Bug Report NeetCode Search In Rotated Sorted Array Solution Analysis And Fix

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Introduction

This article addresses a bug report concerning the "Search in Rotated Sorted Array" problem on NeetCode.io. The bug report highlights a flaw in a submitted solution that passes the existing test cases but fails on specific inputs due to a logical error within the binary search implementation. This article will delve into the details of the bug, provide a clear explanation of the problematic scenario, and suggest improvements for test case coverage to ensure the robustness of solutions.

This exploration is crucial for both aspiring programmers learning from platforms like NeetCode and the platform developers who strive to provide accurate and comprehensive problem sets. By understanding the nuances of this bug and its potential impact, we can collectively enhance the learning experience and improve the quality of algorithmic solutions.

Problem Overview: Search in Rotated Sorted Array

The Search in Rotated Sorted Array problem is a classic algorithmic challenge that tests a candidate's understanding of binary search in a non-standard scenario. The problem statement is as follows:

  • Given a sorted array that has been rotated at some pivot point, and a target value, find the index of the target value in the array. If the target value is not present, return -1.

For instance, the array [4, 5, 6, 7, 0, 1, 2] is a rotated sorted array. The challenge lies in the fact that the traditional binary search algorithm, which relies on the sorted nature of the array, cannot be directly applied due to the rotation. A crucial aspect of solving this problem effectively is to identify the sorted portion of the array in each iteration and adjust the search space accordingly. The submitted solution aims to achieve this, but as the bug report reveals, it contains a logical flaw.

The Bug Report: A Deep Dive

The bug report originates from a user on the NeetCodeDiscussion platform who encountered a specific scenario where a seemingly correct solution failed. The solution, implemented in Python, utilizes a modified binary search approach to locate the target value in the rotated array. Here’s the code snippet from the bug report:

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        l = 0
        r = len(nums) - 1
        while (r > l):
            mid = (r + l - 1) // 2

            if target == nums[mid]:
                return mid
            elif target == nums[l]:
                return l
            elif target == nums[r]:
                return r
            else:
                if nums[l] < target and target < nums[mid]:
                    r = mid
                else:
                    l = mid + 1

        return l if nums[l] == target else -1

The core of the issue lies within the conditional block:

if nums[l] < target and target < nums[mid]:
    r = mid
else:
    l = mid + 1

This block attempts to determine whether the target value lies within the sorted left half of the array. It assumes that if nums[l] is less than the target and the target is less than nums[mid], then the target must be in the left half, and the right boundary r is updated to mid. However, this assumption is flawed when the array is rotated in a specific way, and the target is located in the sorted right half or when the mid value wraps around the pivot.

The Flawed Logic Explained

To fully understand the flaw, let’s break down the problematic scenario. The conditional check nums[l] < target and target < nums[mid] is designed to identify cases where the left portion of the array (from index l to mid) is sorted in ascending order, and the target falls within this range. If this condition holds, the algorithm narrows its search to the left half by setting r = mid. The issue arises when the rotated array's structure disrupts this logic. Specifically, there are two critical cases where this condition fails:

  1. Target in the Sorted Right Half: When the target is actually present in the sorted right half of the array, but the left half appears sorted based on the nums[l] < target and target < nums[mid] condition, the algorithm incorrectly discards the right half from consideration. This leads to the algorithm returning an incorrect result (-1 in this case) because the actual position of the target is never explored.

  2. Mid Value Wraps Around the Pivot: In scenarios where the mid-point mid falls on the left side of the rotation (the higher values), and the target is located on the right side (the lower values), the condition can fail to correctly identify the sorted portion. This is because the left-side condition might seem to hold true, but the actual sorted portion that contains the target is on the other side of the pivot.

Example Demonstrating the Bug

The bug report provides a compelling example to illustrate this flaw:

  • nums = [9, 1, 2, 3, 4, 5, 6, 7, 8]
  • target = 1

The expected output is 1 (the index of the target), but the provided solution incorrectly returns -1. Let’s trace the execution to understand why.

  1. Initial values: l = 0, r = 8
  2. mid = (0 + 8 - 1) // 2 = 3
  3. The target is not equal to nums[mid] (3), nums[l] (9), or nums[r] (8).
  4. The condition nums[l] < target and target < nums[mid] evaluates to 9 < 1 and 1 < 3, which is false.
  5. Therefore, the else block is executed, and l becomes mid + 1 = 4.
  6. In the next iteration, l = 4, r = 8
  7. mid = (4 + 8 - 1) // 2 = 5
  8. The target is not equal to nums[mid] (5), nums[l] (4), or nums[r] (8).
  9. The condition nums[l] < target and target < nums[mid] evaluates to 4 < 1 and 1 < 5, which is false.
  10. Therefore, the else block is executed, and l becomes mid + 1 = 6.
  11. In the next iteration, l = 6, r = 8
  12. mid = (6 + 8 - 1) // 2 = 6
  13. The target is not equal to nums[mid] (6), nums[l] (6), or nums[r] (8).
  14. The condition nums[l] < target and target < nums[mid] evaluates to 6 < 1 and 1 < 6, which is false.
  15. Therefore, the else block is executed, and l becomes mid + 1 = 7.
  16. In the next iteration, l = 7, r = 8
  17. mid = (7 + 8 - 1) // 2 = 7
  18. The target is not equal to nums[mid] (7), nums[l] (7), or nums[r] (8).
  19. The condition nums[l] < target and target < nums[mid] evaluates to 7 < 1 and 1 < 7, which is false.
  20. Therefore, the else block is executed, and l becomes mid + 1 = 8.
  21. Now, l = 8, r = 8, the loop terminates.
  22. The final check nums[l] == target evaluates to nums[8] == 1, which is 8 == 1, is false so -1 is returned

