This is the first post in my Codeforces blog! I plan to use this as a place where I can write down new algorithms I learned, problems that I found interesting, or stupid mistakes I made during contests.
Today, the topic will be the Z Algorithm, which I learned as a result of failing to solve Problem B of Beta Round 93 (http://codeforces.me/contest/126/problem/B). There are some other solutions like binary search + hashing, but this one is quite nice. Anyway, first, a description of the algorithm and why it works; it is simple and makes a lot of sense (as all good algorithms are).
Algorithm
Given a string S of length n, the Z Algorithm produces an array Z where Z[i] is the length of the longest substring starting from S[i] which is also a prefix of S, i.e. the maximum k such that S[j] = S[i + j] for all 0 ≤ j < k. Note that Z[i] = 0 means that S[0] ≠ S[i]. For easier terminology, we will refer to substrings which are also a prefix as prefix-substrings.
The algorithm relies on a single, crucial invariant. As we iterate over the letters in the string (index i from 1 to n - 1), we maintain an interval [L, R] which is the interval with maximum R such that 1 ≤ L ≤ i ≤ R and S[L...R] is a prefix-substring (if no such interval exists, just let L = R = - 1). For i = 1, we can simply compute L and R by comparing S[0...] to S[1...]. Moreover, we also get Z[1] during this.
Now suppose we have the correct interval [L, R] for i - 1 and all of the Z values up to i - 1. We will compute Z[i] and the new [L, R] by the following steps:
- If i > R, then there does not exist a prefix-substring of S that starts before i and ends at or after i. If such a substring existed, [L, R] would have been the interval for that substring rather than its current value. Thus we "reset" and compute a new [L, R] by comparing S[0...] to S[i...] and get Z[i] at the same time (Z[i] = R - L + 1).
- Otherwise, i ≤ R, so the current [L, R] extends at least to i. Let k = i - L. We know that Z[i] ≥ min(Z[k], R - i + 1) because S[i...] matches S[k...] for at least R - i + 1 characters (they are in the [L, R] interval which we know to be a prefix-substring). Now we have a few more cases to consider.
- If Z[k] < R - i + 1, then there is no longer prefix-substring starting at S[i] (or else Z[k] would be larger), meaning Z[i] = Z[k] and [L, R] stays the same. The latter is true because [L, R] only changes if there is a prefix-substring starting at S[i] that extends beyond R, which we know is not the case here.
- If Z[k] ≥ R - i + 1, then it is possible for S[i...] to match S[0...] for more than R - i + 1 characters (i.e. past position R). Thus we need to update [L, R] by setting L = i and matching from S[R + 1] forward to obtain the new R. Again, we get Z[i] during this.
The process computes all of the Z values in a single pass over the string, so we're done. Correctness is inherent in the algorithm and is pretty intuitively clear.
Analysis
We claim that the algorithm runs in O(n) time, and the argument is straightforward. We never compare characters at positions less than R, and every time we match a character R increases by one, so there are at most n comparisons there. Lastly, we can only mismatch once for each i (it causes R to stop increasing), so that's another at most n comparisons, giving O(n) total.
Code
Simple and short. Note that the optimization L = R = i is used when S[0] ≠ S[i] (it doesn't affect the algorithm since at the next iteration i > R regardless).
int L = 0, R = 0;
for (int i = 1; i < n; i++) {
if (i > R) {
L = R = i;
while (R < n && s[R-L] == s[R]) R++;
z[i] = R-L; R--;
} else {
int k = i-L;
if (z[k] < R-i+1) z[i] = z[k];
else {
L = i;
while (R < n && s[R-L] == s[R]) R++;
z[i] = R-L; R--;
}
}
}Application
One application of the Z Algorithm is for the standard string matching problem of finding matches for a pattern T of length m in a string S of length n. We can do this in O(n + m) time by using the Z Algorithm on the string T Φ S (that is, concatenating T, Φ, and S) where Φ is a character that matches nothing. The indices i with Z[i] = m correspond to matches of T in S.
