Printing Longest Common Subsequence

Given two sequences, print the longest subsequence present in both of them.

Examples:
LCS for input Sequences “ABCDGH” and “AEDFHR” is “ADH” of length 3.
LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4.

We have discussed Longest Common Subsequence (LCS) problem in a previous post. The function discussed there was mainly to find the length of LCS. To find length of LCS, a 2D table L[][] was constructed. In this post, the function to construct and print LCS is discussed.

Following is detailed algorithm to print the LCS. It uses the same 2D table L[][].

1) Construct L[m+1][n+1] using the steps discussed in previous post.

2) The value L[m][n] contains length of LCS. Create a character array lcs[] of length equal to the length of lcs plus 1 (one extra to store \0).

2) Traverse the 2D array starting from L[m][n]. Do following for every cell L[i][j]
…..a) If characters (in X and Y) corresponding to L[i][j] are same (Or X[i-1] == Y[j-1]), then include this character as part of LCS.
…..b) Else compare values of L[i-1][j] and L[i][j-1] and go in direction of greater value.

The following table (taken from Wiki) shows steps (highlighted) followed by the above algorithm.

		0	1	2	3	4	5	6	7
		Ø	M	Z	J	A	W	X	U
0	Ø	0	0	0	0	0	0	0	0
1	X	0	0	0	0	0	0	1	1
2	M	0	1	1	1	1	1	1	1
3	J	0	1	1	2	2	2	2	2
4	Y	0	1	1	2	2	2	2	2
5	A	0	1	1	2	3	3	3	3
6	U	0	1	1	2	3	3	3	4
7	Z	0	1	2	2	3	3	3	4

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Following is the implementation of above approach.

C/C++

/* Dynamic Programming implementation of LCS problem */
#include<iostream> 
#include<cstring> 
#include<cstdlib> 
using namespace std; 
  
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
void lcs( char *X, char *Y, int m, int n ) 
{ 
   int L[m+1][n+1]; 
  
   /* Following steps build L[m+1][n+1] in bottom up fashion. Note 
      that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
   for (int i=0; i<=m; i++) 
   { 
     for (int j=0; j<=n; j++) 
     { 
       if (i == 0 || j == 0) 
         L[i][j] = 0; 
       else if (X[i-1] == Y[j-1]) 
         L[i][j] = L[i-1][j-1] + 1; 
       else
         L[i][j] = max(L[i-1][j], L[i][j-1]); 
     } 
   } 
  
   // Following code is used to print LCS 
   int index = L[m][n]; 
  
   // Create a character array to store the lcs string 
   char lcs[index+1]; 
   lcs[index] = '\0'; // Set the terminating character 
  
   // Start from the right-most-bottom-most corner and 
   // one by one store characters in lcs[] 
   int i = m, j = n; 
   while (i > 0 && j > 0) 
   { 
      // If current character in X[] and Y are same, then 
      // current character is part of LCS 
      if (X[i-1] == Y[j-1]) 
      { 
          lcs[index-1] = X[i-1]; // Put current character in result 
          i--; j--; index--;     // reduce values of i, j and index 
      } 
  
      // If not same, then find the larger of two and 
      // go in the direction of larger value 
      else if (L[i-1][j] > L[i][j-1]) 
         i--; 
      else
         j--; 
   } 
  
   // Print the lcs 
   cout << "LCS of " << X << " and " << Y << " is " << lcs; 
} 
  
/* Driver program to test above function */
int main() 
{ 
  char X[] = "AGGTAB"; 
  char Y[] = "GXTXAYB"; 
  int m = strlen(X); 
  int n = strlen(Y); 
  lcs(X, Y, m, n); 
  return 0; 
} 

Java

// Dynamic Programming implementation of LCS problem in Java 
import java.io.*; 
  
class  LongestCommonSubsequence 
{ 
    // Returns length of LCS for X[0..m-1], Y[0..n-1] 
    static void lcs(String X, String Y, int m, int n) 
    { 
        int[][] L = new int[m+1][n+1]; 
   
        // Following steps build L[m+1][n+1] in bottom up fashion. Note 
        // that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]  
        for (int i=0; i<=m; i++) 
        { 
            for (int j=0; j<=n; j++) 
            { 
                if (i == 0 || j == 0) 
                    L[i][j] = 0; 
                else if (X.charAt(i-1) == Y.charAt(j-1)) 
                    L[i][j] = L[i-1][j-1] + 1; 
                else
                    L[i][j] = Math.max(L[i-1][j], L[i][j-1]); 
            } 
        } 
   
        // Following code is used to print LCS 
        int index = L[m][n]; 
        int temp = index; 
   
        // Create a character array to store the lcs string 
        char[] lcs = new char[index+1]; 
        lcs[index] = '\u0000'; // Set the terminating character 
   
