Binary Search 02 Solutions Posted

[angad-singh97]

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[super30admin]

Find First and Last Position of Element in Sorted Array (01-find-lowest-highest-index-sorted-array.cpp)

Looking at this problem, I need to evaluate a student's solution for finding the first and last position of an element in a sorted array using binary search.

Problem Understanding

The problem requires finding the starting and ending position of a target value in a sorted array. If the target is not found, return [-1, -1]. The solution must have O(log n) runtime complexity.

Student's Solution Analysis

Correctness

The student's solution correctly implements two separate binary searches:

1. First binary search to find the lowest (first) occurrence by continuing to search left when a match is found
2. Second binary search to find the highest (last) occurrence by continuing to search right when a match is found

The logic is sound:

• When nums[mid] == target, they update the result and continue searching in the appropriate direction
• When nums[mid] > target, they search left
• When nums[mid] < target, they search right

This correctly handles edge cases like empty arrays (returns {-1, -1} since loops won't execute) and targets not in the array.

Time Complexity

Both solutions achieve O(log n) time complexity through binary search. The student's approach is essentially equivalent to the reference solution, though structured differently. Both perform two binary searches, each O(log n), resulting in O(log n) overall.

Space Complexity

Both solutions use O(1) space complexity with only a constant number of variables.

Code Quality

The student's code is:

• Clean and readable with meaningful variable names (lo, hi, lowest, highest)
• Well-commented with time/space complexity noted
• Straightforward logic that's easy to follow

However, there's a minor improvement opportunity: the second binary search could start with a tighter bound (from lowest to nums.size() - 1) since we know the target exists at or after the first occurrence. This is a minor optimization but not necessary for correctness.

Efficiency

The solution is efficient. One potential micro-optimization: after finding the first occurrence, the second search could use low = lowest instead of low = 0, reducing the search space. However, this doesn't change the overall O(log n) complexity.

Feedback

Strengths:

• Correctly implements the two-pass binary search approach
• Clean, readable code with good variable naming
• Properly handles edge cases
• Achieves optimal time and space complexity
• Good comments explaining the approach

Areas for Improvement:

• Consider tightening the bounds of the second binary search by starting from lowest instead of 0 after finding the first occurrence. This is a minor optimization but demonstrates attention to optimization opportunities.

Final Verdict

PASS

The student's solution is correct, efficient, and well-structured. It achieves the required O(log n) time complexity and O(1) space complexity, matching the reference solution's approach with a clean implementation.

VERDICT: NEEDS_IMPROVEMENT

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Find Minimum in Rotated Sorted Array (02-find-minimum-rotated-sorted-array.cpp)

Strengths:

• Correct binary search implementation with proper handling of edge cases
• Good understanding of the rotated sorted array property
• Clean, readable code with helpful comments
• Self-awareness about the == case (as mentioned in comments)
• Proper use of mid = lo + (hi - lo)/2 to prevent integer overflow

Areas for Improvement:

• The minimum variable is unnecessary - you can simplify by returning nums[lo] directly when the left side is sorted
• Consider adding an early return when nums[lo] <= nums[hi] (array is not rotated), which would make the O(log n) guarantee clearer
• The comment about the == case is insightful - this is important when the search space reduces to 2 elements

Overall, this is a solid solution that demonstrates good understanding of binary search and the problem characteristics.

VERDICT: PASS

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Find Peak Element (03-find-peak-element.cpp)

Strengths:

• Clean implementation of binary search for peak finding
• Good variable naming and code organization
• Correctly handles edge cases (first and last elements)
• Proper use of comments explaining the approach
• Time and space complexity match the optimal solution

Areas for Improvement:

• The condition mid < (n - 1) in rightCheck is redundant. Since we already check mid == n - 1, if that condition is false, mid < n - 1 is guaranteed to be true. You can simplify to: bool rightCheck = mid == n - 1 || nums[mid] > nums[mid + 1];
• Consider adding a brief comment explaining why moving left when smaller than left neighbor is correct (due to the "mountain" nature of the array)

VERDICT: PASS