Z algorithm (Linear time pattern searching Algorithm)

15 Jun, 2022
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.htmThe 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++
Java
Python3
C#
Javascript
// A C++ program that implements Z algorithm for pattern searching#include<iostream>using namespace std; void getZarr(string str, int Z[]); // prints all occurrences of pattern in text using Z algovoid search(string text, string pattern){    // Create concatenated string "P$T"    string concat = pattern + "$" + text;    int l = concat.length();     // Construct Z array    int Z[l];    getZarr(concat, Z);     // now looping through Z array for matching condition    for (int i = 0; i < l; ++i)    {        // if Z[i] (matched region) is equal to pattern        // length we got the pattern        if (Z[i] == pattern.length())            cout << "Pattern found at index "                << i - pattern.length() -1 << endl;    }} // Fills Z array for given string str[]void getZarr(string str, int Z[]){    int n = str.length();    int L, R, k;     // [L,R] make a window which matches with prefix of s    L = R = 0;    for (int i = 1; i < n; ++i)    {        // if i>R nothing matches so we will calculate.        // Z[i] using naive way.        if (i > R)        {            L = R = i;             // R-L = 0 in starting, so it will start            // checking from 0'th index. For example,            // for "ababab" and i = 1, the value of R            // remains 0 and Z[i] becomes 0. For string            // "aaaaaa" and i = 1, Z[i] and R become 5            while (R<n && str[R-L] == str[R])                R++;            Z[i] = R-L;            R--;        }        else        {            // k = i-L so k corresponds to number which            // matches in [L,R] interval.            k = i-L;             // if Z[k] is less than remaining interval            // then Z[i] will be equal to Z[k].            // For example, str = "ababab", i = 3, R = 5            // and L = 2            if (Z[k] < R-i+1)                Z[i] = Z[k];             // For example str = "aaaaaa" and i = 2, R is 5,            // L is 0            else            {                // else start from R and check manually                L = i;                while (R<n && str[R-L] == str[R])                    R++;                Z[i] = R-L;                R--;            }        }    }} // Driver programint main(){    string text = "GEEKS FOR GEEKS";    string pattern = "GEEK";    search(text, pattern);    return 0;}
// A Java program that implements Z algorithm for pattern// searchingclass GFG {     //  prints all occurrences of pattern in text using    // Z algo    public static void search(String text, String pattern)    {         // Create concatenated string "P$T"        String concat = pattern + "$" + text;         int l = concat.length();         int Z[] = new int[l];         // Construct Z array        getZarr(concat, Z);         // now looping through Z array for matching condition        for(int i = 0; i < l; ++i){             // if Z[i] (matched region) is equal to pattern            // length we got the pattern             if(Z[i] == pattern.length()){                System.out.println("Pattern found at index "                              + (i - pattern.length() - 1));            }        }    }     // Fills Z array for given string str[]    private static void getZarr(String str, int[] Z) {         int n = str.length();                 // [L,R] make a window which matches with        // prefix of s        int L = 0, R = 0;         for(int i = 1; i < n; ++i) {             // if i>R nothing matches so we will calculate.            // Z[i] using naive way.            if(i > R){                 L = R = i;                 // R-L = 0 in starting, so it will start                // checking from 0'th index. For example,                // for "ababab" and i = 1, the value of R                // remains 0 and Z[i] becomes 0. For string                // "aaaaaa" and i = 1, Z[i] and R become 5                 while(R < n && str.charAt(R - L) == str.charAt(R))                    R++;                                 Z[i] = R - L;                R--;             }            else{                 // k = i-L so k corresponds to number which                // matches in [L,R] interval.                int k = i - L;                 // if Z[k] is less than remaining interval                // then Z[i] will be equal to Z[k].                // For example, str = "ababab", i = 3, R = 5                // and L = 2                if(Z[k] < R - i + 1)                    Z[i] = Z[k];                 // For example str = "aaaaaa" and i = 2, R is 5,                // L is 0                else{                  // else start from R and check manually                    L = i;                    while(R < n && str.charAt(R - L) == str.charAt(R))                        R++;                                         Z[i] = R - L;                    R--;                }            }        }    }         // Driver program    public static void main(String[] args)    {        String text = "GEEKS FOR GEEKS";        String pattern = "GEEK";         search(text, pattern);    }} // This code is contributed by PavanKoli.
