Vishal 写道: > Hash table should give you O(1) insertion and search complexity; which > is what we need, right? > There is no constraint on space complexity, I believe. > > On 9/24/07, * daizisheng* <[EMAIL PROTECTED] > <mailto:[EMAIL PROTECTED]>> wrote: > > > the problem is you need a hash table to maintain all the keywords,:) > because they do not have to be a single characters, or you can > store them in > array, but then you need binary search,:) > > Vishal 写道: > > How about keeping two pointers - startp and endp. > > Keep a count of frequencies of keywords between startp and endp, > both > > inclusive. We can use an array / hash table for this. > > Now, the minimum length substring has to start with a keyword > and has > > to end with a keyword. > > > > 1. Initially startp=0 and endp=1. L = infinity > > 2. Increment endp till you encounter a keyword or it exceeds the > total > > length. Update frequencies. Check if you have all the keywords. > (This > > can be done in O(1) using a bitmap or similar). If it exceeds the > > total length, QUIT. > > 3. If the str(startp,endp) contains all keywords and length < L, > save > > values of startp and endp. > > 4. Now, increment startp, till you get a keyword. If the > > str(startp,endp) still contains all keywords, update saved values of > > startp and endp. > > 5. Repeat step 4 till str(startp,endp) does NOT contain all > keywords. > > 6. Goto step 2. > > > > The final values of startp and endp should give you the minimum > summary. > > Since both values go from 0 to N-1, its O(N). > > > > ~Vishal > > > > On 9/24/07, *daizisheng* < [EMAIL PROTECTED] > <mailto:[EMAIL PROTECTED]> > > <mailto:[EMAIL PROTECTED] <mailto:[EMAIL PROTECTED]>>> wrote: > > > > > > I think hash method is ok, at lease in expectation way it's > O(n) > > why not use it? it's very effeciently > > > > I think there should be some worst case O(n) algorithm, but > it will be > > more complex and not as effecient as the above one in practise > > > > > > Sticker 写道: > > > Declare: this question is originally from Google. The > original form > > > is: given a document, how to find a shortest summary > containing all > > > the key words. After translated, it will be: given a > sequence, > > how to > > > find a shortest substring that contains all the items > required. > > > Example: a sequence "abaccdefgacel" and a set of key words > "a", "c", > > > "d" and "e". The shortest substring contianing all of the key > > words is > > > "accde". Find one of such shortest substrings. In the original > > > question, there is time complexity boundary O(N), where N > is the > > > length of the sequence. > > > > > > To me, this problem is rather questionable. So far my solution > > > requires a hash table that gives no conflict and can locate a > > key word > > > in O(1) time. Otherwise the time complexity will > definitely be > > related > > > to the number of key words. > > > > > > Does anyone have idea for a O(N) algorithm to solve this > problem? > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > yes, that's we need. but seems the starter of this thread who did not like hash,:)
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