检查一个集合是否包含在另一个集合中的时间复杂度 [英] Time complexity of checking whether a set is contained in another set
问题描述
我正在尝试实现查找包含模式 char
的给定字符串 s
的最短子字符串的示例.我的代码运行良好,但我的目标是达到 O(N)
的时间复杂度,其中 N 是 s
的长度.这是我的代码;
I am trying to implement the example of finding the shortest substring of a given string s
containing the pattern char
. My code is working fine, but my goal is to attain the time complexity of O(N)
where N is length of s
.
Here is my code;
def shortest_subtstring(s,char):
#smallest substring containing char.use sliding window
start=0
d=defaultdict(int)
minimum=9999
for i in range(len(s)):
d[s[i]]+=1
#check whether all the characters from char has been visited.
while set(char).issubset(set([j for j in d if d[j]>0])):
#if yes, can we make it shorter
length=i-start+1
minimum=min(length,minimum)
if length==minimum:
s1=s[start:i+1]
d[s[start]]-=1
start+=1
return (minimum,s1)
我的问题是在线;
while set(char).issubset(set([j for j in d if d[j]>0]))
每次我检查char
的所有字符串是否都保存在我的字典中,使用is.subset
的思想.我可以知道如何在我的代码中找到这一步的时间复杂度吗?它是 O(1)
吗,这对于检查元素是否存在于集合中是正确的.否则时间复杂度会远大于O(N)
.感谢帮助.
Each time I am checking whether all strings of char
are saved in my dictionary or not, using the idea of is.subset
. May I know how can I find the time complexity of this step in my code? Is it O(1)
, which is true for checking wether an element exists in a set or not. Otherwise, the time complexity will be much greater than O(N)
. Help is appreciated.
推荐答案
Per docs s.issubset(t)
相当于 s <= t
意味着在操作期间它将测试 s 中的每个元素是否是在 t.
Per docs s.issubset(t)
is the equivalent of s <= t
meaning that during the operation it will test if every element in s is in t.
最佳场景:如果 s 是 t 的第一个元素 -> O(1)
Best scenario: if s is the first element of t -> O(1)
最坏情况:如果 s 在 t 的最后一个元素 -> O(len(t))
Worst case scenario: if s in the last element of t -> O(len(t))
那是针对 isubset 的.对于列表理解:
That is for isubset. For the list comprehension :
j for j in d
是 O(n) 获取每个键
j for j in d
is O(n) for getting each key
if d[j]>0
是 O(n) 比较字典 d
if d[j]>0
is O(n) for comparing each value of dictionary d
此处您可以找到更多信息.
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