作者:
Leon (Achilles)
2013-03-08 06:33:45※ 引述《Leon (Achilles)》之銘言：
: ※ 引述《RockLee (Now of all times)》之銘言：
: : 原始網址:
: : http://www.careercup.com/question?id=15381730
: : 題目:
: : 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)?
: : 假設 occurrrences lists 是 sorted, 用 TreeMap 應可在 O(N*logK) 解決,
: : 一開始將每個 occurrrences list 最小的丟入 TreeMap 就可得到第一個 window,
: : (key = index in the document, value = the occurrrences list it belonged & its
: : index in that occurrrences list),
: : 之後將 TreeMap 最小的移除並加入同一個 occurrrences list 的下一個即可移動
: : window,
: : 直到任一個 occurrrences list 耗完.
: : 但要在 O(N) 的時間內解決就想不出來了...
: : 不知道板上有沒有人有什麼 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, 剛剛想了一下.
這個題目, 是 KMP 的變形.
難的地方在於, 你怎麼想到正確的方向
(聽起來有點廢話, 哈哈..)
這裡我採用之前的定義.
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 }.
在開始之前, 我們先看幾個非常簡單的作法.
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).
接下來, 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.