Search in Rotated Sorted Array

10 Prerequisite LeetCode Problems

“Search in Rotated Sorted Array” requires an understanding of binary search, along with the ability to handle edge cases and slight variations in the traditional binary search technique. Here are 10 problems to develop these skills:

“Binary Search” (LeetCode Problem #704): As the name suggests, this problem provides the basics of binary search and is a good starting point.
“First Bad Version” (LeetCode Problem #278): This problem applies the concept of binary search in a practical scenario and is an extension of basic binary search problems.
“Guess Number Higher or Lower” (LeetCode Problem #374): This problem further refines your understanding of binary search, with a focus on correctly identifying the high and low boundaries.
“Search Insert Position” (LeetCode Problem #35): This problem involves a variant of binary search where the target may not be present in the array.
“Find Smallest Letter Greater Than Target” (LeetCode Problem #744): This problem presents another variation on the binary search concept and practices finding not just an exact match but a relative match.
“Two Sum II - Input array is sorted” (LeetCode Problem #167): It helps understand the property of sorted arrays which will be beneficial in “Search in Rotated Sorted Array”.
“Find Minimum in Rotated Sorted Array” (LeetCode Problem #153): This problem involves a rotated sorted array, similar to the “Search in Rotated Sorted Array” problem, and practices finding a specific element (the minimum) in it.
“Find Peak Element” (LeetCode Problem #162): This problem provides practice on binary search in scenarios where the sorted order is not strictly increasing or decreasing, which is important for “Search in Rotated Sorted Array”.
“Rotated Sorted Array” (LeetCode Problem #33): This is an introductory problem to deal with search in a rotated sorted array.
“Squares of a Sorted Array” (LeetCode Problem #977): This problem deals with a sorted array and helps understand its properties.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left, right = 0, len(nums)-1

        while left<=right:
            mid = left+(right-left)//2
            if nums[mid] == target:
                return mid
            if nums[mid] < nums[right]:
                if nums[mid] < target<=nums[right]:
                    left=mid+1
                else:
                    right=mid-1
            else:
                if nums[left] <= target < nums[mid]:
                    right=mid-1
                else:
                    left = mid+1
        return -1

Problem Classification

It involves the use of an array (a fundamental data structure), and the algorithm to solve it is expected to have a certain time complexity (O(log n), implying a binary search or similar approach).

The ‘What’ components of the problem are:

Input: The problem takes as input an integer array nums that has been sorted in ascending order (with distinct values), but then possibly rotated at an unknown pivot index. Another input is an integer target.
Output: The output is the index of the target integer in the nums array if it exists, or -1 if it does not.
Constraints: There are specific constraints on the length of nums array and the range of values that elements in nums and target can take. Moreover, the algorithm must have a time complexity of O(log n).
Examples: The problem provides specific examples that clarify the problem statement and demonstrate the expected outputs given certain inputs.

We can classify this problem as a Search problem. The task is to search for a specific target in a rotated, sorted array. This problem requires knowledge of binary search algorithms or similar divide-and-conquer strategies that have logarithmic time complexity, and understanding of array manipulations.

The ‘Search in Rotated Sorted Array’ problem can also be categorized under Array Manipulation because it involves a rotation operation on a sorted array and then searching within that array. It requires understanding the properties of sorted arrays and how they change when they are rotated.

This problem falls under the general class of problems known as “Transformation and Query”, where a data structure is transformed in some way (in this case, a sorted array is rotated), and then queries (in this case, searching for a target value) are performed on the transformed data structure.

Distilling the Problem to Its Core Elements

Fundamental Concept: The key concept behind this problem is Binary Search, a fundamental algorithm in computer science used to efficiently find a target value within a sorted set of values. Despite the array rotation, the properties of sortedness still allow binary search to be used with some modifications.
Simplifying the Problem: Imagine you’ve lined up some numbers in increasing order but then decide to play a little game where you choose a spot in the line, pick up all the numbers to the right of that spot and place them at the beginning. Now, your task is to find a specific number in this jumbled line. If you can’t find it, just say it’s not there.
Core Problem: We have a list of unique numbers that was sorted in increasing order but then potentially rotated (meaning some numbers from the start may have been moved to the end). We need to determine if a specific number (the target) is in the list, and if so, at which position. If it’s not, return -1. The method we use to solve this problem should be fast, specifically, it should not require looking at every number in the list.
Key Components:
- Array: A list of numbers that was sorted but then potentially rotated.
- Target: A number we are trying to find within the array.
- Index: The position of the target within the array, which we need to return.
Minimal Set of Operations:
- Identify the middle element of the array.
- Determine which half of the array is in sorted order (left or right).
- If the target is within the range of the sorted half, perform a binary search on that half. If not, perform the search on the other half.
- Repeat the process with the chosen half of the array until the target is found or until there are no more elements to inspect.
- If the target is found, return its index. If not, return -1.

Visual Model of the Problem

Visualizing this problem can help you understand it better. Here’s how you can represent it:

Consider an array of numbers sorted in ascending order, for example, [0,1,2,4,5,6,7]. This array can be visualized as a line of numbers increasing from left to right.

0 - 1 - 2 - 4 - 5 - 6 - 7

The array might be rotated at an unknown pivot index. For example, if it was rotated at index 3, the number at that index (4 in this case) would move to the start of the array, followed by the rest of the numbers on its right. The numbers that were on its left would move to the end of the array.

The array would then look like this:

4 - 5 - 6 - 7 - 0 - 1 - 2

If we were asked to find the number 0 (the target), we could visualize it as identifying its position in this rotated line of numbers. In this case, 0 is at index 4 (0-indexed), so the output would be 4.

If we were asked to find a number not in the array, like 3, it would not be present in the line of numbers, and so the output would be -1.

Remember that our aim is to find the target using a method faster than looking at every number. Instead, we want to divide our line of numbers into sections in a strategic way that allows us to disregard large portions of it at a time, hence the requirement for O(log n) time complexity, hinting at a binary search approach.

This visualization can help you understand what’s happening when the array is rotated and what your task is when asked to find the target.

Problem Restatement

You’re given a list of unique numbers, initially sorted in ascending order, but possibly twisted at some point. By ’twisted’, we mean that a chunk of numbers from the start of the list might have been moved to the end. For instance, if we twist the list [0,1,2,4,5,6,7] at the 4th position, it turns into [4,5,6,7,0,1,2].

Your task is to find the position of a particular number (let’s call it ’target’) in this possibly twisted list. If the ’target’ number exists in the list, return its position (starting the count from 0). If it doesn’t exist in the list, return -1.

There is a key requirement to solve this task: you need to do it quickly, specifically, you can’t go through each number one by one. Your solution needs to work faster as the size of the list grows, implying that you may need to use a strategy like binary search.

Abstract Representation of the Problem

Abstracting this problem, we can represent it as follows:

We are given a sequence S of distinct elements, initially ordered in an ascending manner but potentially subjected to a rotation operation. A rotation operation means that a prefix sub-sequence of S has been moved to the end of the sequence, resulting in a new sequence R.

We also have a target element T.

The task is to find the position of T in the sequence R. If T is in R, we return its position (with the index starting from 0). If T is not in R, we return a special indicator, -1.

To satisfy the efficiency requirements of the problem, the function to determine the position of T must have a time complexity of O(log n), where n is the length of the sequence R.

This abstract representation focuses on the core structure of the problem and eliminates any specific real-world details, making it easier to approach the problem from an algorithmic perspective.

Terminology

There are several specialized terms and technical concepts that are crucial to understanding this problem:

Array: An array is a data structure that contains a group of elements. In this problem, the array contains integers and serves as the main data that we are working with.
Rotation (of an array): Rotation in the context of an array typically involves moving elements from one end of the array to the other. In this problem, a “rotation” means that some numbers from the start of the array have been moved to the end.
Pivot Index: In the context of array rotations, the pivot index is the position where the array has been rotated. Elements to the right of this position have been moved to the start of the array.
Target: The target is the specific number that we are trying to find within the array.
Binary Search: Binary search is a search algorithm that finds the position of a target value within a sorted array. It compares the target value to the middle element of the array and eliminates half of the array from consideration, thus significantly reducing the search space at each step. It’s important to this problem because the problem requires a logarithmic time complexity, hinting at a binary search or similar approach.
Time Complexity: Time complexity is a concept in computer science that describes the amount of computational time taken by an algorithm to complete its task. It’s usually expressed using Big O notation. In this problem, the requirement is for the solution to have a time complexity of O(log n), which means that the running time increases logarithmically with the size of the input.
Index: In the context of arrays, an index refers to the position of an element within the array. Array indices are typically zero-based, meaning that the first element is at position 0.

Understanding these terms is crucial for understanding the problem and formulating a solution.

Problem Simplification and Explanation

Imagine you have a book with pages numbered in order, and you’re looking for a particular page number. If the book is perfectly ordered, you could open it in the middle, and if the page number you’re looking for is higher than the middle page, you know you should look in the second half of the book. If it’s lower, you look in the first half. This process is similar to a binary search, and it’s quick because with each step you effectively halve the size of the book you need to look through.

Now, let’s change it up. What if your mischievous friend took some pages from the beginning of the book and stuck them at the end? Now, your book might look something like this: 23, 24, 25, 1, 2, 3, 4, …, 22. Despite this prank, some order still exists in this book. If you open the book in the middle and look at the page number, you can figure out which half of the book is “in order” (either the first half or the second half). Then, based on the page you’re looking for, you can figure out if it would be in the ordered half or the disordered half.

This jumbled book is like the rotated array in our problem, where the array is the book, and the page numbers are the array elements. Our goal is to find a specific number (the ’target’) in the array, just like finding a specific page in the book.

The key concept here is understanding that even though the array has been rotated, it still retains some sense of order. We can use this order to our advantage by employing a binary search strategy, similar to how we’d search in the book. Each step of a binary search reduces the size of the problem by half, which makes it a fast and efficient method to find a specific element in a sorted (or partly sorted, like in this case) array.

Constraints

Given the constraints and the problem statement, here are some specific characteristics or conditions that can be exploited to our advantage:

Size of the array: The size of the array nums is limited to a maximum of 5000. This is a relatively moderate size and any algorithm with a logarithmic time complexity like binary search will work efficiently on it.
Unique Elements: The fact that all elements of nums are unique can help simplify the problem. If the target number is found, we can return its index immediately without having to worry about duplicates.
Sorted Array: The array nums is sorted in ascending order before the possible rotation. This property allows us to use techniques like binary search to quickly locate the target value. The sorted nature also allows us to easily identify whether the part of the array we are examining is the rotated part or not, based on comparing the first and last elements of the portion.
Rotation: While rotation might seem to complicate the problem, it’s important to note that even after rotation, one part of the array remains sorted. This can be used to decide whether to search in the first half or the second half of the array.
Range of Elements: The elements of the array and the target are within the range of -10^4 to 10^4. These bounds can be useful in error checking and validation, though they don’t significantly contribute to optimizing the solution for this problem.

By understanding these conditions, we can design a more efficient solution that takes advantage of the sorted and rotated nature of the array.

The key insights from analyzing the constraints are as follows:

Moderate Size of Array: The array size is constrained between 1 and 5000. This indicates that our solution must be optimized for efficiency, but doesn’t necessarily need to handle extremely large inputs. An algorithm with logarithmic complexity, such as a binary search, would be able to process this size of an array efficiently.
Unique Values: All values in the array are unique. This eliminates the need to handle cases with duplicate values, simplifying our search algorithm. We know that if we find a match, it’s the only one.
Initial Ascending Order: The array is initially sorted in ascending order. This key insight should lead us to think about solutions that leverage this sorted nature, such as binary search.
Possible Rotation: The array could be rotated. This adds a level of complexity to the problem but also provides us a clue. If we’re able to determine the pivot point, we can essentially handle the problem as two sorted arrays.
Range of Values: The values in the array and the target are bound by -10^4 and 10^4. While this might not directly impact the approach for this problem, it’s good to keep this in mind to avoid integer overflow issues or for error checking.

Overall, the constraints suggest that a solution could involve a form of binary search adapted to handle the rotation of the array. The algorithm should be efficient to handle an array of size up to 5000, but doesn’t necessarily need to handle larger sizes.

Case Analysis

It’s important to consider different scenarios and edge cases to ensure that the solution is robust and works for all possible inputs. Here are several examples:

Minimum size array
- Input: nums = [1], target = 1
- Output: 0
- Explanation: The array contains just one element which is equal to the target, so the index of the target is 0.
Target not in array
- Input: nums = [1,2,3,4,5], target = 6
- Output: -1
- Explanation: In this case, the target number is not present in the array. The function should return -1.
Array rotated at first element
- Input: nums = [5,1,2,3,4], target = 1
- Output: 1
- Explanation: Here, the array is rotated at the first element. The target number 1 is at index 1 in the rotated array.
Array rotated at last element
- Input: nums = [2,3,4,5,1], target = 1
- Output: 4
- Explanation: Here, the array is rotated at the last element. The target number 1 is at index 4 in the rotated array.
Array with negative numbers
- Input: nums = [3,-1,0,1,2], target = -1
- Output: 1
- Explanation: The array can contain negative numbers as well. The function should work correctly even if the target or the array elements are negative. Here, the target number -1 is at index 1 in the rotated array.
Target in the first half of the array
- Input: nums = [4,5,6,7,0,1,2,3], target = 5
- Output: 1
- Explanation: In this scenario, the target number is in the first half of the rotated array. The target number 5 is at index 1.
Target in the second half of the array
- Input: nums = [4,5,6,7,0,1,2,3], target = 2
- Output: 6
- Explanation: In this scenario, the target number is in the second half of the rotated array. The target number 2 is at index 6.

