Completed Binary search 2 course

[sanjoli97]

No description provided.

[super30admin]

Find the First and Last Position of an Element in given Sorted Array (FindFirstAndLastPosInSortedArray.java)

Strengths:

1. Clean and well-documented code with comprehensive comments
2. Correct implementation of binary search for both first and last positions
3. Proper handling of edge cases (empty array, target not found)
4. Good use of early returns for cleaner code flow
5. Follows the same efficient approach as the reference solution

Areas for Improvement:

1. The condition nums[mid] > nums[mid - 1] in binarySearchForStart can be simplified to nums[mid - 1] != target for better clarity, as this directly checks if the previous element is not the target (which is the actual requirement).
2. Similarly, nums[mid] < nums[mid + 1] in binarySearchForLast could be nums[mid + 1] != target for consistency and clarity.
3. Consider adding comments explaining the boundary conditions more explicitly.

Overall, this is a solid solution that correctly solves the problem with optimal time and space complexity.

VERDICT: PASS

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Find the Minimum Element in a Rotated Array(sorted) (FindMinInRotatedArray.java)

Strengths:

• Clean, well-commented solution with clear explanation of the approach
• Correctly implements binary search with O(log n) time complexity
• Proper handling of edge cases through the main binary search logic
• Good variable naming and code structure

Areas for Improvement:

• The special case if (nums.length == 2) is unnecessary. The main while loop with the nums[low] <= nums[high] check already handles arrays of length 2 correctly.
• The return value return 0; at the end is misleading. While it won't be reached due to the problem constraints, returning -1 like the reference solution would be more appropriate as a fallback indicator.

Minor Suggestion:
Consider removing the 2-element special case to simplify the code. The existing logic already handles this case correctly:

• For [3,1]: low=0, high=1, nums[low] > nums[high], mid=0, nums[mid] >= nums[low] (3>=3), so low=1. Next iteration: nums[low] <= nums[high] (1<=2), returns nums[1]=1 ✓

VERDICT: PASS

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Find the Peak Element (FindPeakElement.java)

Strengths:

• Clean, well-documented code with clear comments
• Correctly implements binary search for peak finding
• Proper handling of edge cases (single element, boundaries)
• Good variable naming and code structure

Areas for Improvement:

• The condition int n = nums.length - 1; is slightly confusing - it would be clearer as int n = nums.length; since n is used as a boundary check for the last index
• The final return 0; could be misleading - while it won't be reached in a valid input, returning -1 would be more conventional for "not found" cases

Minor suggestions:

• Consider using long for mid calculation to avoid potential overflow (though for typical array sizes this isn't an issue)
• The else branch could be more explicit about the condition being checked

Overall, this is a solid solution that correctly solves the problem with optimal time complexity.

VERDICT: PASS