# Python3 program that implements Z algorithm# for pattern searching # Fills Z array for given string str[]def getZarr(string, z):    n = len(string)     # [L,R] make a window which matches    # with prefix of s    l, r, k = 0, 0, 0    for i in range(1, n):         # if i>R nothing matches so we will calculate.        # Z[i] using naive way.        if i > r:            l, r = i, i             # R-L = 0 in starting, so it will start            # checking from 0'th index. For example,            # for "ababab" and i = 1, the value of R            # remains 0 and Z[i] becomes 0. For string            # "aaaaaa" and i = 1, Z[i] and R become 5            while r < n and string[r - l] == string[r]:                r += 1            z[i] = r - l            r -= 1        else:             # k = i-L so k corresponds to number which            # matches in [L,R] interval.            k = i - l             # if Z[k] is less than remaining interval            # then Z[i] will be equal to Z[k].            # For example, str = "ababab", i = 3, R = 5            # and L = 2            if z[k] < r - i + 1:                z[i] = z[k]             # For example str = "aaaaaa" and i = 2,            # R is 5, L is 0            else:                 # else start from R and check manually                l = i                while r < n and string[r - l] == string[r]:                    r += 1                z[i] = r - l                r -= 1 # prints all occurrences of pattern# in text using Z algodef search(text, pattern):     # Create concatenated string "P$T"    concat = pattern + "$" + text    l = len(concat)     # Construct Z array    z = [0] * l    getZarr(concat, z)     # now looping through Z array for matching condition    for i in range(l):         # if Z[i] (matched region) is equal to pattern        # length we got the pattern        if z[i] == len(pattern):            print("Pattern found at index",                      i - len(pattern) - 1) # Driver Codeif __name__ == "__main__":    text = "GEEKS FOR GEEKS"    pattern = "GEEK"    search(text, pattern) # This code is contributed by# sanjeev2552
// A C# program that implements Z// algorithm for pattern searchingusing System; class GFG{ // prints all occurrences of// pattern in text using Z algopublic static void search(string text,                          string pattern){     // Create concatenated string "P$T"    string concat = pattern + "$" + text;     int l = concat.Length;     int[] Z = new int[l];     // Construct Z array    getZarr(concat, Z);     // now looping through Z array    // for matching condition    for (int i = 0; i < l; ++i)    {         // if Z[i] (matched region) is equal        // to pattern length we got the pattern         if (Z[i] == pattern.Length)        {            Console.WriteLine("Pattern found at index " +                             (i - pattern.Length - 1));        }    }} // Fills Z array for given string str[]private static void getZarr(string str,                            int[] Z){     int n = str.Length;     // [L,R] make a window which    // matches with prefix of s    int L = 0, R = 0;     for (int i = 1; i < n; ++i)    {         // if i>R nothing matches so we will        // calculate. Z[i] using naive way.        if (i > R)        {            L = R = i;             // R-L = 0 in starting, so it will start            // checking from 0'th index. For example,            // for "ababab" and i = 1, the value of R            // remains 0 and Z[i] becomes 0. For string            // "aaaaaa" and i = 1, Z[i] and R become 5            while (R < n && str[R - L] == str[R])            {                R++;            }             Z[i] = R - L;            R--;         }        else        {             // k = i-L so k corresponds to number            // which matches in [L,R] interval.            int k = i - L;             // if Z[k] is less than remaining interval            // then Z[i] will be equal to Z[k].            // For example, str = "ababab", i = 3,            // R = 5 and L = 2            if (Z[k] < R - i + 1)            {                Z[i] = Z[k];            }             // For example str = "aaaaaa" and            // i = 2, R is 5, L is 0            else            {                  // else start from R and                // check manually                L = i;                while (R < n && str[R - L] == str[R])                {                    R++;                }                 Z[i] = R - L;                R--;            }        }    }} // Driver Codepublic static void Main(string[] args){    string text = "GEEKS FOR GEEKS";    string pattern = "GEEK";     search(text, pattern);}} // This code is contributed by Shrikant13
<script> // A JavaScript program that implements Z algorithm for pattern// searching //  prints all occurrences of pattern in text using    // Z algofunction search(text,pattern){    // Create concatenated string "P$T"        let concat = pattern + "$" + text;          let l = concat.length;          let Z = new Array(l);          // Construct Z array        getZarr(concat, Z);          // now looping through Z array for matching condition        for(let i = 0; i < l; ++i){              // if Z[i] (matched region) is equal to pattern            // length we got the pattern              if(Z[i] == pattern.length){                document.write("Pattern found at index "                              + (i - pattern.length - 1)+"<br>");            }        }}  // Fills Z array for given string str[]function getZarr(str,Z){    let n = str.length;                  // [L,R] make a window which matches with        // prefix of s        let L = 0, R = 0;          for(let i = 1; i < n; ++i) {              // if i>R nothing matches so we will calculate.            // Z[i] using naive way.            if(i > R){                  L = R = i;                  // R-L = 0 in starting, so it will start                // checking from 0'th index. For example,                // for "ababab" and i = 1, the value of R                // remains 0 and Z[i] becomes 0. For string                // "aaaaaa" and i = 1, Z[i] and R become 5                  while(R < n && str[R - L] == str[R])                    R++;                                  Z[i] = R - L;                R--;              }            else{                  // k = i-L so k corresponds to number which                // matches in [L,R] interval.                let k = i - L;                  // if Z[k] is less than remaining interval                // then Z[i] will be equal to Z[k].                // For example, str = "ababab", i = 3, R = 5                // and L = 2                if(Z[k] < R - i + 1)                    Z[i] = Z[k];                  // For example str = "aaaaaa" and i = 2, R is 5,                // L is 0                else{                    // else start from R and check manually                    L = i;                    while(R < n && str[R - L] == str[R])                        R++;                                          Z[i] = R - L;                    R--;                }            }        }} // Driver programlet text = "GEEKS FOR GEEKS";let pattern = "GEEK"; search(text, pattern);  // This code is contributed by rag2127 </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 
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