Print all longest common sub-sequences in lexicographical order

// C# program to find all LCS of two strings in 
// sorted order. 
using System;
class GFG 
{    
    static int MAX = 100;
    
    // length of lcs 
    static int lcslen = 0; 
      
    // dp matrix to store result of sub calls for lcs 
    static int[,] dp = new int[MAX,MAX]; 
      
    // A memoization based function that returns LCS of 
    // str1[i..len1-1] and str2[j..len2-1] 
    static 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 = Math.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 
    static 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'; 
            Console.WriteLine(new string(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-sequence 
                    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. 
    static void prinlAllLCSSorted(string str1, string str2) 
    { 
        // Find lengths of both strings 
        int len1 = str1.Length, len2 = str2.Length; 
      
        // Find length of LCS 
        for(int i = 0; i < MAX; i++)
        {
            for(int j = 0; j < MAX; j++)
            {
                dp[i, j] = -1;
            }
        }
        lcslen = lcs(str1, str2, len1, len2, 0, 0); 
      
        // Print all LCS using recursive backtracking 
        // data[] is used to store individual LCS. 
        char[] data = new char[MAX]; 
        printAll(str1, str2, len1, len2, data, 0, 0, 0); 
    } 

  // Driver code
  static void Main() 
  {
    string str1 = "abcabcaa", str2 = "acbacba"; 
    prinlAllLCSSorted(str1, str2); 
  }
}

// This code is contributed by divyeshrabadiya07