※ 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. --

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