        // Start from the right-most-bottom-most corner and 
        // one by one store characters in lcs[] 
        int i = m; 
        int j = n; 
        while (i > 0 && j > 0) 
        { 
            // If current character in X[] and Y are same, then 
            // current character is part of LCS 
            if (X.charAt(i-1) == Y.charAt(j-1)) 
            { 
                // Put current character in result 
                lcs[index-1] = X.charAt(i-1);  
                  
                // reduce values of i, j and index 
                i--;  
                j--;  
                index--;      
            } 
   
            // If not same, then find the larger of two and 
            // go in the direction of larger value 
            else if (L[i-1][j] > L[i][j-1]) 
                i--; 
            else
                j--; 
        } 
   
        // Print the lcs 
        System.out.print("LCS of "+X+" and "+Y+" is "); 
        for(int k=0;k<=temp;k++) 
            System.out.print(lcs[k]); 
    } 
      
    // driver program 
    public static void main (String[] args)  
    { 
        String X = "AGGTAB"; 
        String Y = "GXTXAYB"; 
        int m = X.length(); 
        int n = Y.length(); 
        lcs(X, Y, m, n); 
    } 
} 
  
// Contributed by Pramod Kumar 

Python

# Dynamic programming implementation of LCS problem 
  
# Returns length of LCS for X[0..m-1], Y[0..n-1]  
def lcs(X, Y, m, n): 
    L = [[0 for x in xrange(n+1)] for x in xrange(m+1)] 
  
    # Following steps build L[m+1][n+1] in bottom up fashion. Note 
    # that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]  
    for i in xrange(m+1): 
        for j in xrange(n+1): 
            if i == 0 or j == 0: 
                L[i][j] = 0
            elif X[i-1] == Y[j-1]: 
                L[i][j] = L[i-1][j-1] + 1
            else: 
                L[i][j] = max(L[i-1][j], L[i][j-1]) 
  
    # Following code is used to print LCS 
    index = L[m][n] 
  
    # Create a character array to store the lcs string 
    lcs = [""] * (index+1) 
    lcs[index] = "" 
  
    # Start from the right-most-bottom-most corner and 
    # one by one store characters in lcs[] 
    i = m 
    j = n 
    while i > 0 and j > 0: 
  
        # If current character in X[] and Y are same, then 
        # current character is part of LCS 
        if X[i-1] == Y[j-1]: 
            lcs[index-1] = X[i-1] 
            i-=1
            j-=1
            index-=1
  
        # If not same, then find the larger of two and 
        # go in the direction of larger value 
        elif L[i-1][j] > L[i][j-1]: 
            i-=1
        else: 
            j-=1
  
    print "LCS of " + X + " and " + Y + " is " + "".join(lcs)  
  
# Driver program 
X = "AGGTAB"
Y = "GXTXAYB"
m = len(X) 
n = len(Y) 
lcs(X, Y, m, n) 
  
# This code is contributed by BHAVYA JAIN 

C#

// Dynamic Programming implementation  
// of LCS problem in C# 
using System; 
  
class GFG 
{ 
    // Returns length of LCS for X[0..m-1], Y[0..n-1] 
    static void lcs(String X, String Y, int m, int n) 
    { 
        int[,] L = new int[m+1, n+1]; 
  
        // Following steps build L[m+1][n+1] in  
        // bottom up fashion. Note that L[i][j]  
        // contains length of LCS of X[0..i-1] 
        // and Y[0..j-1]  
        for (int i = 0; i <= m; i++) 
        { 
            for (int j = 0; j <= n; j++) 
            { 
                if (i == 0 || j == 0) 
                    L[i, j] = 0; 
                else if (X[i-1] == Y[j-1]) 
                    L[i, j] = L[i-1, j-1] + 1; 
                else
                    L[i, j] = Math.Max(L[i-1, j], L[i, j-1]); 
            } 
        } 
  
        // Following code is used to print LCS 
        int index = L[m, n]; 
        int temp = index; 
  
        // Create a character array  
        // to store the lcs string 
        char[] lcs = new char[index+1]; 
          
        // Set the terminating character 
        lcs[index] = '\0';  
  
        // Start from the right-most-bottom-most corner 
        // and one by one store characters in lcs[] 
        int k = m, l = n; 
        while (k > 0 && l > 0) 
        { 
            // If current character in X[] and Y  
            // are same, then current character  
            // is part of LCS 
            if (X[k-1] == Y[l-1]) 
            { 
                // Put current character in result 
                lcs[index-1] = X[k-1];  
                  
                // reduce values of i, j and index 
                k--;  
                l--;  
                index--;      
            } 
  
