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).
3. Traverse the 2D array starting from L[m][n]. Do following for every cell L[i][j] 
   • 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. 
   • 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

Following is the implementation of the above approach. 

C++14

/* Dynamic Programming implementation of LCS problem */
#include <cstdlib>
#include <cstring>
#include <iostream>
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

Python3

# 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 i in range(n+1)] for j in range(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 range(m+1):
        for j in range(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])

        # Create a string variable to store the lcs string
    lcs = ""

    # 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 += X[i-1]
            i -= 1
            j -= 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

    # We traversed the table in reverse order
    # LCS is the reverse of what we got
    lcs = lcs[::-1]
    print("LCS of " + X + " and " + Y + " is " + lcs)


# Driver program
X = "AGGTAB"
Y = "GXTXAYB"
m = len(X)
n = len(Y)
lcs(X, Y, m, n)

# This code is contributed by AMAN ASATI

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 contributed by ita_c
?>

Javascript

<script>
function ReverseString(str) {
   return str.split('').reverse().join('')
}

function max(a, b)
{
    if (a > b)
        return a;
    else
        return b;
}
function printLCS(str1, str2) {
    var len1 = str1.length;
    var len2 = str2.length;
    var lcs = new Array(len1 + 1);
    for (var i = 0; i <= len1; i++) {
        lcs[i] = new Array(len2 + 1)
    }
    for (var i = 0; i <= len1; i++) {
        for (var j = 0; j <= len2; j++) {
            if (i == 0 || j == 0) {
                lcs[i][j] = 0;
            }
            else {
                if (str1[i - 1] == str2[j - 1]) {
                    lcs[i][j] = 1 + lcs[i - 1][j - 1];
                }
                else {
                    lcs[i][j] = max(lcs[i][j - 1], lcs[i - 1][j]);
                }
            }
        }}

        var n = lcs[len1][len2];
         document.write("Length of common subsequence is: " +
         n + "<br>" + "The subsequence is : ");
        var str="";
       var i = len1;
       var j = len2;
       while(i>0&&j>0)
       {
            if(str1[i - 1] == str2[j - 1])
            {
                str += str1[i - 1];
                i--;
                j--;
            }
            else{
            if(lcs[i][j-1]>lcs[i-1][j])
            {
                j--;
            }
            else
            {
                i--;
            }
            }
        }
       return ReverseString(str);
    }
    var str1 = "AGGTAB";
    var str2 = "GXTXAYB";
    document.write(printLCS(str1, str2));

    // This code is contributed by akshitsaxenaa09
</script>

Output

LCS of AGGTAB and GXTXAYB is GTAB

Time Complexity: O(m*n)
Auxiliary Space: O(m*n)

Top-down approach for printing Longest Common Subsequence:

Follow the steps below for the implementation:

• Check if one of the two strings is of size zero, then we return an empty string because the LCS, in this case, is empty (base case).
• Check if not the base case, then if we have a solution for the current a and b saved in the memory, we return it, else we calculate the solution for the current a and b and store it in memory.
• For calculation of solution, we compare the last two characters of a and b. 
• Check if they are equal, 
  • If true, we add this character to the solution string, then we erase the last character from each string and add to the solution string the string returned from LCS(a, b) after deleting the last characters.
  • Otherwise, if the last two characters don’t equal each other, we call LCS(a_without_last_character, b) and LCS(a, b_without_last_character). Then we compare the two returned strings and add the bigger string to the solution string.
• Store the solution in the memory and return it.
• Reverse the final returned solution, given that our top-down approach generates a reversed string.

Here is the implementation of the above recursive approach.

C++

#include <iostream>
#include <map>
#include <string>
using namespace std;
using mpsss = map<pair<string, string>, string>;
using mpssi = map<pair<string, string>, int>;
using ss = pair<string, string>;

mpsss dp;

// For keep track of visited subproblem or not (0 = not
// visited, 1 = visited)
mpssi vs;

// utiltiy function to reverse a string, we need it because
// our top-down approach return a reversed solution
string reverse(string str)
{
    string ans = str;
    int u = 0;
    int v = ans.length() - 1;
    while (u < v) {
        swap(ans[u], ans[v]);
        u++;
        v--;
    }
    return ans;
}

// utiltiy function that compares two strings and return the
// longer in size.
string max_str(string a, string b)
{
    if (a.length() > b.length())
        return a;
    else
        return b;
}

string LCS_core(string a, string b)
{

    // size of string a
    int n_a = a.length();

    // size of string b
    int n_b = b.length();

    // Base case
    if (n_a == 0 || n_b == 0)
        return "";

    // dp index to access the dp structure
    ss dp_i = make_pair(a, b);

    // ans points to the memory location in the dp
    // structure in which the solution string will be stored
    string& ans = dp[dp_i];

    // if visited return solution from memory.
    if (vs[dp_i] == 1)
        return dp[dp_i];

