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Printing Longest Common Subsequence

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  • Difficulty Level : Medium
  • Last Updated : 23 Jun, 2022

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.

 01234567
ØMZJAWXU
0Ø00000000
1X00000011
2M01111111
3J01122222
4Y01122222
5A01123333
6U01123334
7Z01223334

Following is the implementation of 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)
 


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