These scenarios cover a wide range of possible inputs and edge cases, and they should help in ensuring that your solution is robust and works correctly in all situations.

Analyzing the different test cases provides several key insights into the problem:

Handling Small Arrays: In the case of a single-element array, it’s important to ensure that the algorithm can handle such small inputs. If the element matches the target, it should return the index, which in this case is 0.
Non-existing Target: The solution must also account for scenarios where the target is not present in the array. In such cases, the algorithm needs to end gracefully and return -1, indicating the target wasn’t found.
Location of Rotation Point: The location of the rotation point in the array is a crucial aspect of this problem. The rotation point could be at any location in the array. Whether it is at the beginning, end, or somewhere in the middle, the algorithm should still correctly find the target.
Negative Numbers: The algorithm should handle negative numbers in the array as well as when the target is negative. This can be a point of concern if absolute values or other operations that are sensitive to sign are used in the solution.
Target Location: The target number could be located in either half of the rotated array. Whether the target is in the sorted half or the rotated half of the array, the solution should be able to correctly locate it.
Return Index, Not Value: It’s important to remember that the problem asks for the index of the target value, not the value itself. The solution should be set up to return the correct index position rather than the value.

These insights can guide the development of a solution that handles all possible scenarios and edge cases, and ensure that the solution is robust and accurate.

Identification of Applicable Theoretical Concepts

The problem is closely related to a classic algorithmic concept in computer science called Binary Search. This is due to the fact that the input is a sorted (though rotated) array, and we are looking for a specific element in it.

Here are the key properties of Binary Search and how they apply to this problem:

Divide and Conquer: Binary Search is a divide and conquer algorithm, meaning it breaks the problem into smaller subproblems and combines their results to solve the original problem. In the context of this problem, we can use binary search to effectively narrow down the possible locations of the target value by dividing the array into two halves.
Time Complexity: Binary Search has a logarithmic time complexity of O(log n), which is very efficient and scalable for searching in sorted arrays. This aligns with the requirement of the problem for an algorithm with O(log n) runtime complexity.
Sorted Input: Binary Search operates on sorted arrays. In this problem, although the entire array might not be sorted due to the rotation, at least one half of the array is always sorted, which can be exploited to decide which half of the array to search for the target.

Apart from Binary Search, the problem is also related to Array Manipulation and Access concepts. Understanding how arrays work, how to access elements, and how indices are affected by rotation are key to solving this problem.

To summarize, the concept of Binary Search and understanding of Array Manipulation are crucial in this problem. By applying Binary Search, we can simplify the problem and make it more manageable by efficiently narrowing down the search space for the target value.

Problem Breakdown and Solution Methodology

To solve this problem, we can use a modified version of binary search, taking into account that the array is sorted but rotated. The trick here is to use the property of the sorted and rotated array - at any given point, at least one half of the array (either the left or the right) is always sorted. We use this fact to decide which half to continue our search in.

Here are the detailed steps:

Initialization: Start with defining two pointers, low and high, where low is at the start of the array, and high is at the end of the array.
Middle Point Calculation: In each step, find the middle point of the current search space defined by low and high. This can be done by (low + high)/2.
Determine Sorted Half: Check which part of the array is sorted. If nums[low] <= nums[mid], then the left half is sorted. If nums[mid] <= nums[high], then the right half is sorted.
Decide Which Half to Search: If the left half is sorted, check if the target lies in the range nums[low] to nums[mid]. If it does, continue the search in the left half by setting high to mid. If it doesn’t, the target must be in the right half, so set low to mid + 1. Similarly, if the right half is sorted and the target lies in the range nums[mid] to nums[high], continue the search in the right half by setting low to mid + 1. Otherwise, continue in the left half by setting high to mid.
Repeat the Search: Continue the process until low is not greater than high.
Find Target or Return -1: If the target is found, return the index. If the search space is empty (low > high), return -1, indicating that the target is not in the array.

Consider a librarian trying to find a specific book in a library. The books are arranged in ascending order of their ID numbers, but due to some shifting, some books from the start are moved to the end while keeping their relative order. The librarian uses the binary search technique. She goes to the middle of the library. If the required book’s ID lies in the sorted part, she goes to that part, else she goes to the other part.

Let’s apply this approach to an example:

Given nums = [4,5,6,7,0,1,2], target = 0

Initialize low = 0 and high = 6.
Calculate mid = (0+6)/2 = 3. So, nums[mid] = 7.
Since nums[low] <= nums[mid], the left half is sorted.
As 0 does not lie in between nums[low] and nums[mid], the target must be in the right half. So, update low = mid + 1 = 4.
Repeat the process. Now, mid = (4+6)/2 = 5, nums[mid] = 1. The right half is sorted, and the target lies in the right half. So, again, low = mid + 1 = 6.
Now, low = 6, high = 6, mid = 6, and nums[mid] = nums[6] = 2. The target 0

Apologies for the abrupt cut-off in the last response. Let’s continue from where we left off:

Now, low = 6, high = 6, mid = 6, and nums[mid] = nums[6] = 2. The target 0 does not match nums[mid] but since nums[mid] <= nums[high], we know the right half is sorted. However, 0 does not lie in this half, so we update high to mid - 1 = 5.
Finally, low = 6 and high = 5, which means our search space is empty and we didn’t find the target. In this case, however, that’s a mistake as we do have 0 in our array.

From this example, we can see that an important edge case was not handled correctly, which is when low or high becomes equal to mid. To fix this issue, in the 4th step, we should adjust our ranges to [low, mid] and [mid+1, high] when searching the left and right halves respectively.

With this adjustment, our example will be processed as:

Start with low = 0, high = 6, mid = 3.
As nums[low] <= nums[mid], the left half is sorted. But 0 doesn’t fall between nums[low] and nums[mid], so go to the right half by setting low = mid + 1 = 4.
Now low = 4, high = 6, mid = 5. The right half is sorted, but 0 doesn’t fall between nums[mid] and nums[high]. So, go to the left half by setting high = mid - 1 = 4.
Now low = high = mid = 4 and nums[mid] = 0 matches the target, so return mid.

This gives us the correct answer. It’s important to handle these edge cases when implementing the solution.

A key aspect of this problem is the logarithmic constraint. If we were to simply scan the array linearly, the time complexity would be O(n), which is not acceptable per the problem’s constraint. So, the binary search approach is essential here. Similarly, if the array was not rotated, a regular binary search would suffice. But since it’s rotated, we modify our binary search to suit that condition.

Changes in the problem’s parameters can affect the solution. For example, if the array was not sorted, we could not apply binary search at all. If the array was not distinct, we would have to deal with cases where multiple indices could be valid answers. If negative numbers were not allowed, we wouldn’t need to consider negative edge cases.

Inference of Problem-Solving Approach from the Problem Statement

The problem tells us that we are dealing with a sorted array that has been rotated, and we need to search for a target value in this array. Additionally, the problem asks for a solution with a runtime complexity of O(log n), which is the runtime complexity of a binary search. These clues strongly suggest that we can apply a form of binary search.

Here’s the rationale behind it:

Sorted array: The binary search algorithm is designed to efficiently search in a sorted array. It reduces the search space by half at each step, hence the logarithmic time complexity.
Rotated: Although the array is rotated, it still maintains a form of sorted order. We just have two sorted subarrays instead of one. The pivot point (the point of rotation) divides the array into two halves - both of which are sorted. This is a key property that we can utilize.
Logarithmic complexity: The problem specifically asks for a logarithmic time complexity solution. Binary search is one of the most common algorithms that achieves this time complexity.
Finding a target value: The problem involves searching for a specific value (the target) in the array, which is precisely what binary search is used for. In a standard binary search, we compare the target with the middle element to decide whether to search in the left half or the right half of the array. In this problem, we can adopt a similar strategy. Instead of simply choosing the half based on the comparison of the target and the middle element, we also consider which half is sorted and whether the target lies within the range of that sorted half.

So, putting these clues and properties together, we infer that a modified version of binary search can be used to solve this problem. The modification is necessary to handle the rotation of the sorted array.

Here’s a stepwise refinement of our approach to solving the problem:

Step 1: Understand the Problem Understand the problem statement clearly - we have a rotated sorted array and we need to find the index of a given target value. If the target does not exist, we return -1. We also note that the problem requires a solution with a time complexity of O(log n).

Step 2: Identify the Appropriate Algorithm Identify that the problem can be solved using a variant of the Binary Search algorithm. This is because we’re dealing with a sorted (though rotated) array and we need to find an element within it.

Step 3: Implement the Binary Search Start by implementing a traditional Binary Search algorithm. Define the low and high pointers. Then, in a loop, calculate the mid index, and check if the target is at the mid index.

Step 4: Modify for Rotated Sorted Array Next, modify the Binary Search algorithm to handle the rotated sorted array. For this, in each iteration of the Binary Search, identify which half of the array (from low to mid, or from mid to high) is the sorted half.

Step 5: Decide Which Half to Explore Based on which half is sorted and whether the target lies within the range of that half, decide whether to explore the left half or the right half in the next iteration.

Step 6: Update Pointers Update the low and high pointers accordingly and continue the search until the target is found or the search space becomes empty (i.e., low > high).

Step 7: Return the Result If the target is found, return its index. If not, return -1 as required by the problem.

Step 8: Test the Solution Finally, test the solution with various test cases, including edge cases such as when the array contains only one element or when the target is not in the array.

This stepwise approach should help us to efficiently find the target value in the rotated sorted array.

The high-level solution for this problem is to apply a modified binary search due to the sorted nature of the array which is rotated at an unknown pivot. We can break down this approach into the following granular, actionable steps:

Initialize Variables: Start by initializing two pointers, low and high, to the start and end of the array.
Iterative Binary Search: While low is less than or equal to high, perform the following steps:
- Mid Calculation: Compute the mid index by adding low and high and dividing by 2. This is the point where we split the array into two halves for the binary search.
- Check for Match: If the mid value is equal to the target, return the mid index as the answer.
- Left Sorted Half: If the value at low index is less than or equal to the value at mid index, it means the left half of the array is sorted. Now, check if the target lies in the left sorted half i.e., between low and mid. If it does, update high to mid - 1, else update low to mid + 1.
- Right Sorted Half: If the value at low index is greater than the value at mid index, it implies the right half of the array is sorted. Again, check if the target lies in the right sorted half i.e., between mid and high. If it does, update low to mid + 1, else update high to mid - 1.
Return Result: If we exit the while loop without finding the target, it means the target is not present in the array. In this case, return -1.

By following these granular steps, you can apply the modified binary search algorithm to find the target in the rotated sorted array efficiently. Remember, the aim is to always restrict the search space to the half where the target could possibly be, based on the sorted nature of the halves.

There are several aspects that can be solved independently. Let’s break these down:

Determining the Sorted Half: The problem can be broken down into identifying whether the left half or the right half of the array from the mid-point is sorted. This only requires a single comparison between two elements in the array (i.e., comparing the value at the low index and the mid index).
Determining the Target’s Location: After identifying the sorted half, the next independent part of the problem is determining whether the target lies within this sorted half or the unsorted half. This can be done by checking if the target is greater than the smallest element and less than the largest element of the sorted half.
Implementing Binary Search: Implementing a standard binary search can be solved independently and is used as a fundamental part of the problem. In this case, the binary search is slightly modified to deal with the rotation in the sorted array, but the basic principles remain the same.

These components can be worked on separately and then combined to solve the overall problem. The ability to decompose a problem into independently solvable parts is a useful skill when tackling complex problems, as it allows you to focus on one aspect at a time and build a solution incrementally.

There are several repeatable patterns within our solution for this problem:

Binary Search Pattern: The most prominent pattern in the solution is the binary search, which is an iterative process that repeatedly divides the search space in half. At each step, we calculate a mid-point and then compare the value at this mid-point with the target. Depending on the result of the comparison, we either narrow our search to the left half or the right half of the current search space.
Sorted Half Identification: The way we identify which half of the array is sorted (from low to mid or from mid to high) is another repeatable pattern. It’s based on comparing the value at low and mid indices.
Target’s Position Determination: Once we identify the sorted half, we again follow a pattern to check if the target lies in that half. This involves comparing the target with the values at the low and mid (for left half) or mid and high (for right half) indices.
Update of Boundaries: Lastly, the way we update our low and high pointers depending on whether the target is in the sorted half or the unsorted half is a repetitive pattern. If the target is in the sorted half, we move the high or low pointer to reduce our search space to that half. Otherwise, we adjust our pointers to focus on the other half.

These patterns are repeatedly used until the entire array has been searched or the target has been found.

Solution Approach and Analysis

Let’s break down the approach to solve this problem. We’ll make use of the Binary Search algorithm due to the fact that our array is sorted (although rotated). The key insight here is that despite the rotation, one half of the array (either left or right of the middle point) will always be in sorted order.

Let’s go through the steps:

Initialization: We start by setting two pointers - low at the start of the array and high at the end. We’ll also have our target which we need to find.
Midpoint Calculation: In each iteration of our search process, we calculate a mid point as the average of low and high.
Sorted Half Identification: At each step, we identify which half of our current search space is sorted. We can do this by comparing the elements at the low and mid indices. If nums[low] <= nums[mid], the left half is sorted. If nums[mid] <= nums[high], the right half is sorted.
Target Location Determination: Next, we check if our target lies within the sorted half. If it does, we can discard the other half. If not, we discard the sorted half. For example, if the left half is sorted and nums[low] <= target < nums[mid], we know our target must be in the left half.
Boundary Update: Based on the target location, we update our low or high pointer to the appropriate half of the current search space and repeat the process.
Termination: We continue this process until our low pointer is greater than the high pointer (which indicates the target is not present in the array) or we find our target at the mid index.

