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Binary Solution#2466

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Binary Solution#2466
avcode3 wants to merge 1 commit intosuper30admin:masterfrom
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@avcode3
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@avcode3 avcode3 commented Feb 6, 2026

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@super30admin
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Your solution for the rotated sorted array search (in problem2.py) is well-written and correct. It efficiently handles the problem with O(log n) time and O(1) space. The logic is clear and follows the standard approach for this problem.

Strengths:

  • You correctly identified the need to determine which half of the array is sorted.
  • Your conditions for narrowing the search range are accurate.
  • The code is concise and easy to read.

Areas for improvement:

  • Although your condition in the else branch is correct, consider using nums[mid] < target instead of nums[mid] <= target to be consistent with the fact that we already checked for equality at the start. This might make the code slightly clearer, but it is not a critical issue.
  • Ensure that you are submitting the correct file for the problem. In this case, you have provided multiple files for different problems. While practicing multiple problems is great, when submitting a solution for a specific problem, make sure the code is isolated to that problem.

Note: The solution in problem1.py is for "Search a 2D Matrix" and problem3.py is for "Search in a Sorted Array of Unknown Size". These are different problems and are not relevant to the current evaluation.

Keep up the good work!

@super30admin
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Your solution for the rotated sorted array search (problem2.py) is well-implemented. It correctly uses binary search to achieve O(log n) time complexity and O(1) space complexity. The logic is clear and follows the standard approach.

Strengths:

  • You correctly handle the two cases: when the left half is sorted and when the right half is sorted.
  • You efficiently narrow down the search space based on whether the target is within the sorted range.
  • The code is concise and easy to follow.

Areas for improvement:

  • In the else branch (right sorted half), the condition can be written more intuitively as:
    if nums[mid] <= target <= nums[high]:
    This makes it clearer that you are checking if the target is within the sorted right half.
  • Although not necessary, you might consider adding comments to explain the two cases (left sorted and right sorted) for better readability.

Additionally, please ensure that you are submitting only the relevant solution for the problem. In this case, problem2.py is the correct file for the rotated sorted array problem. The other files (problem1.py and problem3.py) are for different problems and should not be included unless required.

Overall, your solution is correct and efficient. Keep up the good work!

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