Print all longest common sub-sequences in lexicographical order


You are given two strings.Now you have to print all longest common sub-sequences in lexicographical order?

Examples:

Input : str1 = "abcabcaa", str2 = "acbacba"
Output: ababa
        abaca
        abcba
        acaba
        acaca
        acbaa
        acbca

This problem is an extension of longest common subsequence. We first find length of LCS and store all LCS in 2D table using Memoization (or Dynamic Programming). Then we search all characters from ‘a’ to ‘z’ (to output sorted order) in both strings. If a character is found in both strings and current positions of character lead to LCS, we recursively search all occurrences with current LCS length plus 1.

Below is the implementation of algorithm.

Cpp



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// C++ program to find all LCS of two strings in
// sorted order.
#include<bits/stdc++.h>
#define MAX 100
using namespace std;
  
// length of lcs
int lcslen = 0;
  
// dp matrix to store result of sub calls for lcs
int dp[MAX][MAX];
  
// A memoization based function that returns LCS of
// str1[i..len1-1] and str2[j..len2-1]
int lcs(string str1, string str2, int len1, int len2,
                                      int i, int j)
{
    int &ret = dp[i][j];
  
    // base condition
    if (i==len1 || j==len2)
        return ret = 0;
  
    // if lcs has been computed
    if (ret != -1)
        return ret;
  
    ret = 0;
  
    // if characters are same return previous + 1 else
    // max of two sequences after removing i'th and j'th
    // char one by one
    if (str1[i] == str2[j])
        ret = 1 + lcs(str1, str2, len1, len2, i+1, j+1);
    else
        ret = max(lcs(str1, str2, len1, len2, i+1, j),
                  lcs(str1, str2, len1, len2, i, j+1));
    return ret;
}
  
// Function to print all routes common sub-sequences of
// length lcslen
void printAll(string str1, string str2, int len1, int len2,
              char data[], int indx1, int indx2, int currlcs)
{
    // if currlcs is equal to lcslen then print it
    if (currlcs == lcslen)
    {
        data[currlcs] = '\0';
        puts(data);
        return;
    }
  
    // if we are done with all the characters of both string
    if (indx1==len1 || indx2==len2)
        return;
  
    // here we have to print all sub-sequences lexicographically,
    // that's why we start from 'a'to'z' if this character is
    // present in both of them then append it in data[] and same
    // remaining part
    for (char ch='a'; ch<='z'; ch++)
    {
        // done is a flag to tell that we have printed all the
        // subsequences corresponding to current character
        bool done = false;
  
        for (int i=indx1; i<len1; i++)
        {
            // if character ch is present in str1 then check if
            // it is present in str2
            if (ch==str1[i])
            {
              for (int j=indx2; j<len2; j++)
              {
                // if ch is present in both of them and
                // remaining length is equal to remaining
                // lcs length then add ch in sub-sequenece
                if (ch==str2[j] &&
                  lcs(str1, str2, len1, len2, i, j) == lcslen-currlcs)
                {
                  data[currlcs] = ch;
                  printAll(str1, str2, len1, len2, data, i+1, j+1, currlcs+1);
                  done = true;
                  break;
                }
              }
            }
  
            // If we found LCS beginning with current character. 
            if (done)
                break;
        }
    }
}
  
// This function prints all LCS of str1 and str2
// in lexicographic order.
void prinlAllLCSSorted(string str1, string str2)
{
    // Find lengths of both strings
    int len1 = str1.length(), len2 = str2.length();
  
    // Find length of LCS
    memset(dp, -1, sizeof(dp));
    lcslen = lcs(str1, str2, len1, len2, 0, 0);
  
    // Print all LCS using recursive backtracking
    // data[] is used to store individual LCS.
    char data[MAX];
    printAll(str1, str2, len1, len2, data, 0, 0, 0);
}
  
// Driver program to run the case
int main()
{
    string str1 = "abcabcaa", str2 = "acbacba";
    prinlAllLCSSorted(str1, str2);
    return 0;
}

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python3

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# Python3 program to find all LCS of two strings in
# sorted order.
MAX=100
lcslen = 0
  
# dp matrix to store result of sub calls for lcs
dp=[[-1 for i in range(MAX)] for i in range(MAX)]
  
# A memoization based function that returns LCS of
# str1[i..len1-1] and str2[j..len2-1]
def lcs(str1, str2, len1, len2, i, j):
  
    # base condition
    if (i == len1 or j == len2):
        dp[i][j] = 0
        return dp[i][j]
  
    # if lcs has been computed
    if (dp[i][j] != -1):
        return dp[i][j]
  
    ret = 0
  
    # if characters are same return previous + 1 else
    # max of two sequences after removing i'th and j'th
    # char one by one
    if (str1[i] == str2[j]):
        ret = 1 + lcs(str1, str2, len1, len2, i + 1, j + 1)
    else:
        ret = max(lcs(str1, str2, len1, len2, i + 1, j),
                  lcs(str1, str2, len1, len2, i, j + 1))
    dp[i][j] = ret
    return ret
  
# Function to prall routes common sub-sequences of
# length lcslen
def printAll(str1, str2, len1, len2,data, indx1, indx2, currlcs):
      
    # if currlcs is equal to lcslen then prit
    if (currlcs == lcslen):
        print("".join(data[:currlcs]))
        return
  
    # if we are done with all the characters of both string
    if (indx1 == len1 or indx2 == len2):
        return
  
    # here we have to prall sub-sequences lexicographically,
    # that's why we start from 'a'to'z' if this character is
    # present in both of them then append it in data[] and same
    # remaining part
    for ch in range(ord('a'),ord('z') + 1):
  
        # done is a flag to tell that we have printed all the
        # subsequences corresponding to current character
        done = False
  
        for i in range(indx1,len1):
            # if character ch is present in str1 then check if
            # it is present in str2
            if (chr(ch)==str1[i]):
              for j in range(indx2, len2):
  
                # if ch is present in both of them and
                # remaining length is equal to remaining
                # lcs length then add ch in sub-sequenece
                if (chr(ch) == str2[j] and lcs(str1, str2, len1, len2, i, j) == lcslen-currlcs):
                  data[currlcs] = chr(ch)
                  printAll(str1, str2, len1, len2, data, i + 1, j + 1, currlcs + 1)
                  done = True
                  break
  
            # If we found LCS beginning with current character.
            if (done):
                break
  
# This function prints all LCS of str1 and str2
# in lexicographic order.
def prinlAllLCSSorted(str1, str2):
    global lcslen
    # Find lengths of both strings
    len1,len2 = len(str1),len(str2)
  
    lcslen = lcs(str1, str2, len1, len2, 0, 0)
  
    # Prall LCS using recursive backtracking
    # data[] is used to store individual LCS.
    data = ['a' for i in range(MAX)]
    printAll(str1, str2, len1, len2, data, 0, 0, 0)
  
# Driver program to run the case
if __name__ == '__main__':
    str1 = "abcabcaa"
    str2 = "acbacba"
    prinlAllLCSSorted(str1, str2)
  
# This code is contributed by mohit kumar 29

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Output:

ababa
abaca
abcba
acaba
acaca
acbaa
acbca

This article is contributed by Shashak Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : mohit kumar 29