[LeetCode] 1090. Largest Values From Labels 標簽中的最大價值

We have a set of items: the i-th item has value values[i] and label labels[i].

Then, we choose a subset S of these items, such that:

• |S| <= num_wanted
• For every label L, the number of items in S with label L is <= use_limit.

Return the largest possible sum of the subset S.

Example 1:

Input: values = [5,4,3,2,1], labels = [1,1,2,2,3], `num_wanted` = 3, use_limit = 1
Output: 9
Explanation: The subset chosen is the first, third, and fifth item.

Example 2:

Input: values = [5,4,3,2,1], labels = [1,3,3,3,2], `num_wanted` = 3, use_limit = 2
Output: 12
Explanation: The subset chosen is the first, second, and third item.

Example 3:

Input: values = [9,8,8,7,6], labels = [0,0,0,1,1], `num_wanted` = 3, use_limit = 1
Output: 16
Explanation: The subset chosen is the first and fourth item.

Example 4:

Input: values = [9,8,8,7,6], labels = [0,0,0,1,1], `num_wanted` = 3, use_limit = 2
Output: 24
Explanation: The subset chosen is the first, second, and fourth item.

Note:

1. 1 <= values.length == labels.length <= 20000
2. 0 <= values[i], labels[i] <= 20000
3. 1 <= num_wanted, use_limit <= values.length

這道題說是給了一堆物品，每個物品有不同的價值和標簽，分別放在 values 和 labels 數組中，現在讓選不超過 num_wanted 個物品，且每個標簽類別的物品不超過 use_limit，問能得到的最大價值是多少。說實話這道題博主研究了好久才弄懂題意，而且主要是看例子分析出來的，看了看踩比贊多，估計許多人跟博主一樣吧。這道題可以用貪婪算法來做，因為需要盡可能的選價值高的物品，但同時要兼顧到物品的標簽種類。所以可以將價值和標簽種類組成一個 pair 對兒，放到一個優先隊列中，這樣就可以按照價值從高到低進行排列了。同時，由於每個種類的物品不能超過 use_limit 個，所以需要統計每個種類被使用了多少次，可以用一個 HashMap 來建立標簽和其使用次數之間的映射。先遍歷一遍所有物品，將價值和標簽組成 pair 對兒加入優先隊列中。然后進行循環，條件是 num_wanted 大於0，且隊列不為空，此時取出隊頂元素，將其標簽映射值加1，若此時仍小於 use_limit，說明當前物品可以入選，將其價值加到 res 中，並且 num_wanted 自減1即可，參見代碼如下：

class Solution {
public:
    int largestValsFromLabels(vector<int>& values, vector<int>& labels, int num_wanted, int use_limit) {
        int res = 0, n = values.size();
        priority_queue<pair<int, int>> pq;
        unordered_map<int, int> useMap;
        for (int i = 0; i < n; ++i) {
            pq.push({values[i], labels[i]});
        }
        while (num_wanted > 0 && !pq.empty()) {
            int value = pq.top().first, label = pq.top().second; pq.pop();
            if (++useMap[label] <= use_limit) {
                res += value;
                --num_wanted;
            }
        }
        return res;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1090

參考資料：

https://leetcode.com/problems/largest-values-from-labels/

https://leetcode.com/problems/largest-values-from-labels/discuss/312720/C%2B%2B-Greedy

https://leetcode.com/problems/largest-values-from-labels/discuss/313011/Question-Explanation-and-Simple-Solution-or-Java-or-100

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