Lastly, to solve Problem B of Beta Round 93, we simply compute Z for the given string S, then iterate from i to n - 1. If Z[i] = n - i then we know the suffix from S[i] is a prefix, and if the largest Z value we've seen so far is at least n - i, then we know some string inside also matches that prefix. That gives the result.
int maxz = 0, res = 0;
for (int i = 1; i < n; i++) {
if (z[i] == n-i && maxz >= n-i) { res = n-i; break; }
maxz = max(maxz, z[i]);
}
Hmmm, like some sort of... catalog?
Interesting, I will check it out. It didn't exist 11 years ago
Here I post some tutorial link.
If you know any new tutorial blog post please comment , I will add them .
Thanks paladin8 for his awesome tutorial.
Here also by using Z-algorithm 6780044
That site is awesome, if someone could translate it to english, i think that it would be the best algorithm reference ever.
Here is the english version of this site: http://e-maxx-eng.github.io
Also the pdf ebook containing all the tutorials and algo in English can be downloaded from: https://www.dropbox.com/s/j8z839pzamxxqw2/e-maxx_algo.ru.en.pdf?dl=1
Non Russians can use this site to read the tutorials in English: http://e-maxx-eng.github.io
Or if one visit that site using google chrome, then s/he can automaticaly translate in English with the help of google chrome browser.
Also I have a pdf file in English containing all posts on this site. This pdf ebook containing all the tutorials and algo in English can be downloaded from: https://www.dropbox.com/s/j8z839pzamxxqw2/e-maxx_algo.ru.en.pdf?dl=1
import java.io.*;
import java.util.*;
public class D {
public static void main(String[] args) throws IOException{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String s = in.readLine();
char[] c = s.toCharArray();
int n = c.length;
int[] memo = new int[n];
/* memo[i] will store the length of the longest prefix of s that matches the tail of s1...si */
for (int i=1; i<n; i++) {
int j=i;
while (j>0 && c[i] != c[memo[j-1]]) j = memo[j-1];
memo[i] = (j>0) ? memo[j-1]+1 : 0;
}
int max = 0;
for (int i=1; i<n-1; i++) max = Math.max(max, memo[i]);
/* max = Maximum internal match to prefix */
j = memo[n-1];
while (j>max) j = memo[j-1];
System.out.println((j==0)? "Just a legend" : s.substring(0, j));
}
}
may be this will work
for i = 1 to np[i + z[i]] = max(p[i + z[i]], z[i])and one more travesing from n to 1. doing
p[i] = max(p[i+1] - 1, p[i])"and one more travesing from n to 1. doing p[i] = max(p[i+1] — 1, p[i])"
why is this necessary ?
A portion of necroposting.
You can solve 2 problems: 1) Create a sample string with given Z-function over an infinite alphabet. Just follow Z-function generation algorithm and store all the equalities in DST. Next, check the result. 2) Create a sample string with given prefix-function over an infinite alphabet. Just follow prefix function generation algorithm and store all the equalities in DST. Next, check the result.
(in Russian: http://contest.samara.ru/ru/problemset/735/)
Z to KMP —
for i=1;i<n$$$P[i+Z[i]-1] = max(P[i+Z[i]-1],z[i])$$$KMP to Z —
for i=1;i<n$$$Z[i-P[i]+1] = max(Z[i-P[i]+1],P[i])$$$I dont know if your Z<-> prefix transformation is mutual. I've written a post to try to construct z from kmp, http://liruqi.info/post/62511983824/kmp-z (In Chinese). Beside, you may check my code here, https://raw.github.com/liruqi/topcoder/master/HackerRank/SaveHumanity.cpp
Can someone give both transformation implementations? If possible with some explanations. Thanking you in advance. Learn and let learn.
No advantages =)
But if we compare Z-function and Prefix-function then:
1) "monotony"
If z[i] = x, it means substrings [0..j) [i..i+j) are equals for all j from 0 to x.
