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1773. Count Items Matching a Rule
Description
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"
andruleValue == typei
.ruleKey == "color"
andruleValue == colori
.ruleKey == "name"
andruleValue == 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.
Constraints:
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.
Solutions
-
class Solution { public int countMatches(List<List<String>> items, String ruleKey, String ruleValue) { int i = ruleKey.charAt(0) == 't' ? 0 : (ruleKey.charAt(0) == 'c' ? 1 : 2); int ans = 0; for (var v : items) { if (v.get(i).equals(ruleValue)) { ++ans; } } return ans; } }
-
class Solution { public: int countMatches(vector<vector<string>>& items, string ruleKey, string ruleValue) { int i = ruleKey[0] == 't' ? 0 : (ruleKey[0] == 'c' ? 1 : 2); return count_if(items.begin(), items.end(), [&](auto& v) { return v[i] == ruleValue; }); } };
-
class Solution: def countMatches(self, items: List[List[str]], ruleKey: str, ruleValue: str) -> int: i = 0 if ruleKey[0] == 't' else (1 if ruleKey[0] == 'c' else 2) return sum(v[i] == ruleValue for v in items)
-
func countMatches(items [][]string, ruleKey string, ruleValue string) (ans int) { i := map[byte]int{'t': 0, 'c': 1, 'n': 2}[ruleKey[0]] for _, v := range items { if v[i] == ruleValue { ans++ } } return }
-
function countMatches(items: string[][], ruleKey: string, ruleValue: string): number { const key = ruleKey === 'type' ? 0 : ruleKey === 'color' ? 1 : 2; return items.reduce((r, v) => r + (v[key] === ruleValue ? 1 : 0), 0); }
-
impl Solution { pub fn count_matches(items: Vec<Vec<String>>, rule_key: String, rule_value: String) -> i32 { let key = if rule_key == "type" { 0 } else if rule_key == "color" { 1 } else { 2 }; items .iter() .filter(|v| v[key] == rule_value) .count() as i32 } }