This algorithm finds all occurrences of a pattern in a text in linear time. Let length of text be n and of pattern be m, then total time taken is O(m + n) with linear space complexity. Now we can see that both time and space complexity is same as KMP algorithm but this algorithm is Simpler to understand.
In this algorithm, we construct a Z array.
What is Z Array?
For a string str[0..n-1], Z array is of same length as string. An element Z[i] of Z array stores length of the longest substring starting from str[i] which is also a prefix of str[0..n-1]. The first entry of Z array is meaning less as complete string is always prefix of itself.
Example:
Index 0 1 2 3 4 5 6 7 8 9 10 11
Text a a b c a a b x a a a z
Z values X 1 0 0 3 1 0 0 2 2 1 0
More Examples:
str = "aaaaaa"
Z[] = {x, 5, 4, 3, 2, 1}
str = "aabaacd"
Z[] = {x, 1, 0, 2, 1, 0, 0}
str = "abababab"
Z[] = {x, 0, 6, 0, 4, 0, 2, 0}
How is Z array helpful in Searching Pattern in Linear time?
The idea is to concatenate pattern and text, and create a string “P$T” where P is pattern, $ is a special character should not be present in pattern and text, and T is text. Build the Z array for concatenated string. In Z array, if Z value at any point is equal to pattern length, then pattern is present at that point.
Example:
Pattern P = "aab", Text T = "baabaa"
The concatenated string is = "aab$baabaa"
Z array for above concatenated string is {x, 1, 0, 0, 0,
3, 1, 0, 2, 1}.
Since length of pattern is 3, the value 3 in Z array
indicates presence of pattern. How to construct Z array?
A Simple Solution is to run two nested loops, the outer loop goes to every index and the inner loop finds length of the longest prefix that matches the substring starting at the current index. The time complexity of this solution is O(n2).
We can construct Z array in linear time.
The idea is 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 –
1) 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
str[0..] to str[i..] and get Z[i] (= R-L+1).
2) If i <= R then let K = i-L, now Z[i] >= min(Z[K], R-i+1) because
str[i..] matches with str[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
str[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 str[R] onwards and
get new R then we will update interval [L,R] and calculate Z[i] (=R-L+1).For better understanding of above step by step procedure please check this animation – http://www.utdallas.edu/~besp/demo/John2010/z-algorithm.htm
The algorithm runs in linear time because we never compare character less than R and with matching we increase R by one so there are at most T comparisons. In mismatch case, mismatch happen only once for each i (because of which R stops), that’s another at most T comparison making overall linear complexity.
Below is the implementation of Z algorithm for pattern searching.
C++
#include<iostream>
using namespace std;
void getZarr(string str, int Z[]);
void search(string text, string pattern)
{
string concat = pattern + "$" + text;
int l = concat.length();
int Z[l];
getZarr(concat, Z);
for (int i = 0; i < l; ++i)
{
if (Z[i] == pattern.length())
cout << "Pattern found at index "
<< i - pattern.length() -1 << endl;
}
}
void getZarr(string str, int Z[])
{
int n = str.length();
int L, R, k;
L = R = 0;
for (int i = 1; i < n; ++i)
{
if (i > R)
{
L = R = i;
while (R<n && str[R-L] == str[R])
R++;
Z[i] = R-L;
R--;
}
else
{
k = i-L;
if (Z[k] < R-i+1)
Z[i] = Z[k];
else
{
L = i;
while (R<n && str[R-L] == str[R])
R++;
Z[i] = R-L;
R--;
}
}
}
}
int main()
{
string text = "GEEKS FOR GEEKS";
string pattern = "GEEK";
search(text, pattern);
return 0;
}
|
Java
class GFG {
public static void search(String text, String pattern)
{
String concat = pattern + "$" + text;
int l = concat.length();
int Z[] = new int[l];
getZarr(concat, Z);
for(int i = 0; i < l; ++i){
if(Z[i] == pattern.length()){
System.out.println("Pattern found at index "
+ (i - pattern.length() - 1));
}
}
}
private static void getZarr(String str, int[] Z) {
int n = str.length();
int L = 0, R = 0;
for(int i = 1; i < n; ++i) {
if(i > R){
L = R = i;
while(R < n && str.charAt(R - L) == str.charAt(R))
R++;
Z[i] = R - L;
R--;
}
else{
int k = i - L;
if(Z[k] < R - i + 1)
Z[i] = Z[k];
else{
L = i;
while(R < n && str.charAt(R - L) == str.charAt(R))
R++;
Z[i] = R - L;
R--;
}
}
}
}
public static void main(String[] args)
{
String text = "GEEKS FOR GEEKS";
String pattern = "GEEK";
search(text, pattern);
}
}
|
Python3
def getZarr(string, z):
n = len(string)
l, r, k = 0, 0, 0
for i in range(1, n):
if i > r:
l, r = i, i
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
else:
k = i - l
if z[k] < r - i + 1:
z[i] = z[k]
else:
l = i
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
def search(text, pattern):
concat = pattern + "$" + text
l = len(concat)
z = [0] * l
getZarr(concat, z)
for i in range(l):
if z[i] == len(pattern):
print("Pattern found at index",
i - len(pattern) - 1)
if __name__ == "__main__":
text = "GEEKS FOR GEEKS"
pattern = "GEEK"
search(text, pattern)
|
C#
using System;
class GFG
{
public static void search(string text,
string pattern)
{
string concat = pattern + "$" + text;
int l = concat.Length;
int[] Z = new int[l];
getZarr(concat, Z);
for (int i = 0; i < l; ++i)
{
if (Z[i] == pattern.Length)
{
Console.WriteLine("Pattern found at index " +
(i - pattern.Length - 1));
}
}
}
private static void getZarr(string str,
int[] Z)
{
int n = str.Length;
int L = 0, R = 0;
for (int i = 1; i < n; ++i)
{
if (i > R)
{
L = R = i;
while (R < n && str[R - L] == str[R])
{
R++;
}
Z[i] = R - L;
R--;
}
else
{
int k = i - L;
if (Z[k] < R - i + 1)
{
Z[i] = Z[k];
}
else
{
L = i;
while (R < n && str[R - L] == str[R])
{
R++;
}
Z[i] = R - L;
R--;
}
}
}
}
public static void Main(string[] args)
{
string text = "GEEKS FOR GEEKS";
string pattern = "GEEK";
search(text, pattern);
}
}
|
Javascript
<script>
function search(text,pattern)
{
let concat = pattern + "$" + text;
let l = concat.length;
let Z = new Array(l);
getZarr(concat, Z);
for(let i = 0; i < l; ++i){
if(Z[i] == pattern.length){
document.write("Pattern found at index "
+ (i - pattern.length - 1)+"<br>");
}
}
}
function getZarr(str,Z)
{
let n = str.length;
let L = 0, R = 0;
for(let i = 1; i < n; ++i) {
if(i > R){
L = R = i;
while(R < n && str[R - L] == str[R])
R++;
Z[i] = R - L;
R--;
}
else{
let k = i - L;
if(Z[k] < R - i + 1)
Z[i] = Z[k];
else{
L = i;
while(R < n && str[R - L] == str[R])
R++;
Z[i] = R - L;
R--;
}
}
}
}
let text = "GEEKS FOR GEEKS";
let pattern = "GEEK";
search(text, pattern);
</script>
|
Output:
Pattern found at index 0
Pattern found at index 10
Time Complexity: O(m+n), where m is length of pattern and n is length of text.
Auxiliary Space: O(m+n)
This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above