Longest Common Subsequence | DP using Memoization

Given two strings s1 and s2, the task is to find the length of the longest common subsequence present in both of them.

Examples:

Input: s1 = “ABCDGH”, s2 = “AEDFHR” 
Output: 3 
LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4.

Input: s1 = “striver”, s2 = “raj” 
Output: 1  

The naive solution for this problem is to generate all subsequences of both given sequences and find the longest matching subsequence. This solution is exponential in terms of time complexity. The general recursive solution of the problem is to generate all subsequences of both given sequences and find the longest matching subsequence. The total possible combinations will be 2ⁿ. Hence, the recursive solution will take O(2ⁿ). 

Optimal Substructure:

• Let the input sequences are X[0… m-1] and Y[0…n-1] of lengths m and n respectively. And let L(X[0… m-1], Y[0…n-1]) be the length of LCS of the two sequences X and Y. Following is the recursive definition of L(X[0… m-1], Y[0…n-1]).
• If last characters of both sequences match (or X[m-1] == Y[n-1]) then L(X[0… m-1], Y[0…n-1]) = 1 + L(X[0… m-2], Y[0…n-2])
• If last characters of both sequences do not match (or X[m-1] != Y[n-1]) then L(X[0… m-1], Y[0…n-1]) = MAX (L(X[0… m-2], Y[0…n-1]), L(X[0… m-1], Y[0…n-2])

Given below is the recursive solution to the LCS problem: 

Implementation:

C++

// A Naive C++ recursive implementation
// of LCS of two strings
#include <bits/stdc++.h>
using namespace std;

// Returns length of LCS for X[0..m-1], Y[0..n-1]
int lcs(string X, string Y, int m, int n)
{
    if (m == 0 || n == 0)
        return 0;

    if (X[m - 1] == Y[n - 1])
        return 1 + lcs(X, Y, m - 1, n - 1);
    else
        return max(lcs(X, Y, m, n - 1),
                   lcs(X, Y, m - 1, n));
}

// Driver Code
int main()
{
    string X = "AGGTAB";
    string Y = "GXTXAYB";

    // Find the length of string
    int m = X.length();
    int n = Y.length();

    cout << "Length of LCS: " << lcs(X, Y, m, n);

    return 0;
}

Java

// A Naive Java recursive implementation
// of LCS of two strings

class GFG {

// Returns length of LCS for X[0..m-1], Y[0..n-1]
    static int lcs(String X, String Y, int m, int n) {
        if (m == 0 || n == 0) {
            return 0;
        }

        if (X.charAt(m - 1) == Y.charAt(n - 1)) {
            return 1 + lcs(X, Y, m - 1, n - 1);
        } else {
            return Math.max(lcs(X, Y, m, n - 1),
                    lcs(X, Y, m - 1, n));
        }
    }
// Driver Code

    public static void main(String[] args) {
        String X = "AGGTAB";
        String Y = "GXTXAYB";

        // Find the length of String
        int m = X.length();
        int n = Y.length();
        System.out.println("Length of LCS: " + lcs(X, Y, m, n));

    }
}

// This code is contributed by 29AjayKumar

Python3

# A Naive Python recursive implementation
# of LCS of two strings

# Returns length of LCS for X[0..m-1], Y[0..n-1]
def lcs(X, Y, m, n):
    if (m == 0 or n == 0):
        return 0

    if (X[m - 1] == Y[n - 1]):
        return 1 + lcs(X, Y, m - 1, n - 1)
    else:
        return max(lcs(X, Y, m, n - 1),
                   lcs(X, Y, m - 1, n))

# Driver Code
if __name__ == '__main__':
    X = "AGGTAB"
    Y = "GXTXAYB"

    # Find the length of string
    m = len(X)
    n = len(Y)

    print("Length of LCS:",
           lcs(X, Y, m, n))

# This code is contributed by 29AjayKumar

C#

// A Naive recursive C#implementation of
// LCS of two strings
using System;

class GFG
{

// Returns length of LCS for
// X[0..m-1], Y[0..n-1]
static int lcs(String X, String Y,
               int m, int n)
{
    if (m == 0 || n == 0)
    {
        return 0;
    }

    if (X[m - 1] == Y[n - 1])
    {
        return 1 + lcs(X, Y, m - 1, n - 1);
    } else {
        return Math.Max(lcs(X, Y, m, n - 1),
                        lcs(X, Y, m - 1, n));
    }
}

// Driver Code
public static void Main()
{
    String X = "AGGTAB";
    String Y = "GXTXAYB";

    // Find the length of String
    int m = X.Length;
    int n = Y.Length;
    Console.Write("Length of LCS: " +
                    lcs(X, Y, m, n));
}
}

