Complete Two-Pointers-1

[dhruvil15]

No description provided.

[super30admin]

Arrange Colors (Problem34.java)

Strengths:

• The solution correctly implements the Dutch National Flag algorithm.
• The code is clean, readable, and well-commented.
• The time and space complexities are optimal.

Areas for Improvement:

• While the code is correct, it is identical to the reference solution. This might indicate that the student has directly used the standard solution without demonstrating a personal understanding. However, since the problem is well-known and the solution is standard, this is acceptable as long as the student understands the logic.
• The student could consider adding more detailed comments explaining why the algorithm works, especially the conditions for incrementing mid after swapping with low (because the element swapped from low to mid is always 0 or 1, and since low has been traversed, mid can safely advance). This would show deeper understanding.

Overall, the solution is excellent.

VERDICT: PASS

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3 sum (Problem35.java)

Strengths:

• The solution correctly implements the two-pointer technique, which is efficient for this problem.
• The code handles duplicates effectively by skipping over identical elements.
• The comments are clear and explain the approach well.

Areas for improvement:

• The condition if (nums[0] > 0) break; is placed inside the loop but checks nums[0] instead of nums[i]. Since the array is sorted, if nums[i] is greater than 0, then all subsequent numbers will also be positive, and no triplet can sum to zero. Therefore, this condition should be if (nums[i] > 0) break; to break early when appropriate.
• The condition if (i != 0 && nums[i] == nums[i-1]) continue; is correct for skipping duplicates, but it's good practice to note that this avoids duplicate triplets starting with the same value.

Overall, the solution is excellent with only a minor correction needed.

VERDICT: PASS

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Container With Most Water (Problem36.java)

Your solution is excellent. You have implemented the optimal two-pointer approach with O(n) time complexity and O(1) space complexity. The code is well-structured and readable. Your comments explain the approach clearly.

One minor suggestion: you can calculate the area immediately after determining the minimum height without moving the pointer first. Actually, your current code moves the pointer and then uses the stored h and w. This is correct because you stored w before moving. However, some developers might find it confusing. You can consider calculating the area before moving the pointer. For example:

    while (low < high) {
        int w = high - low;
        int h = Math.min(height[low], height[high]);
        area = Math.max(area, w * h);
        if (height[low] < height[high]) {
            low++;
        } else {
            high--;
        }
    }

This version might be slightly more intuitive because it calculates the area with the current pointers and then moves. But both are correct.

Overall, great job!

VERDICT: PASS