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pattern searching using Z algorithm #95

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98 changes: 98 additions & 0 deletions ZAlgorithm/ZAlgorithm.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,98 @@
class ZAlgorithm {

// 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];


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 = "HEY OCTOBER BRINGS HACTOBERFEST";
String pattern = "OBER";

search(text, pattern);
}
}