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| Z Algorithm |
Z algorithm is used to find all occurrence of a pattern P in a string T similar to KMP algorithm but this is easier to understand than others. This is a linear time string matching algorithm which runs in O(m+n)~O(n) complexity, where m and n are lengths of the string T and the pattern stirng P respectively.
Algorithm :
Before learning the main algorithm, we have to know about Z array and how we can build it in a efficient way.
Z array: For every string S of length l, there is a Z array with length l, with Z[i] { i=0 to l-1 } Z[i] = the length of the longest substring starting from S[i] which is also a prefix of S i.e substring starting from S[0].
For example, for the text S = "abbcabbxaagh" , Z array will be Z = [X,0,0,0,3,0,0,0,1,1,0,0].
Note : Z[0] is not necessary to be calculated,as the whole string is always a substring of it.
How this Z array can help us in string matching?
We can achieve this by prefixing the pattern P to be matched to the text string T in which we have to search. We create the P$T string, then build the Z array of this P$T string. [ $ -> the character which is not present in the text and patttern] In the Z array, if Z[i] value at any index i is equal to pattern length, then pattern is present at that index.
Note : The $ separiting between pattern and the text is necessary to do not make the useless comparisions.As we just need the maximum matching substring of the pattern only, once the pattern is deducted, this $ character will stop the comparisions.
Example:
Pattern P = "aaba", String Text T = "abaabaa"
The concatenated string is = "aaba$abaabaab"
Z array for above concatenated string is {x, 1, 0, 0, 0, 1, 0, 4, 1, 0, 3, 1, 0}. Length of pattern is 4, the value 4 is in the Z array, we can say that patter P is present in text T at index i-l-1 where i is the index at which Z[i]=l->length of the patttern. In the example, i=7,l=4, therefore pattern is oresent at index 7-4-1=2.
How can we claculate the Z array in a efficient way?
General, brute force idea can be implemented in O(n^2) time complexity.
Efficient way is as follows:
We have to maintain an interval [L, R] which is the interval with max R such that [L,R] is prefix substring (substring which is also prefix).
Steps for maintaining this interval are as follows –
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If i > R then there is no prefix substring that starts before i and ends after i, so we reset L and R and compute new [L,R] by comparing S[0] to str[i] and get Z[i] (= R-L+1).
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If i <= R then let K = i-L, now Z[i] >= min(Z[K], R-i+1) because S[i..] matches with S[K..] for atleast R-i+1 characters (they are in [L,R] interval which we know is a prefix substring).
Now two sub cases arise – a) If Z[K] < R-i+1 then there is no prefix substring starting at S[i] (otherwise Z[K] would be larger) so Z[i] = Z[K] and interval [L,R] remains same. b) If Z[K] >= R-i+1 then it is possible to extend the [L,R] interval thus we will set L as i and start matching from S[R] onwards and get new R then we will update interval [L,R] and calculate Z[i] (=R-L+1).
Most of you might not understand this at first time.
Following link is of a video which will make you clear of everything. It's worth to watch this video and read this once for better understanding.
Tushar-Roy Z algorithm - Easily Understandable
Following is the C++ implementation of the Z algorithm finding the number of occurrences of patter P in string S
#include<bits/stdc++.h>
using namespace std;
vector<int> calculateZ(string s){
int l=0,r=0,k=0,n=s.size();
vector<int> z(n);
for(int i=1;i<n;i++){
if(i>r){
l=r=i;
while(r<n && s[r]==s[r-l])
r++;
z[i]=r-l;
r--;
}
else{
k=i-l;
if(i+z[k]<=r)
z[i]=z[k];
else{
l=i;
while(r<n && s[r]==s[r-l])
r++;
z[k]=r-l;
r--;
}
}
}
return z;
}
int main(){
string p,s;
cin >> p >> s;
string t = p + "$" + s;
int count=0;;
vector<int> z = calculateZ(t);
for(int i=0;i<z.size();i++){
if(z[i]==p.size())
count++;
}
cout << count;
return 0;
}
References :