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" 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.

 

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

Solution 1

Python Code
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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)

Java Code
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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;
    }
}

C++ Code
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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; });
    }
};

Go Code
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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
}

TypeScript Code
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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);
}

Rust Code
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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
    }
}

C Code
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int countMatches(char*** items, int itemsSize, int* itemsColSize, char* ruleKey, char* ruleValue) {
    int k = strcmp(ruleKey, "type") == 0 ? 0 : strcmp(ruleKey, "color") == 0 ? 1
                                                                             : 2;
    int res = 0;
    for (int i = 0; i < itemsSize; i++) {
        if (strcmp(items[i][k], ruleValue) == 0) {
            res++;
        }
    }
    return res;
}