    // if not visited, set the visit value to be one
    // (meaning its now visited)
    else
        vs[dp_i] = 1;

    // if the last two character match
    if (a[n_a - 1] == b[n_b - 1]) {

        // Add this character to the solution string
        ans += a[n_a - 1];

        // Erase last character from a
        a.erase(n_a - 1, 1);

        // Erase last character from b
        b.erase(n_b - 1, 1);

        // add to the solution string the value of
        // LCS_core(a, b) (the remaining strings after
        // deleting last characters)
        ans += LCS_core(a, b);
        return ans;
    }

    // Return longest string
    ans += max_str(LCS_core(a.substr(0, n_a - 1), b),
                   LCS_core(a, b.substr(0, n_b - 1)));
    return ans;
}

string LCS(string a, string b)
{
    ;

    // Reverse obtained result
    return reverse(LCS_core(a, b));
}

int main()
{
    string a = "AGGTAB";
    string b = "GXTXAYB";
    cout << LCS(a, b);
    return 0;
}

Python3

from typing import Dict, Tuple

# Initialize an empty dictionary to store the solutions of subproblems
dp: Dict[Tuple[str, str], str] = {}

# Initialize an empty dictionary to keep track of visited subproblems
vs: Dict[Tuple[str, str], int] = {}

# Utility function to reverse a string, we need it because our top-down approach
# return a reversed solution
def reverse(s: str) -> str:
    ans = list(s)
    u, v = 0, len(ans) - 1
    while u < v:
        ans[u], ans[v] = ans[v], ans[u] #swap operation
        u += 1
        v -= 1
    return "".join(ans)

# Utility function that compares two strings and return the longer in size.
def max_str(a: str, b: str) -> str:
    return a if len(a) > len(b) else b

# Recursive function that takes two strings as input, and returns the LCS of them
def LCS_core(a: str, b: str) -> str:
    # Base case
    if not a or not b:
        return ""
    # dp index to access the dp structure
    dp_i = (a, b)

    # if visited return solution from memory
    if dp_i in vs:
        return dp[dp_i]
    else:
        vs[dp_i] = 1

    # if the last two character match
    if a[-1] == b[-1]:
        ans = a[-1] + LCS_core(a[:-1], b[:-1])
        dp[dp_i] = ans
        return ans

    # Return longest string
    ans = max_str(LCS_core(a[:-1], b), LCS_core(a, b[:-1]))
    dp[dp_i] = ans
    return ans

# Final wrapper function to call the recursive function and reverse the result
def LCS(a: str, b: str) -> str:
    return reverse(LCS_core(a, b))

a = "AGGTAB"
b = "GXTXAYB"
print(LCS(a, b))

# This code is contributed by Shivam Tiwari

Java

import java.util.HashMap;
import java.util.Map;

class Main {
    public static void main(String[] args) {
        String a = "AGGTAB";
        String b = "GXTXAYB";
        System.out.println(LCS(a, b));
    }

    private static Map<String, Map<String, String>> dp = new HashMap<>();
    private static Map<String, Map<String, Integer>> vs = new HashMap<>();

    private static String reverse(String str) {
        String ans = "";
        for (int i = str.length() - 1; i >= 0; i--) {
            ans += str.charAt(i);
        }
        return ans;
    }

    private static String max_str(String a, String b) {
        return a.length() > b.length() ? a : b;
    }

    private static String LCS_core(String a, String b) {
        if (a.isEmpty() || b.isEmpty()) {
            return "";
        }

        if (vs.containsKey(a) && vs.get(a).containsKey(b)) {
            return dp.get(a).get(b);
        }

        String ans = "";
        if (a.charAt(a.length() - 1) == b.charAt(b.length() - 1)) {
            ans += a.charAt(a.length() - 1);
            a = a.substring(0, a.length() - 1);
            b = b.substring(0, b.length() - 1);
            ans += LCS_core(a, b);
        } else {
            ans += max_str(LCS_core(a.substring(0, a.length() - 1), b),
                           LCS_core(a, b.substring(0, b.length() - 1)));
        }
        if (!dp.containsKey(a)) {
            dp.put(a, new HashMap<>());
        }
        if (!vs.containsKey(a)) {
            vs.put(a, new HashMap<>());
        }
        dp.get(a).put(b, ans);
        vs.get(a).put(b, 1);
        return ans;
    }

    private static String LCS(String a, String b) {
        return reverse(LCS_core(a, b));
    }
}

// This code is contributed by Susobhan Akhuli

Output

GTAB

 Time complexity: O(m*n) where m is the length of the first string and n is the length of the second string. This is because, for each character in both strings, we need to check whether they are equal and then recursively call the LCS_core() function.

Auxiliary space: O(m*n) as well. This is because we are using a map structure to keep track of the visited subproblems and the result of each subproblem. This structure is having a size of O(m*n).