How Parameters Affect the Solution: The size of the input array and the position of the rotation pivot point could affect the number of iterations required to find the target. More elements or a pivot point closer to the middle of the array would increase the iterations. However, since binary search is very efficient (with a time complexity of O(log N)), even large input arrays can be handled efficiently.

Example Case:

Consider an array nums = [4,5,6,7,0,1,2] and a target target = 0. Here are the steps we’d follow:

Initially, low = 0, high = 6, mid = 3. The array from low to mid is sorted and the target is not in this range. Therefore, we update low to mid + 1 (i.e., low = 4).
Now, low = 4, high = 6, mid = 5. The array from mid to high is sorted, and the target is not in this range. So, we update high to mid - 1 (i.e., high = 4).
Now, low = high = mid = 4, and nums[mid] equals the target. So, we return mid, which is 4.

Thought Process

Let’s take the problem statement of “Search in Rotated Sorted Array” step by step and write Python code for it.

Cues in the Problem Statement:

The array is sorted but rotated at an unknown pivot - This indicates that we could use binary search, but the rotation introduces a twist.
We need to find the index of a target value - This indicates that the problem will involve searching the array.
The array contains distinct values - This means we don’t need to handle duplicates.
The array is in ascending order before rotation - This means that we should focus on the order of elements.
We are required to achieve O(log n) runtime complexity - This means binary search should be used as it fulfills this time complexity.

From these cues, we understand that we need to modify the standard binary search algorithm to handle the rotation.

Here’s the Python3 code for this problem:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
class Solution:
    def search(self, nums: List[int], target: int) -> int:
        low = 0
        high = len(nums) - 1
        
        while low <= high:
            mid = low + (high - low) // 2

            if nums[mid] == target:
                return mid
            elif nums[mid] >= nums[low]:
                # left half is sorted
                if nums[low] <= target < nums[mid]:
                    # target is in the left half
                    high = mid - 1
                else:
                    # target is in the right half
                    low = mid + 1
            else:
                # right half is sorted
                if nums[mid] < target <= nums[high]:
                    # target is in the right half
                    low = mid + 1
                else:
                    # target is in the left half
                    high = mid - 1
        return -1

In the code above, we start by defining two pointers, low and high, to hold our current search space within the nums array. We then enter a while loop, which continues until low exceeds high, or we find our target (i.e., nums[mid] == target).

At each step, we calculate the mid-point and compare the element at mid with our target. If they match, we have found our target and return its index.

If not, we need to decide whether to look in the left half or the right half for the target. We do this by checking which half is sorted and if the target lies within the sorted half. Once we know where to look, we update our low or high pointer accordingly and repeat the process.

If we exit the while loop without finding the target, it means the target is not in the array, and we return -1.

From Brute Force to Optimal Solution

Let’s start with the brute force solution for the problem “Search in Rotated Sorted Array”.

Brute Force Solution

In a brute force solution, we would simply iterate over every element in the array until we found the target. If we reach the end of the array without finding the target, we return -1.

Here is what the code would look like in Python:

1
2
3
4
5
def search(nums, target):
    for i in range(len(nums)):
        if nums[i] == target:
            return i
    return -1

Inefficiencies of the Brute Force Solution:

The brute force solution is quite inefficient as it has a time complexity of O(n), where n is the number of elements in the array. This is because in the worst-case scenario we have to iterate over every element in the array.

Optimizing the Solution

Given that the array is sorted (though rotated), we can use binary search to find the target in log(n) time, which is much faster than linear time for large n. However, we have to account for the rotation in the array, which is the twist in this problem.

The optimized solution uses binary search but checks which half of the array is sorted before deciding where to search next.

Here is the Python code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
def search(nums, target):
    low = 0
    high = len(nums) - 1
    
    while low <= high:
        mid = low + (high - low) // 2

        if nums[mid] == target:
            return mid
        elif nums[mid] >= nums[low]:
            # left half is sorted
            if nums[low] <= target < nums[mid]:
                # target is in the left half
                high = mid - 1
            else:
                # target is in the right half
                low = mid + 1
        else:
            # right half is sorted
            if nums[mid] < target <= nums[high]:
                # target is in the right half
                low = mid + 1
            else:
                # target is in the left half
                high = mid - 1
    return -1

This code starts by initializing low and high pointers. We then enter a loop which continues until low is greater than high or we find the target. On each iteration, we calculate the midpoint and compare it to the target. If it’s a match, we return the index. If not, we decide where to search next based on which half is sorted.

Impact on Time and Space Complexity

The time complexity of this optimized solution is O(log n), because each iteration of the loop cuts the search space in half, which is the hallmark of a binary search. This is a significant improvement over the brute force solution.

The space complexity of both the brute force solution and the binary search solution is O(1) because both solutions use a constant amount of space regardless of the size of the input array. This is because we only use a few integer variables and do not use any data structures that scale with the input size.

Establishing Preconditions and Postconditions

Preconditions:
- The input nums must be a list of unique integers that is sorted in ascending order and may be rotated at an unknown pivot.
- The length of nums should be in the range [1, 5000] and each element should be an integer in the range [-104, 104].
- The target must be an integer in the range [-104, 104].
- The method assumes that the input parameters meet these constraints.
Method Functionality:
- The method is designed to perform a modified binary search to find the index of a target value in a potentially rotated sorted array.
- The method uses the unique properties of the sorted and rotated array to determine in which half of the array the target may be present, and continuously narrows down the search space until the target is found or the search space is exhausted.
Postconditions:
- The state of the input list nums remains unchanged after the method execution.
- The method returns the index of the target if it is present in the list, otherwise, it returns -1.
- The return value signifies the location of the target value in the input list or the absence of the target in the list.
Error Handling:
- If the preconditions are not met (for example, if the input is not a list or the list contains non-integer elements), Python would raise an appropriate exception (like TypeError or ValueError).
- Within the method itself, no explicit error checking or exception handling is provided. It is assumed that the input meets the problem’s constraints.

Code Explanation and Design Decisions

Let’s break down the optimized solution for this problem.

Initial Parameters:
The function takes two parameters: nums, which is the possibly rotated, sorted array, and target, the value we’re trying to find in the array. The function also defines low and high variables to represent the lower and upper bounds of the search space within the array.
Primary Loop:
The primary loop is a while loop which continues as long as low is less than or equal to high. Each iteration of the loop effectively narrows down the search space by half. This is due to the binary search strategy that the function employs.
Conditions/Branches within the Loop:
Inside the loop, the function first checks if the middle element nums[mid] is equal to the target. If it is, the function immediately returns mid.
If nums[mid] is not equal to the target, the function checks whether the left half of the current search space (from nums[low] to nums[mid]) is sorted. This is done by comparing nums[mid] with nums[low]. If nums[mid] is greater or equal, the left half is sorted.
Based on which half is sorted and whether the target lies within the sorted half, the function updates low or high to narrow down the search space.
Updates/Modifications within the Loop:
If the left half is sorted and the target lies within this half, the function updates high to mid - 1 because we can be sure that the target does not exist in the right half. If the target does not lie within the sorted left half, low is updated to mid + 1 because the target must be in the right half.
If the right half is sorted (which is the case when the left half isn’t sorted) and the target lies within this half, the function updates low to mid + 1. If the target does not lie within the sorted right half, high is updated to mid - 1.
These modifications help to narrow down the search space on each iteration.
Invariant:
The invariant in this problem is that the target, if it exists in the array, is always within the current search space delimited by low and high. This invariant is maintained by always updating low or high based on sound reasoning about where the target can possibly exist. This is crucial in ensuring the correctness of the solution.
Final Output:
The final output of the function is either the index of the target in the array (if the target is found) or -1 (if the target is not found). This satisfies the requirements of the problem statement. In the context of the problem, this represents the position of the searched value in the input array after a potential rotation. If the value isn’t in the array, the return of -1 signifies that fact.

Coding Constructs

High-Level Problem-Solving Strategies: The code uses a modified binary search algorithm to solve the problem. Binary search is a classic divide-and-conquer algorithm that works by repeatedly dividing the search space in half.
Explanation for a Non-programmer: This program is like a game of “Guess the Number” where the numbers are arranged in a circle. You’re trying to find a particular number in the circle. Instead of checking each number one by one, you pick a point, check if the number you have is higher or lower, and depending on that you decide which half of the circle to look in next. You keep narrowing down your search area until you either find the number or run out of places to look.
Logical Elements: The core logical elements used in the code include:
- Conditional Statements: Used to decide which part of the array to search based on comparing the target value with the middle element and the first element.
- Iteration (While Loop): Used to repeatedly divide the search space until the target is found or the search space is exhausted.
Algorithmic Approach in Plain English: We start with the full array as our search space. We find the middle of this space and check if our target is there. If not, we decide whether to continue our search in the left half or the right half based on whether the target is higher or lower than the midpoint and if the left half or right half is sorted. We repeat this process of dividing and selecting until we either find our target or run out of places to look.
Key Steps:
- Determine the midpoint of the current search space.
- Compare the target value with the value at the midpoint and the first point of the search space.
- Decide whether the target might lie in the left half or the right half of the search space.
- Narrow down the search space to the chosen half and repeat the process.
Algorithmic Patterns: The key pattern here is the binary search. This is an algorithmic pattern that’s commonly used for searching in sorted lists. The unique twist in this problem is that the list might be “rotated”, which adds an extra conditional check to decide which half of the array to search.

Language Agnostic Coding Drills

Dissecting the Code:
The main coding concepts or drills involved in this problem are:
- Variable Initialization and Use: We need to keep track of the left and right boundaries of our search space and use these variables to find the middle point.
- Conditional Statements (if-else): Depending on the result of our comparison between the target value and the value at the midpoint, we will decide which part of the array to continue our search in.
- Looping (while loop): The core part of a binary search is the iteration that repeatedly narrows down the search space until the target is found or the search space is exhausted.
- Array Indexing: We’re dealing with an array of integers, and we need to be able to access individual elements of this array using their indices.
List of Coding Drills (Increasing Difficulty):
- Variable Initialization and Use: This is a very basic concept that is necessary for almost every programming problem. The difficulty level is low.
- Array Indexing: This is also a relatively basic concept, but it requires understanding how arrays work and how they are accessed.
- Conditional Statements (if-else): The use of conditionals is slightly more advanced because it involves making decisions based on certain conditions. It requires logical reasoning and understanding of control flow in a program.
- Looping (while loop): Looping is one of the more complex constructs, especially in the context of this problem where the loop must be correctly controlled to ensure that the search space is being properly narrowed down.
Problem-Solving Approach:
Start by initializing the left and right variables to represent the boundaries of our search space.
Next, we enter the main loop of our binary search. This loop continues as long as our search space is not empty (i.e., as long as left is less than or equal to right).
Inside the loop, we first find the midpoint of the current search space. We use array indexing to check the value at this midpoint and compare it with our target.
Based on this comparison and the relative ordering of the numbers at the left boundary, midpoint, and right boundary, we use conditional statements to decide which half of the array to search next. If our target is not found, we update our left or right boundary accordingly to narrow down the search space.
We repeat this process, continuously halving our search space, until we either find our target or exhaust our search space. Finally, we return the index of the target if it was found, or -1 if it was not.
Understanding Arrays: Understanding what arrays (Lists in Python) are is the basis for this problem. The user should understand how to create an array, access its elements, and iterate over it.
Binary Search Concept: The binary search is a fundamental algorithm in computer science to quickly find an item in a sorted list of items. The algorithm works by repeatedly dividing in half the portion of the list that could contain the item until you’ve narrowed down the possible locations to just one.
Finding the Middle Index: Calculating the middle index of a list or sublist is a common step in binary search. This can be done by adding the start and end indices and dividing by 2.
Accessing Array Elements: The user must understand how to access elements in an array using indices.
Updating Indices in Binary Search: The crux of binary search involves updating the search space, i.e., changing the start and end indices based on the comparison with the element at the middle index.
Conditional Statements: Understanding how to use if, else if, and else statements in Python is crucial for this problem.
Comparisons in Python: Understanding how to use comparison operators (==, <, >, <=, >=) is important in this problem for comparing the target with elements of the array and for determining the logic to update indices.
Returning Values: Understand how to use the return statement to output a result from a function.

Below are the drills in the order of increasing difficulty:

Drill 1: Create and manipulate arrays in Python.
Drill 2: Implement a basic binary search on a sorted array.
Drill 3: Write a function that takes an array and two indices and calculates the middle index.
Drill 4: Write a function that takes an array and an index and returns the array element at that index.
Drill 5: Modify your binary search function to update the start and end indices based on comparison with the middle element.
Drill 6: Write a function that takes three numbers and uses conditional statements to return the smallest number.
Drill 7: Write a function that takes two numbers and compares them using all comparison operators.
Drill 8: Modify your binary search function to return the index of the target, or -1 if the target is not found.

After completing these drills, you should be able to combine them to understand and implement the provided function.

