# Conquer the Binary Search: Mastering LeetCode’s Rotated Sorted Array Challenge

## Unlock the secrets to efficiently finding target elements in rotated sorted arrays using binary search techniques.

### Introduction

Leetcode problems are a staple in coding interviews, challenging developers to optimize their problem-solving skills. One particularly intriguing problem involves finding an element in a rotated sorted array.

This task not only tests your understanding of binary search but also your ability to adapt this algorithm to more complex scenarios.

In this article, we will walk through the problem statement, explore the logic behind the solution, and provide a step-by-step guide to implementing it in Python.

### Understanding the Problem

The problem is as follows:

There is an integer array

`nums`

sorted in ascending order (withdistinctvalues).Prior to being passed to your function,

`nums`

ispossibly rotatedat an unknown pivot index`k`

(`1 <= k < nums.length`

) such that the resulting array is`[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]`

(0-indexed). For example,`[0,1,2,4,5,6,7]`

might be rotated at pivot index`3`

and become`[4,5,6,7,0,1,2]`

.Given the array

`nums`

afterthe possible rotation and an integer`target`

, returnthe index of`target`

if it is in`nums`

, or`-1`

if it is not in`nums`

.You must write an algorithm with

`O(log n)`

runtime complexity.

Example 1:`Input: nums = [4,5,6,7,0,1,2], target = 0 Output: 4`

Example 2:`Input: nums = [4,5,6,7,0,1,2], target = 3 Output: -1`

Example 3:`Input: nums = [1], target = 0 Output: -1`

Constraints:

`1 <= nums.length <= 5000`

`-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup>`

All values of

`nums`

areunique.

`nums`

is an ascending array that is possibly rotated.

`-10<sup>4</sup> <= target <= 10<sup>4</sup>`

### The Approach: Leveraging Binary Search

To solve this problem with `O(log n)`

runtime complexity, we need to leverage the binary search algorithm. The key challenge is to determine which part of the array is properly sorted, allowing us to effectively narrow down our search space.

### Step-by-Step Solution

Let's dive into the implementation details.

#### Initialize Variables

First, we initialize two pointers, `left`

and `right`

, to represent the start and end of the array, respectively.

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

#### Binary Search Loop

Next, we enter a loop that continues as long as `left`

is less than or equal to `right`

.

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

#### Check for Target at Midpoint

We check if the midpoint element is the target. If it is, we return the index.

```
if nums[mid] == target:
return mid
```

#### Determine Which Part is Sorted

We determine which part of the array is sorted by comparing the values at the `left`

, `mid`

, and `right`

indices.

```
if nums[left] <= nums[mid]:
# Left part is sorted
if nums[left] <= target < nums[mid]:
right = mid - 1
else:
left = mid + 1
else:
# Right part is sorted
if nums[mid] < target <= nums[right]:
left = mid + 1
else:
right = mid - 1
```

#### Target Not Found

If the loop exits without finding the target, we return `-1`

.

```
return -1
```

### Full Implementation

Here is the complete code:

```
from typing import List
def search(nums: List[int], target: int) -> int:
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return mid
if nums[left] <= nums[mid]:
# Left part is sorted
if nums[left] <= target < nums[mid]:
right = mid - 1
else:
left = mid + 1
else:
# Right part is sorted
if nums[mid] < target <= nums[right]:
left = mid + 1
else:
right = mid - 1
return -1
# Example usage
nums = [4, 5, 6, 7, 0, 1, 2]
target = 0
print(search(nums, target)) # Output: 4
target = 3
print(search(nums, target)) # Output: -1
nums = [1]
target = 0
print(search(nums, target)) # Output: -1
```

### Common Questions and Answers

#### Q1: Why do we use binary search for this problem?

**A1**: Binary search is used because it allows us to achieve `O(log n)`

time complexity by repeatedly dividing the search space in half. This is essential for handling large datasets efficiently.

#### Q2: How do we determine which part of the array is sorted?

**A2**: We compare the values at the `left`

, `mid`

, and `right`

indices. If `nums[left] <= nums[mid]`

, the left part is sorted. Otherwise, the right part is sorted.

#### Q3: What if the array is not rotated?

**A3**: The algorithm still works because the conditions for determining the sorted part will be true for the entire array, effectively making it a standard binary search.

#### Q4: Can this algorithm handle duplicate values in the array?

**A4**: No, this implementation assumes all values are distinct. Handling duplicates would require additional checks to avoid ambiguity in determining the sorted part.

### Conclusion

Solving the rotated sorted array problem on Leetcode is an excellent way to hone your binary search skills.

By understanding how to identify the sorted part of the array and efficiently narrow down the search space, you can tackle this problem with confidence.

Happy coding!

### Related Articles

"Solving LeetCode's Combination Sum Problem: Optimized Techniques for Efficient Solutions"

"Rotting Oranges: A Comprehensive Guide to Solving with BFS in Python"

"Mastering LeetCode: Solving the "Product of Array Except Self" Problem"

"Getting Started with Studying for Software Engineering Interviews Using LeetCode"