LeetCode 1773. Count Items Matching a Rule



You are given an array items, where each items[i] = [typei, colori, namei] describes the type, color, and name of the ith item. You are also given a rule represented by two strings, ruleKey and ruleValue.

The ith item is said to match the rule if one of the following is true:

  • ruleKey == "type" and ruleValue == typei.
  • ruleKey == "color" and ruleValue == colori.
  • ruleKey == "name" and ruleValue == namei.

Return the number of items that match the given rule.

Example 1:

Input: items = [["phone","blue","pixel"],["computer","silver","lenovo"],["phone","gold","iphone"]], ruleKey = "color", ruleValue = "silver"
Output: 1
Explanation: There is only one item matching the given rule, which is ["computer","silver","lenovo"].

Example 2:

Input: items = [["phone","blue","pixel"],["computer","silver","phone"],["phone","gold","iphone"]], ruleKey = "type", ruleValue = "phone"
Output: 2
Explanation: There are only two items matching the given rule, which are ["phone","blue","pixel"] and ["phone","gold","iphone"]. Note that the item ["computer","silver","phone"] does not match.


  • 1 <= items.length <= 104
  • 1 <= typei.length, colori.length, namei.length, ruleValue.length <= 10
  • ruleKey is equal to either "type""color", or "name".
  • All strings consist only of lowercase letters.


Just count how many elements have the demanding rule key and value.

Python Solution

class Solution:
    def countMatches(self, items: List[List[str]], ruleKey: str, ruleValue: str) -> int:
        rule_mapping = {
            "type": 0,
            "color": 1,
            "name": 2            
        count = 0
        for item in items:
            if item[rule_mapping[ruleKey]] == ruleValue:
                count += 1
        return count
  • Time Complexity: O(N).
  • Space Complexity: O(1).

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