Python 1773.py
class Solution:
def countMatches(self, items: List[List[str]], ruleKey: str, ruleValue: str) -> int:
result = 0
for type, color, name in items:
if ruleKey == "type" and type == ruleValue:
result += 1
elif ruleKey == "color" and color == ruleValue:
result += 1
elif ruleKey == "name" and name == ruleValue:
result += 1
return result
Difficulty
and has an acceptance rate of about 85% (300k+ submissions).
Typical patterns to solve this problem:
- Array
- String
AI Disclaimer
However, two years later I set up this website and used various AI tools to generate most of the walkthrough / explanations below.
See this dedicated page for more details.
Code Walkthrough
Overview
The provided implementation defines a Solution class with a single method countMatches. The method receives a collection of items, each represented as a list of three strings (type, color, name), together with a ruleKey and a ruleValue. Its purpose is to count how many items satisfy the rule expressed by the pair (ruleKey, ruleValue). The algorithm iterates over the input list once, evaluates the rule for each item, and increments a counter when the rule holds. Finally, the accumulated count is returned.
Key Steps
Extraction of Fields
At the start of the loop the code unpacks each inner list into three distinct variables, allowing direct reference to the semantic components of an item.
for type, color, name in items:
This unpacking eliminates the need for index-based access and makes the subsequent condition checks more readable.
Rule Matching Logic
The core decision structure consists of three mutually exclusive conditional branches, each corresponding to one possible value of ruleKey. The branches compare the appropriate field with ruleValue.
if ruleKey == "type" and type == ruleValue:
result += 1
elif ruleKey == "color" and color == ruleValue:
result += 1
elif ruleKey == "name" and name == ruleValue:
result += 1
Only the branch that matches the supplied ruleKey performs a comparison; the others are bypassed, ensuring that exactly one field is examined per iteration.
Result Accumulation
When a comparison succeeds, the method increments a running total stored in result. The increment operation is straightforward and occurs inside each conditional branch.
result += 1
After processing all items, the method returns the final count.
return result
Edge Cases Handled
The algorithm gracefully handles an empty items list. Since the loop body never executes, result remains at its initial value of zero, and the method correctly returns zero matches.
When ruleKey is supplied with a value outside the expected set ("type", "color", "name"), none of the conditional branches evaluate to true, and the counter is never incremented. This behavior effectively treats unknown rule keys as yielding zero matches without raising an exception.
The code also tolerates items that contain exactly three elements, as required by the unpacking pattern. If an inner list were shorter or longer, Python would raise a ValueError, but within the intended contract of the method-where each item is guaranteed to have three attributes-this situation does not arise.
Performance Optimizations
The implementation processes the input in a single pass, performing a constant amount of work per item. By unpacking the item tuple once and using direct variable references, it avoids repeated indexing operations that could add overhead.
The use of elif ensures that once a matching ruleKey is found, the remaining branches are not evaluated. This short-circuiting eliminates unnecessary comparisons, especially when the rule key is known early in the sequence.
Incrementing a simple integer counter (result) is an O(1) operation, and returning the final value after the loop avoids any additional data structures or intermediate collections, keeping memory usage minimal.
ELI5 - Simplified Explanation
Imagine a big box filled with many toys. Each toy has three special labels attached to it, which are its type, its color, and its name. This code acts like a tiny helper with a clipboard who wants to count how many toys match a specific sticker.
The helper starts with a count of zero. Then, it picks up every single toy from the box one by one to look at the three labels. If the helper is looking for a specific type, it checks the type label. If the helper is looking for a specific color, it checks the color label instead. If the helper is looking for a specific name, it checks the name label. Whenever the label on a toy matches the secret sticker the helper is holding, it clicks a counter to add one more to the total. After all the toys have been checked, the counting is finished.
Code Quality
Complexity Analysis
As a rule of thumb, the maximum input size for one second execution time is
- factorial: up to 10 elements
- exponential: up to 20 elements
- polynomial: a few thousand elements
- loglinear: up to 100,000 elements
- linear: a few million elements
- logarithmic: a billion elements
- constant: practically unlimited
Other Solutions
Obviously, I did that after I solved the problem myself.
In general, AlgoMonster and WalkCCC solutions are high-quality and always worth a look.
I haven't checked most of the other solutions, but they might be helpful as well.
And of course, the LeetCode forum is always a good place (or even the best) to look for alternative approaches.
(click on a column header to sort the table)
| author | |||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| author | solutions | ||||||||||
| www | AlgoMonster | C++ | Python | Java | TS | ||||||
| WalkCCC | C++ | Java | |||||||||
| leetcode.ca | C++ | Python | Java | Go | C | Rust | TS | ||||
| Progiez | C++ | Python | Java | ||||||||
| github | black-shadows | C++ | |||||||||
| charles-wangkai | Java | ||||||||||
| DimitrisJim | C | ||||||||||
| everthis | JS | ||||||||||
| fartem | Ruby | ||||||||||
| fsq | C++ | ||||||||||
| haoel | C++ | ||||||||||
| JoshCrozier | JS | ||||||||||
| Krazune | C++ | ||||||||||
| lkwq007 | Python | ||||||||||
| muhammadfarhankt | Go | ||||||||||
| suxrobGM | C# | ||||||||||
| trung-hn | Python | ||||||||||
| varunu28 | Java | ||||||||||
| YuriSpiridonov | Python | ||||||||||
| AlgoMonster | C++, Python, Java, TypeScript | ||||||||||
| WalkCCC | C++, Java | ||||||||||
| leetcode.ca | C, C++, Go, Java, Python, Rust, TypeScript | ||||||||||
| Progiez | C++, Java, Python | ||||||||||
| black-shadows | C++, C++ | ||||||||||
| charles-wangkai | Java | ||||||||||
| DimitrisJim | C | ||||||||||
| everthis | JavaScript | ||||||||||
| fartem | Ruby | ||||||||||
| fsq | C++ | ||||||||||
| haoel | C++ | ||||||||||
| JoshCrozier | JavaScript | ||||||||||
| Krazune | C++ | ||||||||||
| lkwq007 | Python | ||||||||||
| muhammadfarhankt | Go | ||||||||||
| suxrobGM | C# | ||||||||||
| trung-hn | Python | ||||||||||
| varunu28 | Java | ||||||||||
| YuriSpiridonov | Python |
Videos
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- problem 1551: Minimum Operations to Make Array Equal
Extra
- Tech Stack - my development environment and tools
- AI - how AI helped me create this website (NOT solving the problems)
- Progress - progression over time
- Quality - code quality