How do you solve Count Items Matching a Rule on LeetCode?

Count Items Matching a Rule (LeetCode #1773) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Directly mapping ruleKey to indices improves efficiency.

What is the brute force solution for Count Items Matching a Rule?

The brute force approach for Count Items Matching a Rule is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach involves checking each item against the rule one by one. This is straightforward but can be inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Items Matching a Rule?

Count Items Matching a Rule uses the following concepts: Array, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Items Matching a Rule?

Clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as empty arrays or non-matching rules. Explain your thought process as you code; it shows your understanding.

What are common mistakes when solving Count Items Matching a Rule?

Not considering all possible values for ruleKey. Confusing the order of attributes in the items array.

#1773

Count Items Matching a Rule

Easy

ArrayStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses the same logic as brute force but is more efficient by directly indexing into the item based on the ruleKey, which reduces unnecessary checks.

⚙️

Algorithm

5 steps

1Step 1: Create a mapping for ruleKey to the corresponding index in the item array.
2Step 2: Use the index to directly access the relevant attribute of each item.
3Step 3: Initialize a counter to zero.
4Step 4: Iterate through each item and check if the attribute at the mapped index matches ruleValue.
5Step 5: Increment the counter for each match and return it.

solution.py7 lines

1def countMatches(items, ruleKey, ruleValue):
2    keyIndex = {'type': 0, 'color': 1, 'name': 2}
3    count = 0
4    for item in items:
5        if item[keyIndex[ruleKey]] == ruleValue:
6            count += 1
7    return count

ℹ

Complexity note: The time complexity remains O(n) as we still iterate through the items. The space complexity is O(1) since we use a fixed amount of extra space.

1Directly mapping ruleKey to indices improves efficiency.
2Understanding the structure of the data helps in optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.