Targeted Drills in Python

Drill 1: Create and manipulate arrays in Python

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
# Create an array
arr = [1, 2, 3, 4, 5]

# Add an element to the end of the array
arr.append(6)

# Remove an element from the array
arr.remove(1)

# Access elements at specific indices
print(arr[0])
print(arr[-1])

Drill 2: Implement a basic binary search on a sorted array

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
def binary_search(sorted_arr, target):
    left, right = 0, len(sorted_arr) - 1
    while left <= right:
        mid = (left + right) // 2
        if sorted_arr[mid] == target:
            return mid
        elif sorted_arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

print(binary_search([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 7))

Drill 3: Write a function that takes an array and two indices and calculates the middle index

1
2
3
4
def middle_index(left, right):
    return (left + right) // 2

print(middle_index(1, 7))

Drill 4: Write a function that takes an array and an index and returns the array element at that index

1
2
3
4
def access_element(arr, index):
    return arr[index]

print(access_element([1, 2, 3, 4, 5], 3))

Drill 5: Modify your binary search function to update the start and end indices based on comparison with the middle element

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
def binary_search(sorted_arr, target):
    left, right = 0, len(sorted_arr) - 1
    while left <= right:
        mid = (left + right) // 2
        if sorted_arr[mid] == target:
            return mid
        elif sorted_arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

print(binary_search([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 7))

Drill 6: Write a function that takes three numbers and uses conditional statements to return the smallest number

1
2
3
4
5
6
7
8
9
def smallest_of_three(a, b, c):
    if a < b and a < c:
        return a
    elif b < a and b < c:
        return b
    else:
        return c

print(smallest_of_three(3, 2, 1))

Drill 7: Write a function that takes two numbers and compares them using all comparison operators

1
2
3
4
5
6
7
8
def compare_numbers(a, b):
    print(f"a == b: {a == b}")
    print(f"a < b: {a < b}")
    print(f"a > b: {a > b}")
    print(f"a <= b: {a <= b}")
    print(f"a >= b: {a >= b}")

compare_numbers(5, 10)

Drill 8: Modify your binary search function to return the index of the target, or -1 if the target is not found

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
def binary_search(sorted_arr, target):
    left, right = 0, len(sorted_arr) - 1
    while

 left <= right:
        mid = (left + right) // 2
        if sorted_arr[mid] == target:
            return mid
        elif sorted_arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

print(binary_search([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 11))

Variable Initialization and Use:

1
2
3
4
5
6
7
8
9
# Initialize two variables with initial values
left = 0
right = 5

# Use these variables in some way
middle = (left + right) // 2

# Print the results
print(f"left: {left}, right: {right}, middle: {middle}")

Array Indexing:

1
2
3
4
5
# Initialize an array
nums = [1, 3, 5, 7, 9]

# Access an element of the array
print(nums[2])  # This will print 5

Conditional Statements (if-else):

1
2
3
4
5
6
7
8
# Initialize a variable
value = 5

# Use a conditional statement to check the variable's value
if value > 3:
    print("Value is greater than 3")
else:
    print("Value is not greater than 3")

Looping (while loop):

1
2
3
4
5
6
7
# Initialize a variable
count = 0

# Use a while loop to increment the variable until it reaches a certain value
while count < 5:
    print(count)
    count += 1

Now, we need to create a drill for the specific concept of a binary search, as this is key to our problem.

Binary Search:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
def binary_search(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

In the context of our problem, the binary search is essential because it allows us to find the target number in logarithmic time complexity, which is much faster than a linear search, especially for large inputs.

Once we have these individual drills, we can start integrating them to form our solution. The binary search drill forms the core of our solution. We just need to modify it to handle the special condition of a rotated sorted array. In the modified binary search, we use the concepts of variable initialization and use, array indexing, conditional statements, and looping. So, all our drills contribute to building our final solution.

Completing these drills will help solidify understanding of arrays, binary search, accessing and comparing elements, and returning values in Python.

Q&A

Wouldn’t it be better to compare with the leftmost value in the array (step 1), instead of the rightmost. It would be faster in taking care of cases where the array is not pivoted?

Actually, both approaches are equally efficient when it comes to handling the case where the array is not pivoted, or sorted. This is because the binary search is performed based on the relationship between the target and the elements at the left, mid, and right positions, not based on whether the array is sorted or not.

However, comparing with the leftmost element in the array does have an advantage when the array has duplicate values. In case of duplicate values, comparing with the rightmost element might lead to scenarios where we’re unable to decide which half of the array to proceed with, thus degrading the performance to O(n) in the worst-case scenario. On the other hand, comparing with the leftmost element can help us avoid such scenarios, thus maintaining a time complexity of O(log n) even when the array contains duplicate values.

That being said, this approach requires an additional check to determine if the array is pivoted or not. Here’s how the modified approach might look:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
def search(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        # If the left half is sorted
        if nums[left] <= nums[mid]:
            # If target is in the sorted part
            if nums[left] <= target < nums[mid]:
                right = mid - 1
            else:
                left = mid + 1
        # If the right half is sorted
        else:
            # If target is in the sorted part
            if nums[mid] < target <= nums[right]:
                left = mid + 1
            else:
                right = mid - 1
    return -1

As you can see, the key difference is that we’re now comparing the mid element with the leftmost element, rather than the rightmost element. The rest of the logic remains the same.

question clearly states “possibly rotated” , which means you have to include a case where their is no pivot

You’re absolutely correct, the term “possibly rotated” indeed implies that there might be cases where the array isn’t rotated at all. However, the good news is that the solution provided actually accounts for this case without needing any extra code. Here’s why:

The logic within the binary search is designed to handle two cases:

The segment of the array (bounded by our current left and right pointers) is fully sorted.
The segment is rotated (i.e., it contains the pivot point).

For case 1, which includes the situation where the entire array is sorted (i.e., not rotated), the search correctly narrows down to the half that could contain the target.

For case 2, the search identifies which half of the segment is sorted and then checks if the target lies within this sorted half. If it does, the search narrows down to this half; if it doesn’t, it narrows down to the other half, which contains the pivot point.

So, in essence, the logic already accommodates for the “possibly rotated” aspect of the problem, whether that be partially rotated, fully rotated, or not rotated at all. The rotation doesn’t impact the effectiveness of the search, but rather how we decide to narrow down our search space.

The key idea is during the process fo binary search, we compare the number at position mid and position left. If the number at mid is larger, it implies that range [left, mid] is ordered (not rotated), so we can decide whether to move towards that range for further search. Otherwise, the number at mid is smaller, which implies that range [mid, right] is ordered (not rotated).

Yes, that’s a great way to think about it!

By comparing the mid value to the leftmost value, we can quickly identify whether the left-half or the right-half of our current search space is sorted. This in turn helps us decide where to focus our search next.

If the mid value is greater than the leftmost value, it means the left half is sorted. So, if the target value is between the leftmost and mid values, we know to search in the left half. If not, we search in the right half.
Conversely, if the mid value is less than the leftmost value, it means the right half is sorted. So, if the target value is between the mid and rightmost values, we search in the right half. If not, we search in the left half.

This approach leverages the binary search strategy while adapting it to accommodate the possible rotation in the array.

The main idea is to use binary search to find the target in the array. Binary search is a technique that divides the search space into two halves and compares the target with the middle element. If the target is equal to the middle element, we have found the index. If the target is smaller than the middle element, we search in the left half. If the target is larger than the middle element, we search in the right half. This way, we can reduce the search space by half in each iteration and achieve O(log n) runtime complexity.

However, since the array is possibly rotated at some unknown pivot, we cannot directly apply binary search. We need to modify it slightly to handle the rotation. The key observation is that one of the two halves must be sorted, and the other half must contain the pivot. For example, if the array is [4,5,6,7,0,1,2], then the left half [4,5,6,7] is sorted and the right half [0,1,2] contains the pivot. If the array is [7,0,1,2,4,5,6], then the right half [2,4,5,6] is sorted and the left half [7,0,1] contains the pivot.

So, in each iteration of binary search, we first check which half is sorted by comparing the left and middle elements. Then we check if the target is in that sorted half by comparing it with the left and middle elements. If it is in that sorted half, we search in that half as usual. If it is not in that sorted half, we search in the other half that contains the pivot.

For example, if the array is [4,5,6,7,0,1,2] and the target is 0:

We start with left = 0 and right = 6. The middle index is 3 and the middle element is 7. We see that the left half [4,5,6,7] is sorted and the right half [0,1,2] contains the pivot. We check if the target 0 is in the left half by comparing it with 4 and 7. It is not in that half because 0 < 4 < 7. So we search in the right half that contains the pivot by setting left = mid + 1 = 4. We repeat this process until we find the target or exhaust the search space.

I agree with your analysis, and it leads us to a single-pass binary search solution to this problem.

Here is the Python code implementing this:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
def search(nums, target):
    left, right = 0, len(nums) - 1

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

        if nums[mid] == target:
            return mid

        # If nums[left:mid] is sorted
        if nums[left] <= nums[mid]:
            if nums[left] <= target < nums[mid]:
                right = mid - 1
            else:
                left = mid + 1
        # If nums[mid:right+1] is sorted
        else:
            if nums[mid] < target <= nums[right]:
                left = mid + 1
            else:
                right = mid - 1

    return -1

In the code above, we’re first initializing two pointers, left and right, to the start and end of the list. Then, we start a loop which will continue as long as left is less than or equal to right.

Within the loop, we first find the middle element. If the middle element is the target, we’re done and return the index.

If the target isn’t found, we need to decide whether to look in the left half or the right half.

To make that decision, we check if the left half is sorted by comparing nums[left] and nums[mid]. If the left half is sorted, we check if the target lies within this range. If it does, we move right to mid - 1. If it doesn’t, the target must be in the right half, so we move left to mid + 1.

Similarly, if the right half is sorted, we check if the target lies within this range. If it does, we move left to mid + 1. If it doesn’t, the target must be in the left half, so we move right to mid - 1.

If the loop ends without finding the target, we return -1 indicating that the target is not in the list.

Thank you. This is probably the most elegant way to solve this problem. With this solution, we don’t even need to find the pivot’s location, which is quite smart.

Here are the constraints - Constraints:

1 <= nums.length <= 5000 -10^4 <= nums[i] <= 10^4 All values of nums are unique. nums is guranteed to be rotated at some pivot. <— array is guranteed to be sorted -10^4 <= target <= 10^4

and with this this input -

[1,3] <– array is sorted but not rotated 1 <– target element

then how come expected output could be - 0th index, although element is present but array is not rotated.

please suggest if i am understanding the wrong way.

The key to understanding the problem statement lies in the phrase “nums is guaranteed to be rotated at some pivot.” This doesn’t necessarily mean that the array must be rotated in a way that visibly changes the order of the elements.

In fact, an array can be considered to have been “rotated” by any number of places divisible by its length. For instance, if you have an array of length 2, like [1, 3], it can be “rotated” by 2 places, which results in the same array [1, 3].

This is akin to rotating a wheel by 360 degrees - the wheel does rotate, but it ends up in the same position it started in. In the context of this problem, a sorted array like [1, 3] can be considered as an array rotated “0 times”, or “2 times”, or “4 times”, etc.

So, even though the array [1,3] appears to be sorted and not visibly rotated, it’s still considered as adhering to the problem’s constraint of “nums is guaranteed to be rotated at some pivot.”

Therefore, the expected output for the target 1 in array [1,3] would be its index 0, which is the correct answer.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution {
	public int search(int[] nums, int target) {
		int n = -1;
		for(int i = 0; i < nums.length; i++){
			if(nums[i] == target){
				n = i;
			}
		}
		return n;
	}
}

This solution is a brute force solution which essentially traverses the entire array to find the target value. Let’s analyze it in detail:

Correctness: The solution is correct in terms of providing the expected output, as it checks every element in the array and returns the index if it finds a match with the target.
Efficiency: The solution is not efficient. It has a time complexity of O(n) where ’n’ is the number of elements in the array. The problem statement specifically mentions that the array is sorted and possibly rotated. We can take advantage of this property and use a binary search technique to find the target, which would reduce the time complexity to O(log n).
Code Style: The variable ’n’ is initialized with -1 which is a good practice as it serves as a default return value indicating the target was not found in the array. However, the variable name ’n’ could be more descriptive. A better name might be ’targetIndex’ or ‘foundIndex’. Also, once the target is found, the loop could be terminated using a ‘break’ statement. This would avoid unnecessary iterations over the rest of the array after the target has been found.

In summary, while the code is correct, it is not efficient. With a few modifications and a change in approach, the efficiency of this solution could be significantly improved.

This problem does not say the numbers in the array have to be unique.

So maybe you want to try a test like:

nums = [3,0,1,3,3,3,3] target = 0

For the given test case, nums = [3,0,1,3,3,3,3] and target = 0, the modified binary search algorithm to find the target in a rotated sorted array will still work, even if there are duplicates.

However, we need to note a specific edge case for this algorithm when dealing with duplicates. When the numbers at the indices left, mid, and right are equal, it is not clear which half of the array to discard. For instance, in the array [3,0,1,3,3,3,3] if left points to index 0, mid to index 3 and right to index 6, all these are 3. In this case, we can’t decide which half to discard as we don’t know where the pivot is.

To handle this, we can add a specific condition in the binary search algorithm, where if nums[left] == nums[mid] and nums[mid] == nums[right], we just increment left and decrement right and continue with the search. This way, we will eventually either find the target or exhaust the search space.

