For jetwick I needed yet another
string algorithm and stumbled over this cool and common problem: trying
to find the longest substring of two strings. Be sure that you
understand the difference to the LC **sequence** problem.

For example if we have two strings:

*Please, peter go swimming!*

and

*I’m peter goliswi*

The algorithm should print out *‘ peter go’*. The longest common substring algorithm can be implemented in an efficient manner with the help of suffic trees.

But in this post I’ll try to explain the bit less efficient ‘dynamic programming‘ version of the algorithm. Dynamic programming means that you can reuse already calculated information in a later step or you break the algorithm into parts to reuse information. To understand the algorithm you just need to fill the entries of an integer-array with the lengths of the identical substrings. Assume we use i for the horizontal string (please …) and j for the vertical string. Then the algorithm hits at some time i=19 and j=0 for one identical character ‘i’. Then the line

num[i][j] = 1;

is executed and saves the lengths of the 1 length identical substring.

please, peter go swimming

i 0000000000000000000100100

' 0000000000000000000000000

m 0000000000000000000011000

0000000100000100100000000

p 1000000020000000000000000

e 0010010003000000000000000

t 0000000000400000000000000

e 0010010001050000000000000

r 0000000000006000000000000

0000000100000700100000000

g 0000000000000080000000000

o 0000000000000009000000000

l 0100000000000000000000000

i 0000000000000000000100100

s 0001000000000000010000000

w 0000000000000000002000000

i 0000000000000000000300100

Later on it hits the m characters and saves 1 two times to the array but then at i=7 and j=3 it starts our substring and saves 1 for the space character. Then some loops later it reaches i=8 and j=4 Now it reuses the already calculated “identical-length” of 1. It will do:

num[8][4] = 1 + num[7][3];

and we get 2. So, we now know we have a substring with two 2 characters. And with

if (num[i][j] > maxlen)

we make sure that we overwrite the existing longest substring (stored in the StringBuilder) ONLY IF there is a longer substring found and either append the character (if it is the current substring in progress):

sb.append(str1.charAt(i));

or we can start a longer substring. See the java code (mainly taken from wikipedia) for yourself:

public static String longestSubstring(String str1, String str2) {

StringBuilder sb = new StringBuilder();

if (str1 == null || str1.isEmpty() || str2 == null || str2.isEmpty())

return "";

// ignore case

str1 = str1.toLowerCase();

str2 = str2.toLowerCase();

// java initializes them already with 0

int[][] num = new int[str1.length()][str2.length()];

int maxlen = 0;

int lastSubsBegin = 0;

for (int i = 0; i < str1.length(); i++) {

for (int j = 0; j < str2.length(); j++) {

if (str1.charAt(i) == str2.charAt(j)) {

if ((i == 0) || (j == 0))

num[i][j] = 1;

else

num[i][j] = 1 + num[i - 1][j - 1];

if (num[i][j] > maxlen) {

maxlen = num[i][j];

// generate substring from str1 => i

int thisSubsBegin = i - num[i][j] + 1;

if (lastSubsBegin == thisSubsBegin) {

//if the current LCS is the same as the last time this block ran

sb.append(str1.charAt(i));

} else {

//this block resets the string builder if a different LCS is found

lastSubsBegin = thisSubsBegin;

sb = new StringBuilder();

sb.append(str1.substring(lastSubsBegin, i + 1));

}

}

}

}}

return sb.toString();

}

*From http://karussell.wordpress.com/2011/04/14/longest-common-substring-algorithm-in-java/ *

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