Description
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 <= values.length == labels.length <= 20000
- 0 <= values[i], labels[i] <= 20000
- 1 <= num_wanted, use_limit <= values.length
分析
题目的意思是：给定values和labels，然后给出限制条件num_wanted和use_limit,num_wanted限制的是values的数量，use_limit限制的是labels的数量。
我参考了一下别人的思路，对values，labels进行从大到小排序，然后遍历，用freq记录遍历过的label的频率，如果不超过use_limit，则可以更新res值，num_wanted减去1，直到num_wanted减少为0为止。
这样就能达到目的了
代码
class Solution:
def largestValsFromLabels(self, values: List[int], labels: List[int], num_wanted: int, use_limit: int) -> int:
res=0
freq=defaultdict(int)
for value,label in sorted(zip(values,labels),reverse=True):
if(freq[label]<use_limit):
res+=value
num_wanted-=1
if(num_wanted==0):
break
freq[label]+=1
return res