Here is a Python code snippet to illustrate this:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
def search(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        if nums[left] == nums[mid] == nums[right]: # special case for duplicates
            left += 1
            right -= 1
        elif nums[left] <= nums[mid]: # left half is sorted
            if nums[left] <= target < nums[mid]: # target is in left half
                right = mid - 1
            else: # target is in right half
                left = mid + 1
        else: # right half is sorted
            if nums[mid] < target <= nums[right]: # target is in right half
                left = mid + 1
            else: # target is in left half
                right = mid - 1
    return -1 # target not found

Using this code, for the input nums = [3,0,1,3,3,3,3] and target = 0, the output will be 1, as expected.

Is the correct approach to binary search for the pivot and then binary search for the target

Binary searching for the pivot and then for the target is one possible approach, and it can indeed solve the problem. In this approach, you perform two separate binary searches: one to find the pivot (i.e., the minimum element, or the point of rotation), and then another to find the target in the appropriate section of the array.

However, this approach could be simplified by combining the two binary searches into one. Instead of explicitly finding the pivot, you can directly search for the target while accounting for the rotation in the array. You still use the same binary search structure, but with a modified condition for choosing the search space.

In each iteration, you determine whether the left or right half of the array is sorted (one half must be), and then check if the target lies within the sorted half. If it does, you continue your search in that half; otherwise, you search in the other half.

This combined approach is more efficient because it only requires one pass through the array (as opposed to two), and it simplifies the implementation by reducing the amount of code. Its time complexity remains O(log n) since it still uses binary search, but it has a smaller constant factor.

Both approaches are valid, but the combined approach is generally preferable due to its greater efficiency and simplicity.

can this be solved by using a map structure by storing element as key and index as value….?

Yes, it’s possible to solve this problem using a map data structure where the key is the element and the value is its index. This approach involves iterating over the entire array once to populate the map, and then you can find the target in constant time.

Here’s a basic outline of this approach in Python:

1
2
3
def search(nums, target):
    num_to_index = {num: idx for idx, num in enumerate(nums)}
    return num_to_index.get(target, -1)

However, there are a few points to consider:

Time complexity: This approach has a time complexity of O(n) due to the initial iteration over the array, which is less efficient than the binary search approach that has a time complexity of O(log n).
Space complexity: This approach requires O(n) additional space to store the map, whereas the binary search approach only requires O(1) space.
Unnecessary work: This approach involves examining every element in the array, even though the array is sorted (or almost sorted, given the rotation). In many cases, the target could be found without having to look at every element.

Therefore, while it’s possible to use a map to solve this problem, it’s not the most efficient solution. The binary search approach takes advantage of the sorted nature of the array and is generally a better choice for this problem.

What’s wrong with leetcode ? Why is not exiting the while loop when s==l? Below is my solution. Can someone tell me if this is issue with leetcode runtime? I am able to run the code successfully in some other online editor,

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
class Solution {
public:
	int search(vector& nums, int target) {
		int s = 0;
		int l = nums.size() -1;
		while(s < l)
		{
			cout << s << " " << l << " ";
			int m = s+l;
			if(m % 2 == 0)
				m = m/2;
			else m = (m-1)/2;

			if(nums[m] == target)
			{
				cout<< "*"<<m<<"*  ";
				s = m;
				l = m;
			}
			else if(nums[s]<=nums[m])
			{
				if(target>=nums[s] && target<=nums[m]) l = m;
				else s = m;
			}
			else
			{
				if(target>=nums[m] && target<=nums[l]) s = m;
				else l=m;
			}
			if(s==l)
			{
				cout<<"BREAK";
				break;
			}
			cout<<s<<" "<<l<<"
				";
		}
		if(nums[s] == target) return s;
		else return -1;
	}
};

From what I can see, your implementation of binary search seems to have an infinite loop issue when the target is not present in the array. This happens because you are updating your s and l pointers to m without checking if m can make s or l to move.

Let’s look at the case when s and l are consecutive numbers. If nums[s] <= nums[m] is true and target is not in the range of nums[s] to nums[m], s will be updated to m. However, if m is equal to s (which is possible when s and l are consecutive), s will not move. This will cause an infinite loop, as s and l will not change. The same problem can occur with l.

A standard binary search implementation makes sure that s or l always changes:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
class Solution {
public:
    int search(vector<int>& nums, int target) {
        int s = 0;
        int l = nums.size();
        while (s < l) {
            int m = s + (l - s) / 2;
            if (nums[m] == target) {
                return m;
            } else if (nums[s] <= nums[m]) {
                if (target >= nums[s] && target < nums[m]) {
                    l = m;
                } else {
                    s = m + 1;
                }
            } else {
                if (target > nums[m] && target <= nums[l-1]) {
                    s = m + 1;
                } else {
                    l = m;
                }
            }
        }
        return -1;
    }
};

In this code, if the target is not found, s or l will be updated such that it either increments or decrements. This way, the function is guaranteed to exit.

  1
  2
  3
  4
  5
  6
  7
  8
  9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 51
 52
 53
 54
 55
 56
 57
 58
 59
 60
 61
 62
 63
 64
 65
 66
 67
 68
 69
 70
 71
 72
 73
 74
 75
 76
 77
 78
 79
 80
 81
 82
 83
 84
 85
 86
 87
 88
 89
 90
 91
 92
 93
 94
 95
 96
 97
 98
 99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
#define endl '\n'
// #define int long long
#define pb push_back
#define FAST
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#define ppb pop_back
#define lld long double
#define all(a) a.begin(), a.end()
#define vi vector
#define vs vector
#define vb vector
#define vvi vector<vector>
// #define sz(a) a.size()
#define rep(i, a, n) for (int i = a; i < n; i++)

class Solution {
public:

	int binary2(vi arr, int pivotindex, int n, int target)
	{
		int low = pivotindex + 1;
		int high = n - 1;

		while (high >= low)
		{
			int mid = low + (high - low) / 2;
			if (arr[mid] == target)
			{
				return mid;
			}
			else if (arr[mid] > target)
			{
				high = mid - 1;
			}
			else
			{
				low = mid + 1;
			}
		}
		return -1;
	}

	int binary1(vi arr, int pivotindex, int n, int target)
	{
		int high = pivotindex;
		int low = n - 1;

		while (high >= low)
		{
			int mid = low + (high - low) / 2;
			if (arr[mid] == target)
			{
				return mid;
			}
			else if (arr[mid] > target)
			{
				high = mid - 1;
			}
			else
			{
				low = mid + 1;
			}
		}
		return -1;
	}

	int pivot(vi arr, int n)
	{
		int low = 0;
		int high = n - 1;
		while (low <= high)
		{
			int mid = low + (high - low) / 2;
			if (arr[mid] <= arr[n - 1])
			{
				// We are in part 2 of the array
				high = mid - 1;
			}
			else // We are in part 1 of the array
			{
				// We are on the pivot element
				if (arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
				{
					return mid;
				}
				// We are not on pivot element
				else
				{
					low = mid + 1;
				}
			}
		}
		return 0;

	}

	int search(vector<int>& numbers, int target) {
		int n = numbers.size();
		int pivotindex = pivot(numbers, n);
		int index;
		if (target <= numbers[n - 1])
		{
			index = binary2(numbers, pivotindex, n, target);
		}
		else if (target > numbers[n - 1])
		{
			index = binary1(numbers, pivotindex, n, target);
		}
		return index;
	}
};

I am getting proper result when i run the code but getting runtime error while submitting pls correct the code and point out the mistake.

The runtime error may be due to an “array index out of bounds” issue. In the pivot function, you’re comparing arr[mid] with arr[mid + 1] and arr[mid - 1] without checking if mid + 1 and mid - 1 are valid indices. If mid happens to be 0 or n - 1, it will result in an out-of-bounds error.

Also, note that in binary1() function, you’ve set low to n - 1 and high to pivotindex, which might be causing a logical error because low should be less than high in a standard binary search.

Here is the corrected version of your code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
class Solution {
public:
    int binary2(vector<int>& arr, int pivotindex, int n, int target) {
        int low = pivotindex + 1;
        int high = n - 1;

        while (high >= low) {
            int mid = low + (high - low) / 2;
            if (arr[mid] == target) {
                return mid;
            }
            else if (arr[mid] > target) {
                high = mid - 1;
            }
            else {
                low = mid + 1;
            }
        }
        return -1;
    }

    int binary1(vector<int>& arr, int pivotindex, int n, int target) {
        int low = 0;
        int high = pivotindex;

        while (high >= low) {
            int mid = low + (high - low) / 2;
            if (arr[mid] == target) {
                return mid;
            }
            else if (arr[mid] > target) {
                high = mid - 1;
            }
            else {
                low = mid + 1;
            }
        }
        return -1;
    }

    int pivot(vector<int>& arr, int n) {
        int low = 0;
        int high = n - 1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] <= arr[n - 1]) {
                high = mid - 1;
            }
            else {
                if (mid > 0 && mid < n - 1 && arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1]) {
                    return mid;
                }
                else {
                    low = mid + 1;
                }
            }
        }
        return 0;
    }

    int search(vector<int>& numbers, int target) {
        int n = numbers.size();
        int pivotindex = pivot(numbers, n);
        int index;
        if (target <= numbers[n - 1]) {
            index = binary2(numbers, pivotindex, n, target);
        }
        else if (target > numbers[n - 1]) {
            index = binary1(numbers, pivotindex, n, target);
        }
        return index;
    }
};

This code should fix the runtime error you’re encountering. Please test it out on different test cases to ensure that it’s working as expected.

The problem has two important keywords, “Search” and “Sorted”. This gives us the idea of using binary search in this problem. Since the array is rotated, so the idea is to check that which protion of array is sorted. First we check that portion of array and then we check if the target lies in the sorted portion of array or not. This process is repeated, if target element is not present in array we return -1.

Yes, that’s absolutely correct. In a sorted and rotated array, at least one half (may be more than a half if elements are duplicates) will always be sorted. So we can always use binary search in the sorted half of the array. Here is a rough sketch of an algorithm based on this idea:

Initialize two pointers, low and high, to the first and last elements of the array.
While low <= high:
1. Compute mid = (low + high) / 2.
2. If nums[mid] is the target, return mid.
3. If nums[low] <= nums[mid] (the left half is sorted):
  1. If nums[low] <= target < nums[mid], set high = mid - 1 (the target must be in the left half).
  2. Else, set low = mid + 1 (the target must be in the right half).
4. Else (the right half is sorted):
  1. If nums[mid] < target <= nums[high], set low = mid + 1 (the target must be in the right half).
  2. Else, set high = mid - 1 (the target must be in the left half).
If we exit the loop, the target is not in the array. Return -1.

This algorithm works because in a sorted and rotated array, at least one half of the array is always sorted. By identifying the sorted half and checking if the target is in the sorted range, we can decide which half to search in next. This process is repeated until we find the target or exhaust the search space. The time complexity is O(log n), the same as standard binary search.

[3,5,1] the target is 3 why not 1? for all cases if the list was rotated then the most right element is smaller than the middle then the index of most right is the target or in the mid equal to target then it is he target why in this case such logic did not work?

Your logic assumes that the target is always the pivot (the smallest element in the rotated array) or the middle element. However, the problem statement says that the target could be any number in the array, not necessarily the pivot or the middle.

In your example [3,5,1], the target could be 3, 5, or 1. If the target is 3, then the expected output is 0 because 3 is at index 0. If the target is 5, then the expected output is 1 because 5 is at index 1. If the target is 1, then the expected output is 2 because 1 is at index 2.

So, the key to solving this problem is not to find the pivot or the middle, but to find the target. This requires modifying the binary search to handle the rotation.

Need Help! why it’s gives Runtime error on test case nums = [1] and target = 1.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
class Solution {
public:
	int search(vector& nums, int target) {
		int n = nums.size() - 1;
		int s=0, e=n;
		while(s < e) {
			int mid = s + (e-s) / 2;
			if(nums[mid] > nums[e]) {
				s = mid + 1;
			}
			else {
				e = mid;
			}
		}
		int pivotIndex = s;
		s = 0 , e = n;
		if(target >= nums[s] && target <= nums[pivotIndex -1]) {
			e = pivotIndex - 1;
		}
		else {
			s = pivotIndex ;
		}
		int mid = s + (e-s) / 2;
		while (s<=e) {
			if(nums[mid] == target) {
				return mid;
			}
			else if(target > nums[mid]) {
				s = mid + 1;
			}
			else {
				e = mid - 1;
			}
			mid = s + (e-s) / 2;
		}
		return -1;
	}
};

Your code is giving a runtime error for an array of size 1 because of the line if(target >= nums[s] && target <= nums[pivotIndex -1]). In the case when nums = [1] and target = 1, pivotIndex is 0, and nums[pivotIndex - 1] refers to nums[-1], which in C++ gives the last element of the array.

However, as there is only one element in the array, nums[-1] is effectively nums[0], and you’re checking if the target is both greater than and less than nums[0] at the same time, which is impossible, leading to your issue.

Instead of checking the range in this way, you should handle the edge case where your array size is 1 before doing any other operations:

1
2
3
4
5
6
int search(vector<int>& nums, int target) {
    int n = nums.size();
    if (n == 1)
        return nums[0] == target ? 0 : -1;
    // Rest of your code...
}

By doing this, you effectively bypass the rest of your code and return the correct index immediately when the size of your array is 1.