            // If not same, then find the larger of two and 
            // go in the direction of larger value 
            else if (L[k-1, l] > L[k, l-1]) 
                k--; 
            else
                l--; 
        } 
  
        // Print the lcs 
        Console.Write("LCS of " + X + " and " + Y + " is "); 
        for(int q = 0; q <= temp; q++) 
            Console.Write(lcs[q]); 
    } 
      
    // Driver program 
    public static void Main ()  
    { 
        String X = "AGGTAB"; 
        String Y = "GXTXAYB"; 
        int m = X.Length; 
        int n = Y.Length; 
        lcs(X, Y, m, n); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php  
// Dynamic Programming implementation of LCS problem  
  
// Returns length of LCS for X[0..m-1], Y[0..n-1]  
function lcs( $X, $Y, $m, $n ) 
{ 
    $L = array_fill(0, $m + 1,  
         array_fill(0, $n + 1, NULL)); 
      
    /* Following steps build L[m+1][n+1] in bottom  
       up fashion. Note that L[i][j] contains length  
       of LCS of X[0..i-1] and Y[0..j-1] */
    for ($i = 0; $i <= $m; $i++) 
    { 
        for ($j = 0; $j <= $n; $j++) 
        { 
            if ($i == 0 || $j == 0) 
                $L[$i][$j] = 0; 
            else if ($X[$i - 1] == $Y[$j - 1]) 
                $L[$i][$j] = $L[$i - 1][$j - 1] + 1; 
            else
                $L[$i][$j] = max($L[$i - 1][$j],  
                                 $L[$i][$j - 1]); 
        } 
    } 
      
    // Following code is used to print LCS 
    $index = $L[$m][$n]; 
    $temp = $index; 
      
    // Create a character array to store the lcs string 
    $lcs = array_fill(0, $index + 1, NULL); 
    $lcs[$index] = ''; // Set the terminating character 
      
    // Start from the right-most-bottom-most corner  
    // and one by one store characters in lcs[] 
    $i = $m; 
    $j = $n; 
    while ($i > 0 && $j > 0) 
    { 
        // If current character in X[] and Y are same, 
        // then current character is part of LCS 
        if ($X[$i - 1] == $Y[$j - 1]) 
        { 
            // Put current character in result 
            $lcs[$index - 1] = $X[$i - 1]; 
            $i--; 
            $j--;  
            $index--;    // reduce values of i, j and index 
        } 
      
        // If not same, then find the larger of two  
        // and go in the direction of larger value 
        else if ($L[$i - 1][$j] > $L[$i][$j - 1]) 
            $i--; 
        else
            $j--; 
    } 
      
    // Print the lcs 
    echo "LCS of " . $X . " and " . $Y . " is "; 
    for($k = 0; $k < $temp; $k++) 
        echo $lcs[$k]; 
} 
  
// Driver Code 
$X = "AGGTAB"; 
$Y = "GXTXAYB"; 
$m = strlen($X); 
$n = strlen($Y); 
lcs($X, $Y, $m, $n); 
  
// This code is contirbuted by ita_c 
?> 

Output:

LCS of AGGTAB and GXTXAYB is GTAB

References:
http://en.wikipedia.org/wiki/Longest_common_subsequence_problem

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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My Personal Notes

		0	1	2	3	4	5	6	7
		Ø	M	Z	J	A	W	X	U
0	Ø	0	0	0	0	0	0	0	0
1	X	0	0	0	0	0	0	1	1
2	M	0	1	1	1	1	1	1	1
3	J	0	1	1	2	2	2	2	2
4	Y	0	1	1	2	2	2	2	2
5	A	0	1	1	2	3	3	3	3
6	U	0	1	1	2	3	3	3	4
7	Z	0	1	2	2	3	3	3	4

		0	1	2	3	4	5	6	7
		Ø	M	Z	J	A	W	X	U
0	Ø	0	0	0	0	0	0	0	0
1	X	0	0	0	0	0	0	1	1
2	M	0	1	1	1	1	1	1	1
3	J	0	1	1	2	2	2	2	2
4	Y	0	1	1	2	2	2	2	2
5	A	0	1	1	2	3	3	3	3
6	U	0	1	1	2	3	3	3	4
7	Z	0	1	2	2	3	3	3	4

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C/C++

Java

Python

C#

PHP

Recommended Posts:

		0	1	2	3	4	5	6	7
		Ø	M	Z	J	A	W	X	U
0	Ø	0	0	0	0	0	0	0	0
1	X	0	0	0	0	0	0	1	1
2	M	0	1	1	1	1	1	1	1
3	J	0	1	1	2	2	2	2	2
4	Y	0	1	1	2	2	2	2	2
5	A	0	1	1	2	3	3	3	3
6	U	0	1	1	2	3	3	3	4
7	Z	0	1	2	2	3	3	3	4