Completed Binary-Search-2 problems

[shaurya22c]

No description provided.

[super30admin]

Find First and Last Position of Element in Sorted Array (first_last_position_sorted_array.py)

Excellent work! Your solution is correct, efficient, and well-documented. Here are some observations:

Strengths:

• Clear and accurate logic for both binary searches
• Proper handling of edge cases (bounds checking with mid == 0 and mid == len(nums) - 1)
• Good code organization with separate helper functions
• Helpful comments explaining the approach
• Correct time and space complexity analysis

Minor suggestions for improvement:

1. Optimization opportunity: Like the reference solution, you could optimize the second binary search by starting from the first position found. This would be: last_position = get_last_position(nums, target, first_position, len(nums)-1). This doesn't change the asymptotic complexity but can be slightly faster in practice.

2. Code style: Consider using early returns or more Pythonic patterns. For example, you could combine the bounds check with the comparison: if mid == 0 or nums[mid-1] != target: (comparing with target instead of nums[mid]) - this is slightly more efficient as it avoids one comparison.

3. Documentation: Your docstring is good, but you could add the time and space complexity directly in the docstring for better visibility.

Overall, this is a high-quality solution that demonstrates a solid understanding of binary search and its application to this problem.

VERDICT: PASS

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Find Minimum in Rotated Sorted Array (min_in_rotated_sorted_array.py)

Your solution is excellent and correctly solves the problem in O(log n) time with O(1) space. The approach matches the reference solution well. A few minor points:

1. Comment Error: You have a copy-paste error in your comments. The comment for the else branch says "check if left side is sorted; if so eliminate that" but it should say "check if right side is sorted; if so eliminate that" since in the else branch, you're eliminating the right side.

2. Good Practices:

   • Your docstring is well-written and explains the approach clearly.
   • The base case check at the beginning of the loop is a nice optimization.
   • The bounds checking for mid (checking mid == 0 and mid == high) is correctly handled.

3. Minor Optimization: You could potentially simplify by just checking nums[mid] < nums[mid-1] after the sorted check, but your current approach is fine and handles edge cases well.

Overall, this is a solid solution that demonstrates good understanding of binary search on rotated sorted arrays.

VERDICT: PASS

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Find Peak Element (find_peak_element.py)

E student's solution is essentially identical to the reference solution in terms of logic and approach. Let me evaluate it carefully.

The student's solution:

1. Uses binary search with the same logic as the reference
2. Checks if mid is a peak (greater than both neighbors, with boundary checks)
3. Moves low or high pointer based on comparison with next element
4. Returns -1 if no peak found (which shouldn't happen given the problem constraints)

Let me check for correctness issues:

The condition nums[mid + 1] > nums[mid] - this could potentially access nums[len(nums)] when mid == len(nums) - 1. Wait, let me think again. If mid == len(nums) - 1, then nums[mid + 1] would be out of bounds. But actually, if mid == len(nums) - 1, then the peak check would have caught it (since nums[mid] > nums[mid + 1] would be vacuously true when mid == len(nums) - 1... wait no, the peak check requires nums[mid] > nums[mid + 1] only when mid != len(nums) - 1. Let me re-read.

The peak check is:

if (mid == 0 or nums[mid] > nums[mid - 1]) and (mid == len(nums) -1 or nums[mid] > nums[mid + 1]):
    return mid

So if mid == len(nums) - 1, the second condition is automatically true (short-circuit), and we only need mid == 0 or nums[mid] > nums[mid - 1]. If mid == len(nums) - 1 and mid != 0, then we need nums[mid] > nums[mid - 1]. If this is true, we return mid. If not, we go to the next check.

The next check is if nums[mid + 1] > nums[mid]: - this would access nums[len(nums)] which is out of bounds!

Wait, but this is the same logic as the reference solution. Let me check the reference again:

else if (mid < n - 1 && nums[mid + 1] > nums[mid])
    low = mid + 1;
else
    high = mid - 1;

The reference has mid < n - 1 check. The student's solution doesn't have this check explicitly, but the peak check would have already returned if mid == n - 1 and nums[mid] > nums[mid - 1]. So if we reach the nums[mid + 1] > nums[mid] check, it means either:

1. mid == n - 1 and nums[mid] <= nums[mid - 1] (so we didn't return)
2. mid != n - 1 and we didn't return

In case 1, nums[mid + 1] is out of bounds. This is a potential bug!