Can someone tell me why this code/solution doesn’t work:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
class Solution {
	public int search(int[] nums, int target) {
		int low = 0;
		int high = nums.length;
		int pivot = 0;

		while (low <= high) {
			int mid = low + (high - low) / 2;

			if (nums[mid - 1] > nums[mid] && nums[mid + 1] > nums[mid]) {
				pivot = mid;
				break;
			}
			else if (nums[mid - 1] < nums[mid] && nums[mid + 1] > nums[mid]) {
				high = mid - 1;
			}
			else if (nums[mid - 1] > nums[mid] && nums[mid + 1] < nums[mid]) {
				low = mid;
			}
		}

		if (target > pivot) {
			low = pivot;
			high = nums.length;

			while (low <= high) {
				int mid = low + (high - low) / 2;
				if (nums[mid] == target) {
					return mid;
				}
				else if (nums[mid] < target) {
					high = mid - 1;
				}
				else {
					low = mid;
				}
			}
		}
		else if (target < pivot) {
			low = pivot;
			high = 0;

			while (high <= low) {
				int mid = high + (low - high) / 2;
				if (nums[mid] == target) {
					return mid;
				}
				else if (nums[mid] < target) {
					low = mid - 1;
				}
				else {
					high = mid;
				}
			}
		}

		return -1;
	}
}

The code has several issues:

int high = nums.length; - The ‘high’ variable is initialized to the length of the array, which is out of range. It should be nums.length - 1.
The while loop while (low <= high) can result in an ArrayIndexOutOfBoundsException, as you are not checking if ‘mid’ is at the ends of the array before accessing nums[mid - 1] and nums[mid + 1].
if (target > pivot) and if (target < pivot): Here you’re comparing the target with the index ‘pivot’, not with the value of ’nums[pivot]’.
In the second part of your code, you’re doing two binary searches, one for the case where the target is greater than the pivot value and one where it is smaller. However, in both cases, you are not adjusting the mid value correctly in the while loop.

Here is a revised version of your code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
class Solution {
    public int search(int[] nums, int target) {
        int low = 0;
        int high = nums.length - 1;
        int pivot = 0;
        
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (mid > 0 && mid < nums.length - 1 && nums[mid - 1] > nums[mid] && nums[mid + 1] > nums[mid]) {
                pivot = mid;
                break;
            }
            else if (nums[mid] > nums[low]) {
                low = mid + 1;
            }
            else {
                high = mid - 1;
            }
        }
        
        if (nums[pivot] == target) return pivot;

        low = 0;
        high = nums.length - 1;
        
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (nums[mid] == target) {
                return mid;
            }
            else if (nums[mid] < target) {
                low = mid + 1;
            }
            else {
                high = mid - 1;
            }
        }
        return -1;
    }
}

This revised code firstly tries to find the pivot using a modified binary search. Then, it does a binary search on the entire array to find the target. This is not the most efficient solution, as the pivot is not utilized, but it corrects the issues in your original code.

In an optimal solution, once the pivot is found, we should do two binary searches: one from 0 to pivot, and another from pivot+1 to nums.length - 1, depending on the value of the target.

getting heap-buffer-overflow error in this code..can anyone help??

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
int pivot(int arr[], int n)
{
	int s = 0;
	int e = n - 1;
	int mid = (s + e) / 2;
	while (s < e)
	{
		if (arr[mid] >= arr[0])
		{
			s = mid + 1;
		}
		else
		{
			e = mid;
		}
		mid = (s + e) / 2;
	}
	return s;
}
int binary(int arr[], int ele, int s, int e)
{
	int mid = (s + e) / 2;
	while (s <= e)
	{
		if (arr[mid] == ele)
		{
			return mid;
		}
		else if (arr[mid] > ele)
		{
			e = mid - 1;
		}
		else
		{
			s = mid + 1;
		}
		mid = (s + e) / 2;
	}
	return -1;
}
int search(int* nums, int numsSize, int target){
	int pi = pivot(nums, numsSize);
	if (target >= nums[0] && target <= nums[pi - 1])
	{
		return binary(nums, target, 0, pi - 1);
	}
	else
	{
		return binary(nums, target, pi, numsSize - 1);
	}
}

The code provided here seems correct in terms of logic for a rotated sorted array, but the error “heap-buffer-overflow” typically occurs when you attempt to read or write outside the memory bounds of a heap-allocated variable.

The problem could be arising from the fact that pi - 1 in the line if (target >= nums[0] && target <= nums[pi - 1]) can be negative if pi is 0. This would happen if the array is already sorted (i.e., not rotated at all), in which case the pivot function would return 0. Accessing nums[-1] would result in a heap-buffer-overflow error because it tries to access memory outside the bounds of the array.

To fix this issue, you should add a check for the case when the array is not rotated:

1
2
3
4
5
6
7
8
int search(int* nums, int numsSize, int target){
    int pi = pivot(nums, numsSize);
    if (pi > 0 && target >= nums[0] && target <= nums[pi - 1]) {
        return binary(nums, target, 0, pi - 1);
    } else {
        return binary(nums, target, pi, numsSize - 1);
    }
}

Also, if the target element is not in the array, it’s good to handle that case. But this is more related to the problem constraints.

Please try this adjustment and see if it resolves the heap-buffer-overflow error.

1
2
3
4
5
6
7
8
class Solution(object):
	def search(self, nums, target):
		if target in nums:
		for i in range(len(nums)):
			if nums[i] == target:
				return i
			else:
				return -1

The solution you provided will indeed find the target if it’s present in the array and return its index, and will return -1 if it’s not. However, this solution has a time complexity of O(n) since it loops through each element in the array once.

But, this doesn’t take full advantage of the fact that the input array is a rotated sorted array. The problem can be solved using a binary search approach, which will reduce the time complexity to O(log n).

Here’s why your current solution could be improved:

Efficiency: The provided solution searches linearly, so it could be slow if the array is large. Binary search can speed this up.
Under-utilization of given properties: The problem statement mentions that the array is sorted (although rotated). Your solution doesn’t take advantage of this.
Redundant if target in nums check: The for-loop will handle the case where the target is not in the nums array by not returning anything (therefore getting to the else: return -1 line after the loop). The if target in nums line is an unnecessary O(n) operation.

Here’s how an optimized binary search might look:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution(object):
    def search(self, nums, target):
        low = 0
        high = len(nums) - 1

        while low <= high:
            mid = (low + high) // 2
            if nums[mid] == target:
                return mid
            elif nums[mid] >= nums[low]:
                if target >= nums[low] and target < nums[mid]:
                    high = mid - 1
                else:
                    low = mid + 1
            else:
                if target <= nums[high] and target > nums[mid]: 
                    low = mid + 1
                else:
                    high = mid - 1

        return -1

This binary search takes advantage of the sorted nature of the array, and will be faster for larger inputs.

1
2
3
4
5
class Solution:
	def search(self, nums: List[int], target: int) -> int:
		if target in nums:
			return nums.index(target)
	return -1

The code you posted is a simple and straightforward approach to the problem, and it would work correctly. However, the time complexity of this solution is O(n) where ’n’ is the number of elements in the input list ’nums’. The reason for this is that both the ‘in’ operator and the ‘index()’ method in Python scan the list linearly to find the target element.

This solution does not utilize the fact that the input list is a rotated sorted list. The problem can be solved in a more efficient way by using a binary search, which reduces the time complexity to O(log n). This could be a significant improvement when dealing with larger lists.

Here’s how a solution using binary search might look:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution:
    def search(self, nums: List[int], target: int) -> int:
        low = 0
        high = len(nums) - 1

        while low <= high:
            mid = (low + high) // 2
            if nums[mid] == target:
                return mid
            if nums[low] <= nums[mid]:
                if nums[low] <= target < nums[mid]:
                    high = mid - 1
                else:
                    low = mid + 1
            else:
                if nums[mid] < target <= nums[high]:
                    low = mid + 1
                else:
                    high = mid - 1

        return -1

This binary search takes advantage of the sorted nature of the array to divide the list into two halves and only continue searching in the half where the target could be present. This optimizes the solution by reducing the number of elements needed to be checked to find the target.

I covered all the ways that be able to go through left and right side of the array. If we don’t go for right side( left = mid +1 ) of array then there might be a chance target element present in left side of array. I felt that conditions are too long but I don’t find other way.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
public int search(int[] nums, int target) {
	int left = 0;
	int right = nums.length-1;
	int mid = -1;
	boolean found = false;
	while(left <= right){
		mid = (left + right)/2;
		if(nums[mid] == target){
			found = true;
			break;
		}
		else if((target > nums[mid] && nums[left] < nums[right]) || (target < nums[mid] && nums[left] > nums[right] && nums[left] > target && nums[left] <= nums[mid] ) || (target > nums[mid] && nums[left] > target) || (target > nums[mid] && nums[mid] > nums[left])){
			left = mid + 1;
		}
		else{
			right = mid - 1;
		}
	}
	return (found)?mid:-1;
}

Your code seems correct, but it is indeed a bit convoluted. In a coding interview, readability and simplicity are often as important as correctness. This means that the ideal solution is not only correct but also easy to understand.

Let’s critique your solution:

Complex conditions: The conditions inside your if and else if blocks are quite complex. This not only makes the code hard to read and understand, but also prone to errors. It would be better to break these conditions down into separate if else blocks with clear comments explaining the logic.
Use of found flag: The boolean flag “found” is not necessary. You can just return mid from within the loop when you find the target.
Variable Naming: It would be better to name your variables according to what they represent. For example, left and right could be low and high, which might more clearly represent the lower and upper bounds of the search range.
Code Comments: Your code lacks comments, which would be useful in explaining the logic, especially when the condition checks are complex.

Here’s a simpler version of a binary search algorithm on a rotated sorted array:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
public int search(int[] nums, int target) {
    int low = 0, high = nums.length - 1;
    while (low <= high) {
        int mid = (low + high) / 2;
        if (nums[mid] == target) return mid;

        if (nums[low] <= nums[mid]) { // left half is sorted
            if (nums[low] <= target && target < nums[mid]) high = mid - 1; // target is in left half
            else low = mid + 1; // target is in right half
        } 
        else { // right half is sorted
            if (nums[mid] < target && target <= nums[high]) low = mid + 1; // target is in right half
            else high = mid - 1; // target is in left half
        }
    }
    return -1;
}

This code is more readable, shorter, and it accomplishes the same goal. It divides the array into two halves, checks which half is sorted, and determines which half the target could be in. Then, it repeats the process for that half.

Why I am getting ‘-1’ for the input [4,5,6,0,1,2] for the following code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
class Solution {
public:
	int search(vector& nums, int target) {
		int n=nums.size(), ans;

		int start=0, end=n-1;
		if(nums[end]>nums[start]){
			ans=nums[start];
		}
		else{
			bool found=false;
			while(start<=end){
				int mid=start+(end-start)/2;
				if(nums[mid]==target){
					found=true;
					ans=mid;
					break;
				}
				else if(nums[mid]>target){
					start=mid+1;
				}
				else{
					end=mid-1;
				}
			}
			if(!found) ans=-1;
        
		}
		return ans;
	}
};

Your current implementation is wrong because it assumes that the array is always sorted in descending order and that if it’s not, it must be a rotated array.

However, the array can be a regular sorted array (ascending order), or a rotated sorted array. Also, you have mistakenly assigned the ans=nums[start]; when the nums[end]>nums[start] which means the array is in ascending order. In this case, you’re not actually performing a search but returning the first element of the array, which is wrong.

Moreover, the logic of your binary search inside the else statement is flawed. The condition else if(nums[mid]>target) implies that if the middle element is greater than the target, you should look to the right side of the array (increasing start), which is incorrect. When the middle element is greater, we need to look to the left side (decreasing end).

The corrected code should look like:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n=nums.size();

        int start=0, end=n-1;
        while(start<=end){
            int mid=start+(end-start)/2;
            if(nums[mid]==target){
                return mid;
            }
            // Determine which half to go, left half is sorted
            else if(nums[start] <= nums[mid]){
                if(target >= nums[start] && target < nums[mid]){
                    end = mid - 1;
                }else{
                    start = mid + 1;
                }
            }
            // Right half is sorted
            else{
                if(target > nums[mid] && target <= nums[end]){
                    start = mid + 1;
                }else{
                    end = mid - 1;
                }
            }
        }
        return -1;
    }
};

This code performs a binary search checking which half is sorted and whether the target falls into that half or not. Then it adjusts the start and end pointers accordingly and repeats until the target is found or the search space is empty.

Idea here is that if we keep breaking the array into two halves, then either the left array will be ordered or right array will be ordered. If the target value lies in the ordered half, then our further finding is restricted to this portion, otherwise it will get restricted to the other portion which is unordered. Doing this recursively, we will be able to find the index of target.

Indeed, your description aptly describes the approach used to solve this problem. The algorithm takes advantage of the sorted nature of one half of the array after rotation. In summary, here’s the detailed approach based on your description:

We start by implementing a binary search where we find the mid element of the array.
Then, we check which half of the array is sorted.
a. If the left half is sorted, we then check if our target is in the sorted range. If it is, we change our binary search to only consider this left half, otherwise, we consider the right half.
b. If the right half is sorted, we check if our target is in the sorted range. If it is, we change our binary search to only consider the right half, otherwise, we consider the left half.
We repeat this process until we either find our target (in which case we return the index) or we exhaust all possibilities (in which case we return -1, as the target is not present in the array).

This strategy guarantees that we check fewer elements each iteration, giving us a time complexity of O(log n), which is a significant improvement over a linear search.

why is this not working?

