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completed binary search 1#2464

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shinjaneegupta wants to merge 1 commit intosuper30admin:masterfrom
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Open

completed binary search 1#2464
shinjaneegupta wants to merge 1 commit intosuper30admin:masterfrom
shinjaneegupta:master

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@shinjaneegupta
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@super30admin
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Your solution for the rotated sorted array search (rotated.py) is excellent. It correctly implements the binary search algorithm for a rotated array by determining which half is sorted and then adjusting the search range based on whether the target lies within the sorted half. The code is efficient, with O(log n) time and O(1) space complexity, and it is well-written.

Strengths:

  • You have correctly identified the core idea: check which half is sorted and then decide which half to discard.
  • The code is concise and easy to understand.
  • You have used integer division without overflow by using low + ((high - low) // 2).

Areas for improvement:

  • While your solution is correct, note that the problem is specifically about the rotated sorted array, and you have provided two additional solutions for other problems. In the future, ensure that you only submit the relevant code for the problem at hand. For this evaluation, we are only considering rotated.py.
  • The condition in the right sorted part: if nums[mid] <= target and nums[high] >= target is correct, but you can also write it as if target >= nums[mid] and target <= nums[high] for clarity. However, your current version is acceptable.

Overall, your solution is correct and efficient.

@super30admin
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Your solution for the rotated sorted array problem is correct and efficient. You have successfully implemented the binary search algorithm that checks which half is sorted and adjusts the search range accordingly. The time and space complexity are optimal.

Strengths:

  • The code is clean and easy to read.
  • The logic is correct and handles all cases.
  • The comments are helpful and explain the approach well.

Areas for Improvement:

  • In the condition for the right sorted part, consider writing the condition as if nums[mid] <= target <= nums[high] for better readability. This is a minor point, but it makes the intention clearer.
  • Ensure that you only submit the relevant code for the problem being evaluated. The inclusion of other files (infinite.py and matrix.py) might be confusing if not required.

Overall, you have done a great job. Keep up the good work!

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