[Leet Code] Count Items Matching a Rule

  • ruleKey == "type" and ruleValue == typei.
  • ruleKey == "color" and ruleValue == colori.
  • ruleKey == "name" and ruleValue == namei.
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"].
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.
class Solution(object):
def countMatches(self, items, ruleKey, ruleValue):
"""
:type items: List[List[str]]
:type ruleKey: str
:type ruleValue: str
:rtype: int
"""

dictionary = {"type": 0, "color": 1, "name" : 2}
indexPos = dictionary[ruleKey]
answer = 0

for i,v in enumerate(items):
if v[indexPos] == ruleValue:
answer += 1

return answer

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