            else:
                left = mid + 1
        else:
            if nums[mid] &lt; target &lt;&equals; nums[right]:
                left = mid + 1
            else:
                right = mid - 1

    return -1

This code implements the binary search algorithm, considering the cases where the target can be in the left or right sorted portions of the rotated array.

We gradually reduce the search range until we either find the target or conclude that it’s not in the array.

Time and Space Complexity

The time complexity of this solution is O(log n), as it narrows down the search range in each iteration.

The space complexity is O(1) since we only use a few variables for tracking.

Conclusion

In this blog post, we explored the LeetCode problem 33: Search in Rotated Sorted Array.

We discussed the problem’s requirements, and constraints, and presented an efficient Python solution using the binary search algorithm.

Solving this problem efficiently is a great exercise for understanding binary search and working with rotated arrays.

We encourage you to practice this algorithm, leave your questions, suggestions, and comments below, and share this content with others who might find it helpful.

LeetCode Problem Link

Now you have a clear understanding of how to tackle this problem.

Happy coding!