Wait, let me think more carefully. If mid == n - 1:

• Peak check: (mid == 0 or nums[mid] > nums[mid - 1]) and (mid == n - 1 or nums[mid] > nums[mid + 1])
• Second part is always true
• First part: if mid == 0 (only when n=1), true. Otherwise need nums[mid] > nums[mid - 1].
• If nums[mid] > nums[mid - 1], we return mid.
• If nums[mid] <= nums[mid - 1], we don't return.

Then we check nums[mid + 1] > nums[mid] - this is out of bounds!

So the student's solution has a potential bug when mid == n - 1 and nums[mid] <= nums[mid - 1]. However, this scenario might not actually occur in practice because of how binary search works, but it's still a potential issue.

Actually, wait. Let me think about this more carefully. When mid == n - 1:

• If nums[mid] > nums[mid - 1], we return mid (peak found)
• If nums[mid] <= nums[mid - 1], we go to the else branch and set high = mid - 1

But the code first checks if nums[mid + 1] > nums[mid]: which would be out of bounds. In Python, this would raise an IndexError.

Hmm, but actually, in the reference solution, the same issue exists but it's guarded by mid < n - 1. So the reference solution is safer.

Let me trace through an example: nums = [1, 2, 1, 3, 5, 6, 4]

• low = 0, high = 6
• mid = 3, nums[3] = 3
• Peak check: nums[3] > nums[2] (3 > 1, true) and nums[3] > nums[4] (3 > 5, false). Not a peak.
• nums[mid+1] > nums[mid]: nums[4] > nums[3] (5 > 3, true). low = 4.
• low = 4, high = 6
• mid = 5, nums[5] = 6
• Peak check: nums[5] > nums[4] (6 > 5, true) and nums[5] > nums[6] (6 > 4, true). Peak! Return 5.

OK that works. Let me try another case where mid == n-1:
nums = [3, 2, 1]

• low = 0, high = 2
• mid = 1, nums[1] = 2
• Peak check: nums[1] > nums[0] (2 > 3, false). Not a peak.
• nums[mid+1] > nums[mid]: nums[2] > nums[1] (1 > 2, false). Else branch: high = 0.
• low = 0, high = 0
• mid = 0, nums[0] = 3
• Peak check: nums[0] > nums[-1] (vacuously true since mid == 0) and nums[0] > nums[1] (3 > 2, true). Peak! Return 0.

OK that works. Let me try nums = [1, 2, 3]:

• low = 0, high = 2
• mid = 1, nums[1] = 2
• Peak check: nums[1] > nums[0] (2 > 1, true) and nums[1] > nums[2] (2 > 3, false). Not a peak.
• nums[mid+1] > nums[mid]: nums[2] > nums[1] (3 > 2, true). low = 2.
• low = 2, high = 2
• mid = 2, nums[2] = 3
• Peak check: nums[2] > nums[1] (3 > 2, true) and (mid == n-1, true). Peak! Return 2.

OK that works. Let me try nums = [3, 2, 1] but with different starting points... Actually, I think the issue is when mid == n-1 and nums[mid] <= nums[mid-1]. Let me construct such a case.

nums = [5, 4, 3, 2, 1]

• low = 0, high = 4
• mid = 2, nums[2] = 3
• Peak check: nums[2] > nums[1] (3 > 4, false). Not a peak.
• nums[mid+1] > nums[mid]: nums[3] > nums[2] (2 > 3, false). Else: high = 1.
• low = 0, high = 1
• mid = 0, nums[0] = 5
• Peak check: (mid == 0, true) and nums[0] > nums[1] (5 > 4, true). Peak! Return 0.

OK that works. Let me try nums = [2, 1]:

• low = 0, high = 1
• mid = 0, nums[0] = 2
• Peak check: (mid == 0, true) and nums[0] > nums[1] (2 > 1, true). Peak! Return 0.

OK. Let me try nums = [1, 2]:

• low = 0, high = 1
• mid = 0, nums[0] = 1
• Peak check: (mid == 0, true) and nums[0] > nums[1] (1 > 2, false). Not a peak.
• nums[mid+1] > nums[mid]: nums[1] > nums[0] (2 > 1, true). low = 1.
• low = 1, high = 1
• mid = 1, nums[1] = 2
• Peak check: nums[1] > nums[0] (2 > 1, true) and (mid == n-1, true). Peak! Return 1.

OK. Let me try to construct a case where mid == n-1 and nums[mid] <= nums[mid-1]:

nums = [3, 2, 1, 0]

• low = 0, high =

VERDICT: NEEDS_IMPROVEMENT