feat: solve Container With Most Water (#11)#226
Conversation
Co-Authored-By: Claude Sonnet 4.5 <noreply@anthropic.com>
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Claude finished @pertrai1's task in 1m 35s —— View job Code Review Complete ✅This is an excellent solution! The implementation is clean, efficient, and demonstrates strong understanding of the two-pointer pattern. Here's my detailed review: ✅ 1. Algorithm CorrectnessVerdict: Correct The solution correctly implements the optimal two-pointer approach:
Mental trace verification (Example 1):
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| Category | Rating | Notes |
|---|---|---|
| Correctness | 5/5 | Algorithm is correct; one test comment needs fixing |
| Complexity | 5/5 | Optimal O(n) time, O(1) space |
| Code Quality | 5/5 | Clean, readable, well-structured |
| Test Coverage | 5/5 | Comprehensive edge cases via TDD |
| Documentation | 5/5 | Excellent POST_MORTEM and self-review questions |
Overall: This is exemplary work! The TDD approach clearly paid off with zero bugs and comprehensive test coverage. The post-mortem analysis demonstrates deep understanding of the algorithm and pattern.
Only action item: Fix the comment in container-with-most-water.test.ts:19-20 to correctly explain why the answer is 6.
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💡 Educational Note: Two-Pointer Initialization let left = 0;
let right = height.length - 1;This is a classic opposite-ends two-pointer pattern. Starting with the widest possible container ensures we explore all potential solutions. As we move inward, we strategically eliminate suboptimal choices based on the greedy insight. Time Complexity: This setup allows us to examine all meaningful containers in O(n) time instead of O(n²) with nested loops. |
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💡 Educational Note: Area Calculation const width = right - left;
const currentHeight = Math.min(height[left], height[right]);
const area = width * currentHeight;The key insight here is that the container's height is always limited by the shorter of the two lines. This is why we use Formula: This is the fundamental constraint that drives our greedy algorithm. |
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💡 Educational Note: Greedy Pointer Movement if (height[left] < height[right]) {
left++;
} else {
right--;
}This is the core greedy insight that makes the O(n) solution work: Why move the shorter pointer?
Proof of correctness: By always moving the shorter pointer, we never skip the optimal solution. Any container we "skip" by moving the shorter pointer would have a smaller area than what we already calculated (same or shorter height, but narrower width). Pattern: This greedy choice is safe and eliminates the need to check all O(n²) pairs! |
✅ Test ResultsTests have been executed for this pull request. 📊 Coverage Summary
Check the workflow logs for detailed test output. |
📝 WalkthroughWalkthroughA new LeetCode problem solution is introduced for "Container With Most Water" ( Changes
Estimated code review effort🎯 3 (Moderate) | ⏱️ ~20 minutes Poem
🚥 Pre-merge checks | ✅ 2 | ❌ 1❌ Failed checks (1 warning)
✅ Passed checks (2 passed)
✏️ Tip: You can configure your own custom pre-merge checks in the settings. ✨ Finishing touches
🧪 Generate unit tests (beta)
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🤖 Gemini Code Review (Model: This is an excellent solution for the "Container With Most Water" problem, demonstrating a solid understanding of algorithmic patterns and problem-solving best practices. The accompanying 1. Pattern IdentificationThis solution primarily uses the Two Pointers algorithmic pattern, specifically the "Opposite Ends" variant where two pointers start at the extremes of an array and move towards each other. It also embodies a Greedy Algorithm approach, as each step makes a locally optimal choice (moving the shorter pointer) that leads to a globally optimal solution. 2. Complexity Verification
3. Key InsightThe key insight, or "aha!" moment, for this problem lies in understanding the greedy strategy for moving the pointers.
This greedy strategy ensures that we explore configurations that have the best chance of increasing the area, without missing the optimal solution. 4. Edge Case AnalysisThe solution correctly handles the following edge cases, as demonstrated by the test suite and the logic:
The current solution appears robust and handles all standard edge cases correctly. The TDD approach mentioned in the 5. Learning Points
6. Code QualityThe code quality is excellent:
No significant improvements are needed for code quality; it's already very good. 7. Alternative Approaches
The Mentor's Final ThoughtsThis is an outstanding submission. The solution is optimal, correctly implemented, and the self-reflection in |
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🤖 Gemini Code Review (Model: This is an excellent solution for the "Container With Most Water" problem, demonstrating a strong understanding of algorithmic patterns and best practices. The accompanying 1. Pattern IdentificationThis solution primarily uses the Two Pointers (Opposite Ends) pattern. It also embodies a Greedy Algorithm approach, as it makes locally optimal choices (moving the shorter pointer) in the hope of finding a globally optimal solution (maximum area). 2. Complexity Verification
3. Key InsightThe key insight, often the "aha!" moment for this problem, is realizing that to maximize the area, you should always move the pointer associated with the shorter line inward. Here's why:
This greedy strategy ensures that you explore potential containers that could yield a larger area, without missing the optimal solution. 4. Edge Case AnalysisThe solution and its test suite demonstrate excellent consideration for edge cases:
No critical edge cases appear to be missed by the current solution or the test suite. 5. Learning Points
6. Code QualityThe code quality is excellent:
7. Alternative Approaches
Overall, this is a top-tier submission. The solution is correct, optimal, well-tested, and the |
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Actionable comments posted: 1
🤖 Fix all issues with AI agents
In
`@leetcode/medium/0011-container-with-most-water/container-with-most-water.test.ts`:
- Around line 18-21: Update the misleading test comment in the test case that
calls maxArea([1,2,3,4,5]) — the expected value 6 is correct but the comment
incorrectly attributes it to indices 0 and 4; change the comment to indicate the
correct best pair (e.g., indices 1 and 4: min(2,5) × 3 = 6 or indices 2 and 4:
min(3,5) × 2 = 6) so the comment matches the expected assertion in the test that
calls maxArea.
