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1773. 统计匹配检索规则的物品数量

题目描述

给你一个数组 items ，其中 items[i] = [typei, colori, namei] ，描述第 i 件物品的类型、颜色以及名称。

另给你一条由两个字符串 ruleKey 和 ruleValue 表示的检索规则。

如果第 i 件物品能满足下述条件之一，则认为该物品与给定的检索规则 匹配 ：

  • ruleKey == "type" 且 ruleValue == typei 。
  • ruleKey == "color" 且 ruleValue == colori 。
  • ruleKey == "name" 且 ruleValue == namei 。

统计并返回 匹配检索规则的物品数量 。

 

示例 1：

输入：items = [["phone","blue","pixel"],["computer","silver","lenovo"],["phone","gold","iphone"]], ruleKey = "color", ruleValue = "silver"
输出：1
解释：只有一件物品匹配检索规则，这件物品是 ["computer","silver","lenovo"] 。

示例 2：

输入：items = [["phone","blue","pixel"],["computer","silver","phone"],["phone","gold","iphone"]], ruleKey = "type", ruleValue = "phone"
输出：2
解释：只有两件物品匹配检索规则，这两件物品分别是 ["phone","blue","pixel"] 和 ["phone","gold","iphone"] 。注意，["computer","silver","phone"] 未匹配检索规则。

 

提示：

  • 1 <= items.length <= 104
  • 1 <= typei.length, colori.length, namei.length, ruleValue.length <= 10
  • ruleKey 等于 "type"、"color" 或 "name"
  • 所有字符串仅由小写字母组成

解法

方法一：计数模拟

由于 ruleKey 只可能是 "type"、"color" 或 "name"，我们可以直接取 ruleKey 的第一个字符来确定 item 的下标 $i$。然后遍历 items 数组，统计 item[i] == ruleValue 的个数即可。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为 items 的长度。

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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)

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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;
    }
}

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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; });
    }
};

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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
}

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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);
}

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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
    }
}

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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;
}

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