Блог пользователя AdHocMan

Автор AdHocMan, 4 года назад, По-английски

Today I was trying to solve the problem CF R1600 — Camp Schedule. I realized that I will need the longest prefix of a string which is also a suffix of that string. I searched it on google and found that it is called 'Prefix Function'. I tried to understand this concept from the CP Algorithms article, but I am too dumb to understand it.

Though I got the code from there what I needed, I can't understand how it works. Can anyone please help me to understand this concept? (with some examples?)

Prefix Function Code

Note: I am very new to this type of algorithms.

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4 года назад, # |
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Go to youtube and type "tushar roy kmp algorithm"

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4 года назад, # |
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Lets get the basic things clear. $$$\pi[i] - \pi[i-1] \le 1$$$, because each new character, can add only one more match.

In my amazing drawing, The substrings covered by the same color are the same, and we have processed upto $$$i$$$.

First we check, if we can extend red. That requires $$$C_2 = C_3$$$. If that is not true, we need to check with the next lower match. The next lower match, is cyan. to check if cyan works, we have to check $$$C_1 = C_3$$$.

Now if cyan doesnt work either, we want such a matching inside cyan. Which is $$$\pi[\pi[\pi[i]]]$$$. And we continuously keep on going down, to the next largest, until the next character matches. This is $$$O(n)$$$, because checking characters takes $$$O(1)$$$. Each character can add most one to $$$\pi$$$. and each, time we go back, the matching decreases by at least $$$1$$$. So we can deduce that the number of times the while loop runs cannot be more than $$$O(n)$$$.

Also keep in mind, is that the substrings may intersect, but it doesn't really matter. I didnt draw intersecting substrings, to not make the diagram too messy.

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    4 года назад, # ^ |
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    while(j > 0 && s[j] != s[i]) j = pi[j - 1];

    Isn't this that part which is searching the next lower match?

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      4 года назад, # ^ |
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      Well, read the loop like this.

      while there is some matching left before it
               if Characters at the positions are equal : break
               else : go to next smaller matching.