Binary Search - 2

[KartavyaBhatt]

No description provided.

[super30admin]

Find the First and Last Position of an Element in given Sorted Array (Problem_1.py)

Strengths:

• Well-documented code with clear explanations of the algorithm
• Correct implementation of binary search for both first and last positions
• Good use of early returns when target is not found
• The approach of starting searchLast from the first index is a smart optimization

Areas for Improvement:

• In searchLast, the return statement returns first when target is not found, but this should return -1 for consistency with the problem requirements. However, in practice, since searchLast is only called after confirming first exists, this won't cause issues.
• Consider using elif for the comparison conditions to avoid redundant checks after finding a match.

Minor Style Note:
The code would benefit from importing List from typing for type hints, though this doesn't affect functionality.

VERDICT: PASS

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Find the Minimum Element in a Rotated Array(sorted) (Problem_2.py)

Strengths:

• Correctly implements binary search for the rotated sorted array problem
• Good use of boundary checks (m == 0 and m == len(nums)-1)
• Clean and readable code with helpful comments
• Efficient O(log n) time complexity

Areas for Improvement:

• Variable names could be more descriptive (e.g., 'low', 'high', 'mid' instead of 'l', 'r', 'm')
• Could add an early termination check when low == high to avoid the final iteration
• The comment about "ignoring the right side since the mid will always be on the left" is slightly misleading - the algorithm actually ignores the left side when nums[m] < nums[r] because the right side contains the minimum

VERDICT: PASS

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Find the Peak Element (Problem_3.py)

Strengths:

• Correctly implements the binary search strategy for finding peak elements
• Good handling of boundary conditions (first and last elements)
• Clean, readable code with helpful comments
• Optimal time and space complexity

Areas for Improvement:

• The condition if m != 0 and nums[m-1] > nums[m] can be simplified. Since the peak check already handles m=0, if we reach the else clause, we know nums[m+1] > nums[m], so elif nums[m-1] > nums[m] would suffice.
• Consider adding type hints for better code documentation (e.g., nums: List[int] with proper import)

VERDICT: PASS