# 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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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.

```// 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;
}
```

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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