Recursive solution to common longest substring between two strings

Question

I am trying to return the length of a common substring between two strings. I'm very well aware of the DP solution, however I want to be able to solve this recursively just for practice.

I have the solution to find the longest common subsequence...

def get_substring(str1, str2, i, j):
    if i == 0 or j == 0:
        return
    elif str1[i-1] == str2[j-1]:
        return 1 + get_substring(str1, str2, i-1, j-1)
    else:
        return max(get_substring(str1, str2, i, j-1), get_substring(str1, str2, j-1, i))

However, I need the longest common substring, not the longest common sequence of letters. I tried altering my code in a couple of ways, one being changing the base case to...

if i == 0 or j == 0 or str1[i-1] != str2[j-1]:
    return 0

But that did not work, and neither did any of my other attempts.

For example, for the following strings...

X = "AGGTAB"
Y = "BAGGTXAYB"
print(get_substring(X, Y, len(X), len(Y)))

The longest substring is AGGT.

My recursive skills are not the greatest, so if anybody can help me out that would be very helpful.

Nishant · Accepted Answer · 2018-08-02 13:16:29Z

2

package algo.dynamic;

public class LongestCommonSubstring {

public static void main(String[] args) {
    String a = "AGGTAB";
    String b = "BAGGTXAYB";
    int maxLcs = lcs(a.toCharArray(), b.toCharArray(), a.length(), b.length(), 0);
    System.out.println(maxLcs);
}

private static int lcs(char[] a, char[] b, int i, int j, int count) {
    if (i == 0 || j == 0)
        return count;
    if (a[i - 1] == b[j - 1]) {
        count = lcs(a, b, i - 1, j - 1, count + 1);
    }
    count = Math.max(count, Math.max(lcs(a, b, i, j - 1, 0), lcs(a, b, i - 1, j, 0)));
    return count;
}

}

answered Aug 2, 2018 at 13:16

Nishant

811 silver badge6 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

btilly · Accepted Answer · 2018-06-26 18:07:56Z

1

You need to recurse on each separately. Which is easier to do if you have multiple recursive functions.

def longest_common_substr_at_both_start (str1, str2):
    if 0 == len(str1) or 0 == len(str2) or str1[0] != str2[0]:
        return ''
    else:
        return str1[0] + longest_common_substr_at_both_start(str1[1:], str2[1:])

def longest_common_substr_at_first_start (str1, str2):
    if 0 == len(str2):
        return ''
    else:
        answer1 = longest_common_substr_at_both_start (str1, str2)
        answer2 = longest_common_substr_at_first_start (str1, str2[1:])
        return answer2 if len(answer1) < len(answer2) else answer1

def longest_common_substr (str1, str2):
    if 0 == len(str1):
        return ''
    else:
        answer1 = longest_common_substr_at_first_start (str1, str2)
        answer2 = longest_common_substr(str1[1:], str2)
        return answer2 if len(answer1) < len(answer2) else answer1

print(longest_common_substr("BAGGTXAYB","AGGTAB") )

answered Jun 26, 2018 at 18:07

btilly

47.8k3 gold badges69 silver badges101 bronze badges

1 Comment

Malu Over a year ago

Check geeksforgeeks.org/longest-common-substring-dp-29 The second part of the article has an easier recursive solution.

fp_mora · Accepted Answer · 2018-07-17 23:41:02Z

I am so sorry. I didn't have time to convert this into a recursive function. This was relatively straight forward to compose. If Python had a fold function a recursive function would be greatly eased. 90% of recursive functions are primitive. That's why fold is so valuable.

I hope the logic in this can help with a recursive version.

(x,y)= "AGGTAB","BAGGTXAYB"
xrng=  range(len(x)) # it is used twice

np=[(a+1,a+2) for a in xrng] # make pairs of list index values to use

allx = [ x[i:i+b] for (a,b) in np for i in xrng[:-a]] # make list of len>1 combinations

[ c for i in range(len(y)) for c in allx if c == y[i:i+len(c)]] # run, matching x & y

...producing this list from which to take the longest of the matches

['AG', 'AGG', 'AGGT', 'GG', 'GGT', 'GT']

I didn't realize getting the longest match from the list would be a little involved.

ls= ['AG', 'AGG', 'AGGT', 'GG', 'GGT', 'GT']
ml= max([len(x) for x in ls])
ls[[a for (a,b) in zip(range(len(ls)),[len(x) for x in ls]) if b == ml][0]]

"AGGT"

Collectives™ on Stack Overflow

Recursive solution to common longest substring between two strings

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related