LeetCode解题分享:1090. Largest Values From Labels

本文介绍了一种算法,用于从具有特定值和标签的项目集合中,选择满足数量限制和标签使用限制条件下的最大价值子集。通过实例演示了算法的运行过程,并提供了Python实现代码。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Problem

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
解题思路

   根据题意可以知道,我们需要在每个对应的labellabellabel中选择不超过use_limituse\_limituse_limit个数字,总的数字数目不超过num_wantednum\_wantednum_wanted,求这些数字的可以组成的最大的和。那么我们只需要对每个labellabellabel对应的数据进行排序,挑选出最大的 use_limituse\_limituse_limit 个数字,将这些数字放进同一个数组中,进行排序,在挑选出最大的num_wantednum\_wantednum_wanted个数字进行求和即可。由于这里需要进行排序挑选,因此也可以使用heap进行求解,原理是一致的。

   代码如下:

class Solution:
    def largestValsFromLabels(self, values: List[int], labels: List[int], num_wanted: int, use_limit: int) -> int:
        record = collections.defaultdict(list)

        for v, l in zip(values, labels):
            record[l].append(v)

        seq = []

        for vs in record.values():
            seq += heapq.nlargest(min(len(vs), use_limit), vs)

        return sum(heapq.nlargest(min(len(seq), num_wanted), seq))

   或者:

class Solution:
    def largestValsFromLabels(self, values: List[int], labels: List[int], num_wanted: int, use_limit: int) -> int:
        record = collections.defaultdict(list)

        for v, l in zip(values, labels):
            record[l].append(v)

        seq = []

        for vs in record.values():
            seq += sorted(vs, reverse=True)[:min(len(vs), use_limit)]

        seq.sort(reverse=True)

        return sum(seq[:min(len(seq), num_wanted)])
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值