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 C++ 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] = '\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.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] = "\0"

# 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

```

Output:
`LCS of AGGTAB and GXTXAYB is GTAB`

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