added all 3 files by megharaykar · Pull Request #1885 · super30admin/Two-Pointers-1

megharaykar · 2026-09-13T07:05:07Z

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super30admin · 2026-09-13T08:06:35Z

Sort Colors (Problem1.py)

EVERTS: PASS

EVERTS: swap(nums, mid, high) - this swaps the element at mid with the element at high, then decrements high. This is correct because we don't increment mid since the swapped element from high could be 0, 1, or 2 and needs to be processed.

swap(nums, mid, low) - this swaps the element at mid with the element at low, then increments both low and mid. This is correct because the element at low was either 0 or 1 (already processed), so the new element at mid is guaranteed to be 1 or 0 (already in correct position).
else case - mid increments, which is correct for when nums[mid] == 1.

The solution is correct.

Time Complexity: O(n) - same as reference solution
Space Complexity: O(1) - same as reference solution
Code Quality: The code is well-structured and readable. The student has included comments explaining the approach and time complexity. The swap method is properly defined as an instance method.

Efficiency: The solution is optimal for this problem.

One minor issue: The student uses self.swap which means the swap method needs to be called on an instance. This is fine for LeetCode since they instantiate the class, but it's worth noting that this is a Python-specific consideration.

Overall, this is a correct and efficient solution that matches the reference solution in terms of algorithm and complexity.

EVERTS: PASS

EVERtd: PASS

EVERtd:

VERDICT: PASS

3Sum (Problem2.py)

Strengths:

Your solution demonstrates a strong understanding of the "Two Sum" pattern and how to extend it to "Three Sum."
The time complexity of O(n²) is excellent and significantly better than the brute force O(n³) approach.
Using tuple(sorted(...)) to handle duplicates is a clever and Pythonic technique.
The code is concise and readable.

Areas for Improvement:

Variable naming: st is not very descriptive. Consider renaming it to target or needed_sum for clarity.
You could add a brief comment explaining the algorithm's approach for future readers.
Alternative approach: A sorting-based solution (sort the array, then use two pointers) is also O(n²) and avoids the overhead of hash sets. It's worth knowing both approaches.
Minor optimization: You could sort the array first and skip duplicate values for nums[i] to avoid unnecessary iterations, though this is a minor optimization.

VERDICT: PASS

Container With Most Water (Problem3.py)

It appears that you have not submitted a solution to this problem. The file is empty. To receive a passing evaluation, you need to:

Understand the problem: You need to find two vertical lines in the array that, together with the x-axis, form a container holding the maximum amount of water. The area is calculated as min(height[i], height[j]) * (j - i).
Implement a solution:
- A brute-force O(n²) approach would check every pair of lines (this is what the reference solution does, though it would TLE on large inputs).
- A more efficient O(n) approach uses the two-pointer technique: start with one pointer at the beginning and one at the end, calculate the area, then move the pointer pointing to the shorter line inward. This works because moving the longer line inward can never increase the area (the width decreases and the height is bounded by the shorter line).
Test your solution: Make sure your code handles edge cases like arrays of length 2, arrays with all zeros, and arrays with varying heights.

Please submit a complete solution so it can be properly evaluated.

VERDICT: NEEDS_IMPROVEMENT

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megharaykar commented Sep 13, 2026

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super30admin commented Sep 13, 2026

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megharaykar commented Sep 13, 2026

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super30admin commented Sep 13, 2026

Sort Colors (Problem1.py)

3Sum (Problem2.py)

Container With Most Water (Problem3.py)

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