LeetCode 11: Container with most water #113
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WalkthroughNew LeetCode solution for problem 11 (Container With Most Water) added with a two-pointer algorithm implementation. Solution class includes maxArea method that computes the maximum water container area. Corresponding parametric tests added to verify correctness across multiple test cases. Changes
Estimated code review effort🎯 2 (Simple) | ⏱️ ~10 minutes
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Pre-merge checks and finishing touches❌ Failed checks (1 warning)
✅ Passed checks (2 passed)
✨ Finishing touches
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Actionable comments posted: 1
🧹 Nitpick comments (1)
tests/test_leetcode.py (1)
472-486: Solid parametrized coverage; consider a couple more edge cases (optional).The test data here correctly exercises the standard examples for LeetCode 11 and validates the implementation as expected. If you want to harden coverage further, you could optionally add cases like:
- A strictly increasing or strictly decreasing array where the optimal container uses non‑adjacent indices (e.g.,
[1, 2, 3, 4, 5]).- An array where the maximum area is formed by two interior bars, not involving index
0orlen(height) - 1.Not required for correctness, but it would make regressions in pointer movement logic less likely.
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📒 Files selected for processing (2)
LeetCode/medium/container_with_most_water_11.py(1 hunks)tests/test_leetcode.py(1 hunks)
🧰 Additional context used
🧬 Code graph analysis (1)
tests/test_leetcode.py (1)
LeetCode/medium/container_with_most_water_11.py (2)
Solution(5-32)maxArea(6-32)
🪛 Ruff (0.14.7)
LeetCode/medium/container_with_most_water_11.py
12-12: Docstring contains ambiguous × (MULTIPLICATION SIGN). Did you mean x (LATIN SMALL LETTER X)?
(RUF002)
🔇 Additional comments (1)
LeetCode/medium/container_with_most_water_11.py (1)
22-32: Canonical two-pointer implementation looks correct and efficient.The
maxArealogic matches the standard optimal solution:
- Single pass with
left/rightpointers.- Correct area formula
(right - left) * min(height[left], height[right]).- Pointer movement based on the shorter height, ensuring O(n) time and O(1) space.
- Safe behavior even for edge lengths (returns
0whenlen(height) < 2).No changes needed here.
| """ | ||
| Pseudocode: | ||
| The problem is we dont know the max area where we we can store most of the water | ||
| So we can store from the start to end of the container but if the starting height of container is 1 then | ||
| max we can store height of 1 as more then that will spile the water regardless the end height | ||
| so if len is 9 and height is 1 then container can contain Area = length × width | ||
|
|
||
| So, what we can do it use two pointer solution to find the max area | ||
| - initialize two pointer left and right, left with value of index 0 and | ||
| right index len(arr)-1 and res which contain most largest value | ||
| - we will continue until our left and right point cross their path | ||
| - we will be moving our smallest pointer as we know smallest length contain less water | ||
| - each time we will check which one is largest previous area which stored in res | ||
| - at the end we return the res as res will be containing the largest area value we got | ||
| """ |
There was a problem hiding this comment.
Fix non-ASCII × in docstring and optionally tighten wording.
Ruff is flagging the × character in "Area = length × width" as ambiguous; this can break linting/CI. Consider switching to plain ASCII (e.g., length * width or length x width) and fixing a few typos for clarity.
For example:
- so if len is 9 and height is 1 then container can contain Area = length × width
+ so if len is 9 and height is 1 then the container can contain
+ area = length * heightYou could also rename len(arr) in the comments to len(height) to match the actual parameter name, but that’s purely cosmetic.
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| """ | |
| Pseudocode: | |
| The problem is we dont know the max area where we we can store most of the water | |
| So we can store from the start to end of the container but if the starting height of container is 1 then | |
| max we can store height of 1 as more then that will spile the water regardless the end height | |
| so if len is 9 and height is 1 then container can contain Area = length × width | |
| So, what we can do it use two pointer solution to find the max area | |
| - initialize two pointer left and right, left with value of index 0 and | |
| right index len(arr)-1 and res which contain most largest value | |
| - we will continue until our left and right point cross their path | |
| - we will be moving our smallest pointer as we know smallest length contain less water | |
| - each time we will check which one is largest previous area which stored in res | |
| - at the end we return the res as res will be containing the largest area value we got | |
| """ | |
| """ | |
| Pseudocode: | |
| The problem is we dont know the max area where we we can store most of the water | |
| So we can store from the start to end of the container but if the starting height of container is 1 then | |
| max we can store height of 1 as more then that will spile the water regardless the end height | |
| so if len is 9 and height is 1 then the container can contain | |
| area = length * height | |
| So, what we can do it use two pointer solution to find the max area | |
| - initialize two pointer left and right, left with value of index 0 and | |
| right index len(arr)-1 and res which contain most largest value | |
| - we will continue until our left and right point cross their path | |
| - we will be moving our smallest pointer as we know smallest length contain less water | |
| - each time we will check which one is largest previous area which stored in res | |
| - at the end we return the res as res will be containing the largest area value we got | |
| """ |
🧰 Tools
🪛 Ruff (0.14.7)
12-12: Docstring contains ambiguous × (MULTIPLICATION SIGN). Did you mean x (LATIN SMALL LETTER X)?
(RUF002)
🤖 Prompt for AI Agents
In LeetCode/medium/container_with_most_water_11.py around lines 7 to 21, the
docstring uses a non-ASCII multiplication sign "×" and has a few
wording/identifier mismatches that trip linters; replace "Area = length × width"
with an ASCII form like "Area = length * width" (or "length x width"), correct
typos (e.g., "dont" → "don't", "spile" → "spill"), and change references from
len(arr) to len(height) to match the function parameter; keep the two-pointer
explanation the same but tighten wording for clarity.
Summary by CodeRabbit
New Features
Tests
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