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
class Solution {
public:
	int search(vector& nums, int target) {
		int high = nums.size()-1;
		int low = 0;
		while(low < high){
			int mid = low + (high - low)/2;
			if(nums[mid] == target) return mid + 1;
			else if(nums[mid] > nums[mid + 1]){
				if(nums[mid] > target)
					low = mid + 1;
				else 
					high = mid - 1;
			}
			else if(nums[mid] < nums[mid + 1]){
				if(nums[mid] >= target && nums[low] <= target)
					high = mid - 1;
				else 
					low = mid + 1;
			}
		}
		return -1;
	}
};

There are a couple of issues with your code. Here are the corrections:

Your binary search loop should continue while low <= high, not while low < high.
The check nums[mid]==target should return mid, not mid+1.
Your conditions for checking which half to look in are incorrect. You want to determine if the left half is sorted and if the target is in that range, or if the right half is sorted and if the target is in that range.

Here is a corrected version of your code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
class Solution {
public:
    int search(vector<int>& nums, int target) {
        int high = nums.size() - 1;
        int low = 0;
        while(low <= high){
            int mid = low + (high - low) / 2;
            if(nums[mid] == target) 
                return mid;
            if(nums[low] <= nums[mid]) {  // Left half is sorted.
                if(nums[low] <= target && target < nums[mid]) 
                    high = mid - 1;  // Target is in the left half.
                else
                    low = mid + 1;  // Target is in the right half.
            }
            else {  // Right half is sorted.
                if(nums[mid] < target && target <= nums[high])
                    low = mid + 1;  // Target is in the right half.
                else
                    high = mid - 1;  // Target is in the left half.
            }
        }
        return -1;
    }
};

This code works by first finding the middle of the current search range. If the middle element is the target, we’re done. Otherwise, we check if the left half of the range is sorted. If it is, and the target is in the range of the left half, we continue searching in the left half; otherwise, we search in the right half. If the left half is not sorted, that means the right half must be sorted. We then check if the target is in the range of the right half, and if so we continue searching in the right half; otherwise, we search in the left half.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
def count_rotations_binary(nums):
lo = 0
hi = len(nums)-1

while hi>lo:
    mid = lo+hi//2
    mid_number = nums[mid]

    print("lo :",lo ,"hi :", hi,  "mid :",mid ,"mid number :", mid_number)

    if mid > 0 and nums[mid] < nums[mid-1]:
        return mid

    elif nums[mid] < nums[hi]:
        hi = mid-1

    else:
        lo = mid+1

input nums : test1 = { ‘input’ : { ’nums’ : [4,5,6,7,8,1,2,3] # after rotating 5 times }, ‘output’ : 5 # rotated 5 times means output = 5 }

Test failed on second division where mid + 1 = 4. following the same lo+hi//2. expected answer: 4+7=11. 11//2 = 5.

I am getting 7 for division of 11//2 as mid. what’s the problem in this code ?

The issue lies in the order of operations in this line:

1
mid = lo + hi // 2

In Python, the division operator // has higher precedence than the addition operator +. This means that your code is actually performing (hi // 2) first, and then adding lo to the result. This will cause mid to be calculated incorrectly.

You want to divide the sum of lo and hi by 2, so you should use parentheses to ensure the correct order of operations:

1
mid = (lo + hi) // 2

This line will first add lo and hi, and then divide the result by 2 (performing integer division), which will correctly calculate the middle index. Here’s the corrected code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
def count_rotations_binary(nums):
    lo = 0
    hi = len(nums)-1

    while hi > lo:
        mid = (lo + hi) // 2
        mid_number = nums[mid]

        print("lo :", lo, "hi :", hi, "mid :", mid, "mid number :", mid_number)

        if mid > 0 and nums[mid] < nums[mid-1]:
            return mid
        elif nums[mid] < nums[hi]:
            hi = mid - 1
        else:
            lo = mid + 1

This will now correctly calculate the middle index and should produce the expected result.

if the target is different for every list the function has to be separate for every list. or passing two elements in parenthesis nums (list as input) and output(target number as input), where function should return target number index position will work ?

Yes, you can certainly pass the target number as an additional argument to the function. This is a common practice and will allow you to use the same function to search for different targets in different lists. The function signature would look something like this:

1
2
def binary_search(nums, target):
    # Implementation here

In the body of this function, you would implement your binary search (or whatever searching algorithm you’re using). This function would take a list of numbers (nums) and a target number (target), and it would return the index of target within nums. If target is not in nums, you could return -1 or some other sentinel value.

Here’s an example of how you might call this function:

1
2
3
nums = [4, 5, 6, 7, 8, 1, 2, 3]
target = 6
print(binary_search(nums, target))  # This would print 2, assuming the list is zero-indexed

By parameterizing the target number, you can easily reuse the same function to search for different targets in different lists.

uses a while loop to repeatedly divide the search space in half until the target value is found or until the search space is exhausted. The loop starts with the left index set to 0 and the right index set to the last element of the array. The mid index is calculated as the average of the left and right indices. The function then checks if the element at the mid index is equal to the target value, and if so, it returns the mid index as the result. If the element at the mid index is not equal to the target value, the function checks if the left side of the array is strictly increasing or not. If it is, it checks if the target value is between the left and mid elements, and if so, it updates the right index to be one less than the mid index, otherwise it updates the left index to be one more than the mid index. If the left side is not strictly increasing, the function checks if the target value is between the mid and right elements, and if so, it updates the left index to be one more than the mid index, otherwise it updates the right index to be one less than the mid index. If the search space is exhausted without finding the target value, the function returns -1.

The paragraph you provided describes a binary search algorithm that operates on a rotated sorted array. This is a standard approach for solving search problems on sorted or partially sorted structures in logarithmic time complexity. It works by repeatedly dividing the array into two halves and updating the search space based on comparisons with the middle element. If the middle element is not the target and the array is rotated (meaning it is sorted, but starts from a point other than the smallest or largest element), the algorithm identifies which half of the array is sorted and checks if the target could lie within that half. If so, the search continues in that half; otherwise, it continues in the other half. If the target element is not found, -1 is returned.

using the concept the earlier solved provblem number 33 in which we were asked to find the minimin element in the roated sorted arrray usign that concept we find the index of the minimum element of the arrray and the we use the binary search to search the target in the start to index and apply another search on the index+1 to end and how ever return the answer greater than -1 will be final answer

This paragraph you provided describes an approach to solve a variant of the “search in a rotated sorted array” problem. In problem number 33, we are asked to find the minimum element in a rotated sorted array. In this variant, we need to find a specific target value. The approach described in your paragraph does this by:

First, finding the index of the minimum element of the array (which is also the pivot of rotation in the rotated sorted array).
Then, dividing the original problem into two standard binary search problems: one for the subarray from the start of the array to the pivot, and another for the subarray from the pivot to the end of the array.
The final answer is determined by whichever search returns an index greater than -1 (since -1 is used to indicate that the target is not found).

Here the main goal is to divide the array recursively in such a way that will get the target in sorted sub-part of array, and point to be note that at least one part should be sorted, after every division step of binary search Steps that need to follow recursively to find the sorted subarray to search for an element

1). If mid>=a[start], it means first part is sorted try to search in it, if target exist in this (first) half range, then end should be shifted to mid-1, otherwise start should point to mid+1 i.e. second half start. 2). If mid < a[start], it means second half will be sorted try to search in it, if target exist in this half range, then start should be shifted to mid+1, otherwise end should point to mid-1, i.e. first half end.

Points to be noted:

In every step we are discarding one half of remained subarray or original array, same like binary search that’s why here the complexity still remains O(logn). In between division of subparts one of two steps stated above will come, and that’s why recursion is there or these steps should be under while loop.

Your explanation provides a clear description of how to modify a binary search to find a target value in a rotated sorted array.

This method takes advantage of the fact that in a rotated sorted array, at least one half of the array will always be sorted. By checking the middle value against the start value, we can determine which half is sorted. Once the sorted half is identified, we can check if the target falls within that range. If it does, we can continue our binary search within that half. If it doesn’t, we know the target must be in the other half, so we can continue the binary search there.

Here’s how we might summarize the steps in this algorithm:

Compare the middle element of the array to the start element.
If the middle element is greater than or equal to the start element, the first half of the array is sorted.
- If the target is within the range of the first half, continue the binary search there (set the end to mid - 1).
- If the target is not within the range of the first half, continue the binary search in the second half (set the start to mid + 1).
If the middle element is less than the start element, the second half of the array is sorted.
- If the target is within the range of the second half, continue the binary search there (set the start to mid + 1).
- If the target is not within the range of the second half, continue the binary search in the first half (set the end to mid - 1).
Continue this process until the target is found or the search space is exhausted.

The time complexity for this algorithm remains O(logn) as at each step, we are effectively halving the search space, just like in a traditional binary search.

I am getting heap buffer overflow without using any dynamic allocation in heap in the code this

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
int search(vector& nums, int target)
{
	int start=0,end=nums.size()-1,mid,key;
	for(mid=start+(end-start)/2;start<=end;mid=start+(end-start)/2)
	{

		if(nums[mid]<=nums[mid+1])
		{
			end=mid-1;
		}
		else
		{
			key=mid+1;
			break;
		}
	}
	if(nums[nums.size()-1]<target)
	{
		start=0,end=key-1;
	}
	else
	{
		start=key,end=nums.size()-1;
	}
	for(mid=start+(end-start)/2;start<=end;mid=start+(end-start)/2)
	{
		if(nums[mid]>target)
		{
			end=mid-1;
		}
		else if(nums[mid]<target)
		{
			start=mid+1;
		}
		else
		{
			return mid;
		}
	}
	return -1;
}

The error is because you are trying to access the index “mid+1” of the vector “nums” without checking whether “mid+1” is a valid index for the vector. If “mid” is the last index of the vector, “mid+1” will be out-of-bound, and trying to access such an index will result in undefined behavior such as heap buffer overflow.

The problematic line is:

1
if(nums[mid]<=nums[mid+1])

To fix this, you need to ensure that “mid+1” does not exceed the size of the vector before accessing it. Here’s how you could do it:

1
if(mid < nums.size()-1 && nums[mid] <= nums[mid+1])

Please ensure to apply similar changes to any other parts of the code that may attempt to access out-of-bound indices. Always remember that when working with arrays or vectors, it’s crucial to ensure that you’re not accessing indices outside of their size. This is a common source of bugs and can lead to serious problems such as buffer overflows.

Why we do not use this -> sort(nums.begin(), nums.end()); I know then the time complexity will increase but just ignore that part why we do not use this?

Sorting an array is often a useful operation when you want to organize the data in a certain order, typically either ascending or descending. However, there are several reasons why you might not want to sort the array in the context of a particular problem.

Preserving original order: Many problems require you to preserve the original order of elements in the array. Sorting the array would destroy this original order.
Time complexity: As you mentioned, sorting an array has a time complexity of at least O(n log n) for most standard sorting algorithms. This could make your solution too slow if the input array is large.
Space complexity: Sorting algorithms often require additional space. For example, the commonly used quicksort and mergesort algorithms are not in-place and need O(n) extra space.
Problem Constraints: Sometimes, the problem statement specifically asks you not to sort the array or assumes that you’re working with an unsorted array.
Changing the nature of the problem: Sometimes sorting an array could fundamentally change the nature of the problem you’re trying to solve.

In the context of the problem of searching in a rotated sorted array, sorting the array first would defeat the purpose of the problem. The challenge is to leverage the fact that the array is rotated and sorted to achieve a faster search time, ideally O(log n) which is possible with binary search. If you sort the array first, you would destroy the rotation and any information it might provide. Furthermore, you would still have to search through the sorted array, which would take O(log n) time with binary search, and when combined with the sorting step, would increase the overall time complexity of the solution.

why is my solution not working for [3,1] nums vector ?

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
int search(vector& nums, int target) {
	int start=0;
	int end=nums.size()-1;
	while(start <= end)
	{
		int mid=(start + end)/2;
		if(target == nums[mid])
		{
			return mid;
		}
		if(nums[start] < nums[mid])
		{
			if(target >= nums[start] && target<nums[mid])
			{
				end=mid-1;
			}
			else
			{
				start=mid+1;
			}
		}
		else
		{
			if(target > nums[mid] && target <= nums[end])
			{
				start=mid+1;
			}
			else
			{
				end=mid-1;
			}
		}
	}
	return -1;
}

The code you provided is designed to handle a rotated sorted array. In the array [3, 1], the pivot point is at index 0, thus making the condition nums[start] < nums[mid] fail when start and mid are both 0.

This occurs because the array is only of length 2, and the element at the start and mid index is larger than the element at the end index. Your code, however, expects that for the first part of the array (nums[start] to nums[mid]), it will be in sorted order. But due to the pivot at index 0, this is not the case.

To fix this, you can include a check for when start and mid are equal:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
int search(vector<int>& nums, int target) {
    int start=0;
    int end=nums.size()-1;
    while(start <= end)
    {
        int mid=(start + end)/2;
        if(target == nums[mid])
        {
            return mid;
        }
        if(nums[start] <= nums[mid]) // change < to <=
        {
            if(target >= nums[start] && target<nums[mid])
            {
                end=mid-1;
            }
            else
            {
                start=mid+1;
            }
        }
        else
        {
            if(target > nums[mid] && target <= nums[end])
            {
                start=mid+1;
            }
            else
            {
                end=mid-1;
            }
        }
    }
    return -1;
}

This modification handles the case where the pivot is at the start index. Now, if start equals mid and nums[start] is not less than nums[mid], it will still check if the target is within the first half of the array. This is suitable for arrays like [3,1] where the pivot is at the start of the array.

Is this considered a O(log n) solutuion or a O(n/2) ?