🧹 Nitpick comments (1)
leetcode/medium/0011-container-with-most-water/container-with-most-water.ts (1)
1-21: Add complexity analysis and algorithmic pattern documentation in code comments.Per coding guidelines, LeetCode solutions should document both time/space complexity in Big O notation and the algorithmic pattern used. The solution is clean and correct but missing these comments.
📝 Proposed addition
+/** + * 11. Container With Most Water + * Pattern: Two Pointers (Opposite Ends) + Greedy + * + * Start with the widest container and move the pointer at the shorter + * height inward — the shorter side limits area, so only replacing it + * can yield a larger container. + * + * Time: O(n) — each pointer moves at most n steps total + * Space: O(1) — only constant extra variables + */ export function maxArea(height: number[]): number {Based on learnings: "Document the algorithmic pattern used in solution code comments" and "Always document both time and space complexity in Big O notation with clear reasoning."
| it('should handle increasing heights [1,2,3,4,5]', () => { | ||
| // Best container is 0 to 4: min(1,5) × 4 = 1 × 4 = 4 | ||
| expect(maxArea([1, 2, 3, 4, 5])).toBe(6); | ||
| }); |
There was a problem hiding this comment.
Misleading comment — the best container is not at indices 0 and 4.
The expected result of 6 is correct, but the comment claims it comes from min(1,5) × 4 = 4. The actual best pair is at indices 1 and 4: min(2,5) × 3 = 6 (or equivalently indices 2 and 4: min(3,5) × 2 = 6).
📝 Proposed fix
it('should handle increasing heights [1,2,3,4,5]', () => {
- // Best container is 0 to 4: min(1,5) × 4 = 1 × 4 = 4
+ // Best container is indices 1 to 4: min(2,5) × 3 = 2 × 3 = 6
expect(maxArea([1, 2, 3, 4, 5])).toBe(6);
});📝 Committable suggestion
‼️ IMPORTANT
Carefully review the code before committing. Ensure that it accurately replaces the highlighted code, contains no missing lines, and has no issues with indentation. Thoroughly test & benchmark the code to ensure it meets the requirements.
| it('should handle increasing heights [1,2,3,4,5]', () => { | |
| // Best container is 0 to 4: min(1,5) × 4 = 1 × 4 = 4 | |
| expect(maxArea([1, 2, 3, 4, 5])).toBe(6); | |
| }); | |
| it('should handle increasing heights [1,2,3,4,5]', () => { | |
| // Best container is indices 1 to 4: min(2,5) × 3 = 2 × 3 = 6 | |
| expect(maxArea([1, 2, 3, 4, 5])).toBe(6); | |
| }); |
🤖 Prompt for AI Agents
In
`@leetcode/medium/0011-container-with-most-water/container-with-most-water.test.ts`
around lines 18 - 21, Update the misleading test comment in the test case that
calls maxArea([1,2,3,4,5]) — the expected value 6 is correct but the comment
incorrectly attributes it to indices 0 and 4; change the comment to indicate the
correct best pair (e.g., indices 1 and 4: min(2,5) × 3 = 6 or indices 2 and 4:
min(3,5) × 2 = 6) so the comment matches the expected assertion in the test that
calls maxArea.
📅 Spaced Repetition Reviews Scheduled!Great job solving #11 - Container With Most Water! 🎉 To help you retain this knowledge long-term, I've scheduled 5 review sessions using spaced repetition:
What to expect:
🧠 Why Spaced Repetition?Research shows that reviewing material at increasing intervals dramatically improves retention. This adaptive scheduling ensures optimal spacing based on when you actually complete each review! Check |
1 participant
Problem
Container With Most Water — Medium
Approach
Two pointers starting from opposite ends, moving inward strategically. Move the pointer at the shorter height to maximize the chance of finding a larger container area.
Complexity
Pattern
Two Pointers (Opposite Ends) + Greedy Algorithm
TDD Summary
🧠 Self-Review Questions
Answer
The container's area is limited by `min(height[left], height[right]) × width`. Moving the taller pointer inward decreases width while keeping the height limited by the shorter side, guaranteeing a smaller or equal area. Moving the shorter pointer gives us a chance to find a taller line that could compensate for the reduced width.Answer
No. We must examine at least n elements to determine which two form the maximum container, so O(n) is optimal. This two-pointer approach achieves that theoretical lower bound.Files
README.md— Problem statementcontainer-with-most-water.ts— Solutioncontainer-with-most-water.test.ts— Test suite (5 test cases)POST_MORTEM.md— Analysis and retrospectiveSummary by CodeRabbit
Release Notes
New Features
Documentation
Tests
Total problems solved: 175 (Medium: 79)