If p[i] = x, it means [0..j) = (i-j..i] for all j = x, p[x], p[p[x]] and so on.
2) LCP (Largest Common Prefix)
Z-function in fact calculates LCP[0,j] for all j. It can be used for not only substring searching.
I also have two examples of problems which, I hope, show advantages Z-function over Prefix-function.
1) Determine number (No.) of the string in its suffix array in O(n).
2) Determine number (amount) of substrings, which have at least two occuarances in the string in O(n2) (of course, you can solve it in O(n) using suffix tree).
Your comment is valuable. But you have some error, I will show them, to prevent others from misunderstanding. (Someone reopen the post, so I read your comment)
1) "monotony"
If z[i] = x, it means substrings [0..j) [i..i+j) are equals for all j from 0 to x-1.
If p[i] = x, it means [0..j) = (i-j..i] for all j = x, p[x], p[p[x]] and so on.
Thanks. Fixed.
KMP requires only the smaller string to be known to allow for computing the prefix function. It is a stream based algorithm,you can work even if characters are given you one by one for the main string. This may be an advantage in certain scenarios.
Hi, I think that assumption is wrong, take this test case:
Is it right or I misunderstood what he said? :o
(sorry for editing, I am new to Markdown :p)
Amazing tutorial. Thanks to you and e-maxx I have learned a new algorithm today :D
Thanks for this amazing tutorial. Keep writing! :)
Very nice explanation of the algorithm , thanks !
One of my favorite posts. Quick question: the algorithm seems to ignore
z[0]; shouldn't it be the case thatz[0] = n? Thanks.By the algorithm we consider that Z[0] is underfined
you can add it end of your code, if necessary :)
ghabool dari nemikhaym z[0] ?
can you please elaborate on what S[i] stores .
Sis the string.S[i]is the symbol at positioni. IfS = "abcd", thenS[0] = 'a',S[1] = 'b'and so on.Just a silly optimization. In the line
while (R < n && s[R-L] == s[R]) R++;, isn't theR < nredundant for strings? Becauses[R-L] == s[R]will be false when we hit the\0character. This could also work if we are using a sequence of numbers instead of strings (we have to add a number at the end that is different than all others).Not everyone uses C++ char arrays to store strings
that's the point. I'm saying that you can use sequences of any type provided that you insert a sentinel element at the end of your sequence. Then you can save the
R < n.afair, std::string allows s[s.size()] and returns 0
it's guaranteed only for C++11 or later
Guys, this link at youtube might come in handy for a deeper insight and intuition into the algorithm. https://www.youtube.com/watch?v=MFK0WYeVEag
Thank you for this awesome tutorial.
this link contains animation for Z-Algorithm, it may be helpful.
awesome animation...... Thank you eagle93
Thanks for the link! :)
can any one tell me how to solve Problem B of Beta Round 93 with KMP by building Longest common prefix/suffix array :D
If you want a much more detailed explanation with examples and complexity analysis, this link will be quite helpful!
Now we are computing the value at
i = k. The values L and R are as shown in the image. Now in the caseZ[k] ≥ R - i + 1, sincer + 1andr' + 1values are not same(if they were to be same, the interval would have beenLandR + 1instead of the current values), why do we need to check the values afterr + 1?Helpful.
Can anyone answer my question above?
Can the longest sub-string and prefix overlap?
The O(n) complexity made me interested. Such a beautiful algorithm. Thank you :)
Does anyone have more problems related to Z Algorithm?I am appreciate if anyone could offer some.