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
 * @param  {number[]} nums
 * @param  {number} target
 * @return {number}
 */
var search = function( nums, target ){
    let first = 0,
        middle = Math.floor( nums.length / 2 ),
        last = nums.length - 1,
        indexOfTarget = ( start, end ) => {
            for( let index = start; index <= end; index++ ) if( nums[index] == target ) return index
            return -1
        },
        // [7,8,1,2,3,4,5,6] => right[ 3, 4, 5, 6 ] => [4, 7]
        sorted_indexes = ( nums[first] < nums[middle] ) ? [first, middle] : [middle + 1, last],
        // left[ 7, 8, 1, 2 ] => [0, 3]
        unsorted_indexes = sorted_indexes[0] != first && sorted_indexes[0] != middle ? [first, middle] : [middle + 1, last]
        // target = 2 in right[ 3, 4, 5, 6 ] ? false
        inSortedSide = nums[ sorted_indexes[0] ] <= target && target <= nums[ sorted_indexes[1] ]

    if( nums[first] == target )         return first
    else if( nums[middle] == target )   return middle
    else if( nums[last] == target )     return last
    else if( inSortedSide )             return indexOfTarget( sorted_indexes[0], sorted_indexes[1] )
    else /* !inSortedSide */            return indexOfTarget( unsorted_indexes[0], unsorted_indexes[1] )
};

This solution is neither O(log n) nor O(n/2). This solution is O(n) in the worst case scenario.

The reason for this is due to the indexOfTarget function. In the worst-case scenario, this function will iterate through roughly half of the elements in the array, which gives a time complexity of O(n).

For example, consider a case where the target value is at the end of the unsorted half of the array. The indexOfTarget function will need to iterate over all elements in the unsorted half, which can be up to n/2 elements. Because we drop constants in Big O notation, the time complexity of this function is O(n), not O(n/2).

Remember, O(log n) complexity typically implies that you’re dividing the problem in half (or any constant fraction) with each step, such as in a binary search. O(n) complexity implies that you have to potentially check every element, which is what the indexOfTarget function is doing. Despite the initial checks and divisions of the array, the linear search performed by indexOfTarget dominates the time complexity of your solution.

To achieve O(log n) complexity for this problem, you would typically perform a modified binary search, adjusting the high and low points of your search based on the target and the median value, without scanning the array linearly.

Help me to debug this code .

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
class Solution {
public:
	int search(vector& nums, int target) {
		int l = 0,h = nums.size() - 1;
		while(l <= h){
			int m = (l + (h - l))>>1;
			if(nums[m] == target)
				return m;
			//left half is sorted.
			if(nums[l] <= nums[m])
			{
				if(target >= nums[l] && target <= nums[m]) //lies in range ..
					h = m - 1;
				else
					l = m + 1;
			}
			//right half is sorted.
			else
			{
				if(target >= nums[m] && target <= nums[h]) // lies in range..
					l = m + 1;
				else
					h = m - 1;
			}
		}
		return - 1;
	}
};

Your code appears to be correct for a binary search within a rotated sorted array. If the function doesn’t work as expected, there could be a few reasons why:

The vector might not be sorted or properly rotated: This code assumes that the input vector is a sorted array that has been rotated. If it’s not, the code might not work as expected.
The target number might not exist in the array: If the number you’re looking for isn’t in the array, the function will correctly return -1, which is the expected behavior.
Compiler, environment, or runtime error: There could be a problem with your development environment or a runtime error causing the code to fail.

I would suggest first verifying that your input is a rotated sorted array and the target number exists in it. If you’re still having problems, please provide more information about the specific error or issue you’re encountering.

Remember, the provided function’s time complexity is O(log n) as it utilizes the concept of Binary Search, making it a very efficient solution for large arrays.

The logic I am using is finding the position where the trend changes and then putting the limits of i and j and then applying binary search on it ,yet it is giving me the wrong answer? any suggestions? My code below——->

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
class Solution {
public:
	int search(vector& nums, int target) {
		int low=0,high=nums.size()-1;
		while(low < high)
		{
			int mid = (low + high)/2;
			if(nums[mid] > nums[high])
				low = mid + 1;
			else
				high = mid - 1;
		}

		int i = 0,j = 0;
		if(target > nums[high]){
			i= 0; j = low -1;
		}
			else{
				i = low; j = nums.size()-1;
			}
				while(i < j) {
					int mids = (i + j) / 2;
					if(nums[mids] == target)
						return mids;
					else if(nums[mids] > target)
						j = mids - 1;
					else
						i = mids + 1;
				}
				return j;
			}
		};

Your approach to solve the problem using binary search is correct but there are some issues with your implementation:

After finding the smallest element, you need to check whether the target is in the first half or the second half of the rotated array and search in the correct half. However, in your code, you’re defining the boundaries i and j incorrectly. You should compare the target with nums[0] instead of nums[high] because nums[0] would be the starting point of either of the sorted halves of the array.
In the second while loop, you’re using while(i<j). But you should use while(i<=j) to ensure the loop runs when i and j point to the same element.
In the return statement, you’re returning j, which is not correct. If the target is not found, you should return -1.

Here is your corrected code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
class Solution {
public:
    int search(vector<int>& nums, int target) {
        int low=0, high=nums.size()-1;
        while(low < high)
        {
            int mid=(low+high)/2;
            if(nums[mid]>nums[high])
                low=mid+1;
            else
                high=mid;
        }
        // low is now pointing to smallest element
        int i=0, j=0;
        if(target>=nums[0]){ // target is in the first half
            i=0; j=low;
        }
        else{ // target is in the second half
            i=low; j=nums.size()-1;
        }
        while(i<=j){ // modified from i < j to i <= j
            int mid = (i + j) / 2;
            if(nums[mid] == target)
                return mid;
            else if(nums[mid] > target)
                j = mid - 1;
            else
                i = mid + 1;
        }
        return -1; // return -1 if target not found
    }
};

With these changes, your code should work correctly to find the target in the rotated sorted array.

Common reasons for making these mistakes include:

Misunderstanding the Problem: The problem of searching in a rotated sorted array is more complex than searching in a regular sorted array. You need to first identify the pivot point where the array is rotated, then determine which half of the array to search in, and then perform a binary search within that half. Misunderstanding any part of this process can lead to incorrect code.
Off-By-One Errors: These are common in code that deals with array indices or boundaries of loops. In your case, using while(i<j) instead of while(i<=j) is an off-by-one error because it overlooks the case where i and j point to the same element.
Incorrect Return Value: Returning j instead of -1 when the target is not found is likely due to a misunderstanding of what the function should return in this situation. Always make sure you understand the problem statement and what each function or piece of code should return.
Not Testing Enough: Testing your code with various input cases (including edge cases) can help you catch and correct mistakes. In your case, it seems the code might have been tested with input where the target is in the second half of the array but not the first half, leading to the mistake in defining i and j.
Not Using Descriptive Variable Names: Using descriptive variable names can make the code easier to read and understand, reducing the chance of errors. In your code, i and j are used as the boundaries for the binary search, but more descriptive names like start and end might help avoid confusion.
Lack of Experience: Lastly, simply not having enough experience with similar problems or programming in general can lead to mistakes. The more you practice, the better you’ll get at avoiding these types of errors.

pls tell me what’s the error in code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
int search_pivot(vector & nums)
{
	int s=0;
	int e=nums.size()-1;
	int mid=s+ (e-s)/2;
	while(s < e)
	{
		if(nums[mid]>=nums[0])
		{
			s=mid+1;
		}
		else
		{
			e=mid;
		}
		mid=s+ (e-s)/2;
	}
	return s;
}
int binary_search(vector& nums,int s,int e,int key)
{
	int start=s;
	int end=e;
	int mid=start + (end-start)/2;
	while(start<=end)
	{
		if(nums[mid]==key)
		{
			return mid;
		}
		else if(nums[mid] < key)
		{
			start=mid+1;
		}
		else
		{
			end=mid-1;
		}
		mid=start + (end-start)/2;
	}
	return -1;
}

class Solution {
public:
	int search(vector& nums, int target) {
		int n=nums.size()-1;
		int pivot=search_pivot(nums);
		if(target>=pivot && target<=nums[n-1] )
		{
			return binary_search(nums,pivot,n-1,target);
		}
		else
		{
			return binary_search(nums,0,pivot-1,target);
		}
	}
};

The error is in the line where you are checking if(target>=pivot && target<=nums[n-1]).

Here, pivot is an index, not a number. You need to compare target with the values of nums at the indexes pivot and n-1, not with the indexes themselves. So, you should replace the condition as if(target>=nums[pivot] && target<=nums[n-1]).

Here is your corrected code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
int search_pivot(vector<int> & nums)
{
    int s=0;
    int e=nums.size()-1;
    int mid=s+ (e-s)/2;
    while(s < e)
    {
        if(nums[mid] >= nums[0])
        {
            s=mid+1;
        }
        else
        {
            e=mid;
        }
        mid=s+ (e-s)/2;
    }
    return s;
}

int binary_search(vector<int>& nums,int s,int e,int key)
{
    int start=s;
    int end=e;
    int mid=start + (end-start)/2;
    while(start<=end)
    {
        if(nums[mid]==key)
        {
            return mid;
        }
        else if(nums[mid] < key)
        {
            start=mid+1;
        }
        else
        {
            end=mid-1;
        }
        mid=start + (end-start)/2;
    }
    return -1;
}

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n=nums.size()-1;
        int pivot=search_pivot(nums);
        if(target>=nums[pivot] && target<=nums[n-1] )
        {
            return binary_search(nums,pivot,n,target);
        }
        else
        {
            return binary_search(nums,0,pivot,target);
        }
    }
};

Remember that you need to pass n instead of n-1 to binary_search when the target is in the second half of the array, because n is the last index of the array. Similarly, you need to pass pivot instead of pivot-1 when the target is in the first half, because pivot is the first index of the second half.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
class Solution {
	public int search(int[] nums, int target) {
		int max=Integer.MIN_VALUE;
		for(int i=0; i < nums.length;i++)
		{
			max=Math.max(max,nums[i]);
		}
		int index=0;
		for(int i=0;i < nums.length;i++)
		{
			if(nums[i]==max)
			{
				index=i;
			}
		}
		if(target==nums[index])
		{
			return index;
		}
		else if(index==(nums.length-1))
		{
			int search=binarySearch(nums,0,nums.length,target);
			return search;
		}
		else if(target < nums[nums.length-1])
		{
			int search=binarySearch(nums,index,nums.length,target);
			return search;
		}
		else
		{
			int search=binarySearch(nums,0,index,target);
			return search;
		}
	}
	public int binarySearch(int[]nums,int low,int high,int target)
	{
		if(low<=high)
		{
			int mid=low+(high-low)/2;
			System.out.println(mid);
			if(nums[mid]==target)
			{
				return mid;
			}
			else if(target>nums[mid])
			{
				return binarySearch(nums,mid+1,high,target);
			}
			else{
				return binarySearch(nums,low,mid-1,target);
			}
		}
		return -1;

	}
}

This is my code ,for this test case[4,5,6,7,0,1,2] I am getting the mid value as 3.can someone check

I reviewed your code and I believe the issue is related to the boundaries you set for your binary search function.

The binary search function in your code takes as parameters the lower and upper boundaries of the subarray in which it is going to search. However, in Java, the second parameter of the Arrays.copyOfRange function represents the upper boundary exclusive. In other words, the element at the index represented by the second parameter is not included in the new array.

When you are calling the binary search function in the else if(target<nums[nums.length-1]) block, you are passing index as the lower boundary and nums.length as the upper boundary. This means you are trying to search for the target in the subarray that starts at the index of the maximum element and ends at the end of the array. However, the maximum element is not included in this subarray because you are setting its index as the lower boundary.

Similarly, in the else block, you are passing 0 as the lower boundary and index as the upper boundary. This means you are trying to search for the target in the subarray that starts at the beginning of the array and ends at the index of the maximum element. But again, the maximum element is not included in this subarray because you are setting its index as the upper boundary.

You should adjust the boundaries to include the maximum element in the subarray, depending on the case. For example, in the else if(target<nums[nums.length-1]) block, you should decrement index by one before passing it as the lower boundary, and in the else block, you should increment index by one before passing it as the upper boundary.

Also, in your binary search function, you are not updating the mid value after each recursive call. You should update the mid value inside each if and else if block.

I would also suggest refactoring your code to find the pivot point (the point of rotation) instead of the maximum element, and then performing binary search on the appropriate half. This approach would be more efficient and easier to implement.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution {
public:
	int search(vector& nums, int target) {

		if(binary_search(nums.begin(),nums.end(),target) || 0)
			return target;

		else
			return -1;
	} 
};

//Failed testcase nums = [4,5,6,7,0,1,2] target = 0 Output -1 Expected 4

BUT WHY???

The binary_search() function from the C++ Standard Library is used to determine if an element exists in a range (a sorted range, to be precise). It does not return the index of the element if it exists, rather it returns a boolean value indicating whether the element exists in the given range or not.

So, in your case, the binary_search() function is returning true (converted to 1 in the context of your if statement) when the target is in the array and false (converted to 0) when it’s not. As a result, your function returns the target value when it’s found, not the index of the target value.

In the case of your failed test, the binary_search() function is returning true because 0 exists in the array. Then, in your if statement, true is converted to 1, so your function returns the target value 0, not its index 4.

In this case, you need to implement a specific variant of the binary search algorithm that deals with sorted but rotated arrays and returns the index of the target value. The standard binary_search() function cannot solve this problem as it assumes the input range is sorted in a normal (non-rotated) manner.