※ 引述《Leon (Achilles)》之铭言：
: ※ 引述《RockLee (Now of all times)》之铭言：
: : 原始网址:
: : http://www.careercup.com/question?id=15381730
: : 题目:
: : Given a document and a query of K words,
: : how do you find the smallest window that covers all the words at least once
: : in that document?
: : (given you know the inverted lists of all K words, that is, for each word,
: : you have a list of all its occurrrences).
: : Could someone propose an algorithm in O(n)?
: : 假设 occurrrences lists 是 sorted, 用 TreeMap 应可在 O(N*logK) 解决,
: : 一开始将每个 occurrrences list 最小的丢入 TreeMap 就可得到第一个 window,
: : (key = index in the document, value = the occurrrences list it belonged & its
: : index in that occurrrences list),
: : 之后将 TreeMap 最小的移除并加入同一个 occurrrences list 的下一个即可移动
: : window,
: : 直到任一个 occurrrences list 耗完.
: : 但要在 O(N) 的时间内解决就想不出来了...
: : 不知道板上有没有人有什么 idea?
: Sorry, I can't understand your writing.
: Also, the link seems to be wrong...
: Can you describe it, step by step.
: For example, there are only 3 words, (a,b,c)
: how are you going to find the window, for the document
: [b b a c b b b b c b b a] ?
嗯, 刚刚想了一下.
这个题目, 是 KMP 的变形.
难的地方在于, 你怎么想到正确的方向
(听起来有点废话, 哈哈..)
这里我采用之前的定义.
For a sentance, we are given the occurance index.
Take above example, say, only 3 words {a, b, c}.
The sentance is [b b a c b a].
The the occurance index will be
a = {3, 6}.
b = {1, 2, 5}.
c = {4 }.
在开始之前, 我们先看几个非常简单的作法.
1. Naive: Arbitary choose 2 position from [1, N],
then check the condition.
2. Choose one position as the left boundary, then,
use the max-min of the rest set the decide the right boundary.
For example, if I choose 2 as my left boundary,
then I need to choose the right boundary by the max of
d(2,3) and d(2,4) which cover set {a} and set {c}
The result window is [2, 4].
And the complexity should be O(N* K).
接下来, 就是有趣的地方了.
3. Follow the previous algorithm, inspired by KMP.
When we start from 3, actually we don't need to compare all K groups.
Because from previous step, we know [2] covers {b} for sure.
And [3,4] covers {a,c} for sure.
In this case, we only need to check the condition for set {b} !
which is d(3,5).
Then the computational complexity is O(N).
I'll leave the detail for you.