Return Subsequence Algorithm
They have 5 length-2 common subsequences: (AB), (AC), (ad), (BD), and (CD); 2 length-3 common subsequences: (ABD) and (ACD); and no longer common subsequences. It differs from the longest common substring problem: unlike substrings, subsequences are not need to occupy consecutive positions within the original sequences.
package Others;
import java.util.Scanner;
public class ReturnSubsequence {
public static void main(String[] args) {
System.out.println("Enter String: ");
Scanner s = new Scanner(System.in);
String givenString = s.next(); //given string
String[] subsequence = returnSubsequence(givenString); //calling returnSubsequence() function
System.out.println("Subsequences : ");
//print the given array of subsequences
for (int i = 0; i < subsequence.length; i++) {
System.out.println(subsequence[i]);
}
s.close();
}
/**
* @param givenString
* @return subsequence
*/
private static String[] returnSubsequence(String givenString) {
if (givenString.length() == 0) // If string is empty we will create an array of size=1 and insert "" (Empty string) in it
{
String[] ans = new String[1];
ans[0] = "";
return ans;
}
String[] SmallAns = returnSubsequence(givenString.substring(1)); //recursive call to get subsequences of substring starting from index position=1
String[] ans = new String[2 * SmallAns.length];// Our answer will be an array off string of size=2*SmallAns
int i = 0;
for (; i < SmallAns.length; i++) {
ans[i] = SmallAns[i]; //Copying all the strings present in SmallAns to ans string array
}
for (int k = 0; k < SmallAns.length; k++) {
ans[k + SmallAns.length] = givenString.charAt(0) + SmallAns[k]; // Insert character at index=0 of the given substring in front of every string in SmallAns
}
return ans;
}
}
