Printing Longest Common Subsequence (original) (raw)

Last Updated : 23 Jul, 2025

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 #include #include 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

JavaScript

PHP

`

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 #include #include 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;

// utility 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; }

// utility 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; }

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

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

C#

using System; using System.Collections.Generic namespace LongestCommonSubsequence {

public class GFG{

    static void Main(string[] args)
    {
        string a = "AGGTAB";
        string b = "GXTXAYB";
        Console.WriteLine(LCS(a, b));
    }

    // Utility function to reverse a string
    static string Reverse(string str)
    {
        char[] charArray = str.ToCharArray();
        Array.Reverse(charArray);
        return new string(charArray);
    }

    // Utility function to return the larger string
    static string MaxStr(string a, string b)
    {
        return a.Length > b.Length ? a : b;
    }

    // Main function that returns the LCS
    static string LCS_Core(string a, string b, Dictionary<string, string> dp, Dictionary<string, int> vs)
    {
        int n_a = a.Length;
        int n_b = b.Length;

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

        // dp index to access the dp dictionary
        string dp_i = a + "," + b;

        // ans points to the memory location in the dp
        // dictionary in which the solution string will be stored
        string ans;
        if (dp.TryGetValue(dp_i, out ans))
        {
            // If visited return solution from memory
            return ans;
        }

        // If not visited, set the visit value to be one
        // (meaning it's now visited)
        vs[dp_i] = 1;

        // If the last two characters match
        if (a[n_a - 1] == b[n_b - 1])
        {
            // Add this character to the solution string
            ans = a[n_a - 1] + LCS_Core(a.Substring(0, n_a - 1), b.Substring(0, n_b - 1), dp, vs);
            dp[dp_i] = ans;
            return ans;
        }

        // Return longest string
        ans = MaxStr(LCS_Core(a.Substring(0, n_a - 1), b, dp, vs), LCS_Core(a, b.Substring(0, n_b - 1), dp, vs));
        dp[dp_i] = ans;
        return ans;
    }

    static string LCS(string a, string b)
    {
        Dictionary<string, string> dp = new Dictionary<string, string>();
        Dictionary<string, int> vs = new Dictionary<string, int>();

        string ans = LCS_Core(a, b, dp, vs);

        // Reverse the obtained result
        return Reverse(ans);
    }
}

}

` JavaScript ``

// Initialize an empty dictionary to store the solutions of subproblems let dp = {}; // Initialize an empty dictionary to keep track of visited subproblems let vs = {};

// Utility function to reverse a string, we need it because our top-down approach // return a reversed solution function reverse(s) { let ans = s.split(""); let u = 0; let v = ans.length - 1; while (u < v) { [ans[u], ans[v]] = [ans[v], ans[u]]; // swap operation u++; v--; } return ans.join(""); }

// Utility function that compares two strings and return the longer in size. function maxStr(a, b) { return a.length > b.length ? a : b; }

// Recursive function that takes two strings as input, // and returns the LCS of them function LCS_core(a, b) { // Base case if (!a || !b) return "";

// dp index to access the dp structure
let 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[a.length - 1] === b[b.length - 1]) {
    let ans = a[a.length - 1] + LCS_core(a.slice(0, -1), b.slice(0, -1));
    dp[dp_i] = ans;
    return ans;
}

// Return longest string
let ans = maxStr(LCS_core(a.slice(0, -1), b), LCS_core(a, b.slice(0, -1)));
dp[dp_i] = ans;
return ans;

}

// Final wrapper function to call the // recursive function and reverse the result function LCS(a, b) { return reverse(LCS_core(a, b)); }

// Driver Code const a = "AGGTAB"; const b = "GXTXAYB"; console.log(LCS(a, b));

``

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).