How to print string involved in Longest common sub sequence using Recursion? - c++

Question

I have a code for "Longest Common Sub sequence" but I want not only longest length but also string involved in it using recursion.
Any help will be appreciated.
The below code will give only Longest length.
#include<iostream>
#include<string>
using namespace std;
int longest_seq( string s1, string s2, int n1, int n2)
{
if( n1 < 0 || n2 < 0)
return 0;
// If last char of both the string matches
if( s1[n1] == s2[n2])
return ( 1 + longest_seq(s1, s2, n1-1, n2-1));
else
return max(longest_seq(s1, s2, n1-1, n2),longest_seq(s1, s2, n1, n2-1));
}
int main()
{
string s1 = "AGGTAB";
string s2 = "GXTXAYB";
int n1 = s1.length();
int n2 = s2.length();
cout <<longest_seq(s1, s2, n1-1, n2-1);
return 0;
}
current output :
4
Expected output:
4
GTAB

Answer 1

One crude way of doing thing would be to pass an empty string in the longest_seq() function which will populate the string with common sub sequence.
Basically you do it like this -
int main()
{
string s1 = "AGGTAB";
string s2 = "GXTXAYB";
string out;
int n1 = s1.length();
int n2 = s2.length();
// Pass out as an input param
cout <<longest_seq(s1, s2, n1-1, n2-1, out) << endl;
cout << out << endl;
return 0;
}
You need to change longest_seq() too.
int longest_seq( string s1, string s2, int n1, int n2, string& out)
{
if( n1 < 0 || n2 < 0)
return 0;
// If last char of both the string matches
if( s1[n1] == s2[n2]) {
out.insert(0, 1, s1[n1]); // Push the matching character in front
return ( 1 + longest_seq(s1, s2, n1-1, n2-1, out));
} else {
string out1, out2; // temporary strings
int len1 = longest_seq(s1, s2, n1-1, n2, out1);
int len2 = longest_seq(s1, s2, n1, n2-1, out2);
out.insert(0, len1>len2 ? out1 : out2); // Add the correct one to the front
return max(len1, len2);
}
}
Output:
4
GTAB
You can also change the function to return the string instead of passing the output string as an argument. Then after the function returns you can easily check the length of the output of the function to get longest common subsequence length. You can easily make the necessary changes for that.