As seen in the trace, the algorithm incorrectly narrows the search space, missing the actual location of the target. This clearly illustrates the flaw in the logic.

Proposed Solution and Improved Logic

A correct solution must accurately identify the sorted portion of the array and adjust the search boundaries accordingly. One robust approach involves the following steps:

  1. Find the Midpoint: Calculate the middle index (mid) of the current search space.

  2. Check for Target: If nums[mid] equals the target, return mid.

  3. Determine Sorted Portion: Check if the left half (nums[l] to nums[mid]) is sorted. This can be determined by checking if nums[l] <= nums[mid].

  4. Search in Sorted Portion:

    • If the left half is sorted, check if the target lies within the left half (nums[l] <= target < nums[mid]). If so, search the left half by setting r = mid - 1. Otherwise, the target must be in the right half (if it exists), so set l = mid + 1.
    • If the left half is not sorted, then the right half (nums[mid] to nums[r]) must be sorted. Check if the target lies within the right half (nums[mid] < target <= nums[r]). If so, search the right half by setting l = mid + 1. Otherwise, the target must be in the left half, so set r = mid - 1.

  5. Repeat: Continue steps 1-4 until the target is found or the search space is exhausted (l > r).

Here is the corrected version of the code:

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        l, r = 0, len(nums) - 1

        while l <= r:
            mid = (l + r) // 2

            if nums[mid] == target:
                return mid

            # Left sorted portion
            if nums[l] <= nums[mid]:
                if nums[l] <= target < nums[mid]:
                    r = mid - 1
                else:
                    l = mid + 1
            # Right sorted portion
            else:
                if nums[mid] < target <= nums[r]:
                    l = mid + 1
                else:
                    r = mid - 1

        return -1

This corrected approach explicitly checks the sorted portions and ensures that the correct search space is explored, resolving the identified bug.

Improving Test Case Coverage

The bug report underscores the importance of comprehensive test case coverage. The initial solution passed the existing test cases on NeetCode but failed when presented with a specific input that exposed the flaw in the logic. To prevent such issues in the future, it is crucial to add more test cases that cover various scenarios, especially edge cases and those that involve specific rotations.

Suggested Test Cases

  1. Arrays with the target in the right sorted portion:

    • nums = [9, 1, 2, 3, 4, 5, 6, 7, 8], target = 1
    • nums = [5, 1, 3], target = 3

  2. Arrays with the target on the boundary:

    • nums = [3, 1], target = 1
    • nums = [4, 5, 6, 7, 0, 1, 2], target = 0

  3. Arrays with duplicate elements:

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

  4. Single-element arrays:

    • nums = [5], target = 5
    • nums = [5], target = 3

  5. Empty arrays:

    • nums = [], target = 5

By incorporating these test cases, the platform can ensure that solutions are thoroughly tested and that only correct implementations are accepted. This enhances the learning experience for users and maintains the integrity of the problem set.

Conclusion

This bug report highlights the critical importance of rigorous testing and a deep understanding of algorithmic nuances. The Search in Rotated Sorted Array problem, while seemingly straightforward, can present subtle challenges that require careful consideration. The flawed solution, though passing initial tests, failed due to an incorrect assumption about the sorted portion of the array. By identifying and rectifying this bug, we gain a clearer understanding of the problem and the importance of comprehensive test case coverage.

The corrected solution, along with the suggested test cases, provides a more robust approach to solving the problem. This collaborative effort between users and platform developers contributes to the overall quality of the learning resources available on platforms like NeetCode.io. By continually refining our understanding and testing methodologies, we can ensure that algorithmic solutions are both efficient and correct, fostering a more effective learning environment for all.

  • Bug in the solution for "Search in Rotated Sorted Array" on NeetCode.
  • Incorrect logic in the binary search implementation.
  • The solution fails for the input nums = [9, 1, 2, 3, 4, 5, 6, 7, 8] and target = 1.
  • The condition nums[l] < target and target < nums[mid] is flawed.
  • Need to improve test case coverage for rotated sorted array problems.
  • How to correctly implement binary search in a rotated sorted array?

Bug Report NeetCode Search in Rotated Sorted Array Solution Analysis and Fix