try this one
another one
explanation by my friend,
consider this example
s: A B A A B A B A A B B A B A B
z: 0 0 1 3 0 5 0 1 2 0 0 3 0 2 0
here's how we compute it
first, Z[0] = 0
our rightmost border is 0, starting from index 0
onto index 1. since we passed our border 0, we will compute it "normally" (brute force). we get 0 length. now our rightmost border is still not far enough (still 0), and its index is 1
index 2. compute by brute force, we get 1. now our right border is index 2, starting from index 2 (S[2...2] = S[0...0])
index 3. again we compute by brute force, we get 3. now out right border is index 5, starting from index 3 (S[3...5] = S[0...2])
index 4. this time 4 <= right border, so we use past computations. since 3 is the index of start, we put Z[1] into Z[4]. notice that if this makes the range of equality for 4 exceed the right border, we block it by the right border (so for example, if Z[1] = 3, then the range is 4...6 (length 3), and 6 exceeds the right border (5) so we change it to Z[4] = 2 instead of Z[4] = 3). so now, Z[4] = 0. we try extending by brute force, but it doesn't change our farthest is still starting from index 3 till index 5
index 5. using (3, 5), since 5 <= right border, we can use Z[2] for Z[5]. Z[2] is 1, and it just fits for Z[5] with 1, not to exceed our right border. we then try to extend by brute force, and we get Z[5] = 5. now our start index is 5 and border is 9
index 6. using (5, 9), we can use Z[1] for Z[6]. this results in Z[6] = 0. we then try to extend by brute force, but it keeps Z[6] = 0.
index 7. we use Z[2] for Z[7], getting Z[7] starting with value 1. we can try to extend it by brute force, but it will keep at 1.
index 8. we use Z[3] for Z[8], making Z[8] = 3. however, notice that Z[8] = 3 makes it exceed the right border, 9. therefor we block Z[8] by 2, making it Z[8] = 2. we then try to extend, but it doesn't change.
index 9. we use Z[4] for Z[9], making Z[9] = 0. we can try to extend, but it won't work
index 10. now we don't use past values, since we passed our border 9. so we try brute force, and get Z[10] = 0
index 11. again we can't use past values, so we do brute force. we get Z[11] = 3, and now our border is 13 with start 11.
index 12. using (11, 13), we do Z[12] = Z[1] = 0. brute force from here changes nothing.
index 13. using (11, 13), we do Z[13] = Z[2] = 1. it also doesn't exceed our border 13. we then try brute force to extend, and get Z[13] = 2, with new border 14
index 14. using (13, 14), we do Z[14] = Z[1] = 0. we cannot extend anymore.
O(n) reasoning: the first computation (using past values) is O(1) per index. also observe, that when we manage to extend by brute force, then we also extend our right border. since our right border only increases from 0 to n — 1, it changes O(n) times overall, O(n) computation
i thought like for i<=R case Z[i]>=Z[k] where k=i-L but here is a counter example
s: A A A B A A C
z: 0 2 1 0 2 1 0
as we finish index 4, our right border is 5, so we get the pair (4, 5) now for Z[5] so we use Z[1] for it Z[5] = 2 but Z[5] cannot be 2 or more, as you can see from the string the border means we currently have no information about equalities beyond the border.
now you can get the feel for this inequality z[i]>=min(z[k],R-i+1)
moral of this algorithm : try to use the past computational information by keeping track of the [L,R] (largest R basically and it's corresponding L ) values and using them and continuing the brute force comparision from new position to extend it to as maximum as possible , proof of time complexity is well explained by my friend .
the code mentioned by the author is correct but it is doing redundant work for the else case where Z[k]>=R-i+1 instead as we already know Z[i]>=R-i+1 we can just start our R from R+1 itself so the code now changes to this (we need to add this line R=R+1 that's it . )
int L = 0, R = 0; for (int i = 1; i < n; i++) { if (i > R) { L = R = i; while (R < n && s[R-L] == s[R]) R++; z[i] = R-L; R--; }else{ int k = i-L; if (z[k] < R-i+1) z[i] = z[k]; else { L = i; R = R + 1 while (R < n && s[R-L] == s[R]) R++; z[i] = R-L; R--; } }}
this should make more sense now
Here's a question which uses nice DP and zFunction .. https://codeforces.me/gym/102465/problem/K Basic one .. but a learning one
If needed u can have a look at my code : https://pastebin.com/Nrrb1QR0