// This code is contributed by 29AjayKumar

PHP

<?php
// A Naive PHP recursive implementation
// of LCS of two strings

// Returns length of LCS for
// X[0..m-1], Y[0..n-1]
function lcs($X, $Y, $m, $n)
{
    if ($m == 0 || $n == 0)
        return 0;

    if ($X[$m - 1] == $Y[$n - 1])
        return 1 + lcs($X, $Y,
                       $m - 1, $n - 1);
    else
        return max(lcs($X, $Y, $m, $n - 1),
                   lcs($X, $Y, $m - 1, $n));
}

// Driver Code
$X = "AGGTAB";
$Y = "GXTXAYB";

// Find the length of string
$m = strlen($X);
$n = strlen($Y);

echo "Length of LCS: " . lcs($X, $Y, $m, $n);

// This code is contributed by ita_c
?>

Javascript

<script>
// A Naive Javascript recursive implementation
// of LCS of two strings

    // Returns length of LCS for X[0..m-1], Y[0..n-1]
    function lcs(X,Y,m,n)
    {
        if (m == 0 || n == 0) {
            return 0;
        }

        if (X[m-1] == Y[n-1]) {
            return 1 + lcs(X, Y, m - 1, n - 1);
        } else {
            return Math.max(lcs(X, Y, m, n - 1),
                    lcs(X, Y, m - 1, n));
        }
    }

    // Driver Code
    let X = "AGGTAB";
    let Y = "GXTXAYB";
    // Find the length of String
    let m = X.length;
    let n = Y.length;
    document.write("Length of LCS: " + lcs(X, Y, m, n));


// This code is contributed by rag2127
</script>

Output

Length of LCS: 4

Dynamic Programming using Memoization

Considering the above implementation, the following is a partial recursion tree for input strings “AXYT” and “AYZX”

                       lcs("AXYT", "AYZX")
                       /                 \
         lcs("AXY", "AYZX")            lcs("AXYT", "AYZ")
         /           \                   /               \
lcs("AX", "AYZX") lcs("AXY", "AYZ")   lcs("AXY", "AYZ") lcs("AXYT", "AY")

In the above partial recursion tree, lcs(“AXY”, “AYZ”) are being solved twice. On drawing the complete recursion tree, it has been observed that there are many subproblems that are solved again and again. So this problem has Overlapping Substructure property and recomputation of same subproblems can be avoided by either using Memoization or Tabulation. The tabulation method has been discussed here.
A common point of observation to use memoization in the recursive code will be the two non-constant arguments M and N in every function call. The function has 4 arguments, but 2 arguments are constant which does not affect the Memoization. The repetitive calls occur for N and M which have been called previously. Following the below steps will help us to write the DP solution using memoization. 

• Use a 2-D array to store the computed lcs(m, n) value at arr[m-1][n-1] as the string index starts from 0.
• Whenever the function with the same argument m and n are called again, do not perform any further recursive call and return arr[m-1][n-1] as the previous computation of the lcs(m, n) has already been stored in arr[m-1][n-1], hence reducing the recursive calls that happen more than once.

Below is the implementation of the above approach: 

C++

// C++ program to memoize
// recursive implementation of LCS problem
#include <bits/stdc++.h>
using namespace std;

const int maximum = 1000;

// Returns length of LCS for X[0..m-1], Y[0..n-1] */
// memoization applied in recursive solution
int lcs(string X, string Y, int m, int n, int dp[][maximum])
{
    // base case
    if (m == 0 || n == 0)
        return 0;

    // if the same state has already been
    // computed
    if (dp[m - 1][n - 1] != -1)
        return dp[m - 1][n - 1];

    // if equal, then we store the value of the
    // function call
    if (X[m - 1] == Y[n - 1]) {

        // store it in arr to avoid further repetitive
        // work in future function calls
        dp[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1, dp);

        return dp[m - 1][n - 1];
    }
    else {

        // store it in arr to avoid further repetitive
        // work in future function calls
        dp[m - 1][n - 1] = max(lcs(X, Y, m, n - 1, dp),
                               lcs(X, Y, m - 1, n, dp));

        return dp[m - 1][n - 1];
    }
}

// Driver Code
int main()
{

    string X = "AGGTAB";
    string Y = "GXTXAYB";
    int m = X.length();
    int n = Y.length();

    int dp[m][maximum];

    // assign -1 to all positions
    memset(dp, -1, sizeof(dp));

    cout << "Length of LCS: " << lcs(X, Y, m, n, dp);

    return 0;
}

Java

import java.util.Arrays;

// Java program to memoize
// recursive implementation of LCS problem
class GFG {

    static final int maximum = 1000;

// Returns length of LCS for X[0..m-1], Y[0..n-1] */
// memoization applied in recursive solution
    static int lcs(String X, String Y, int m, int n, int dp[][]) {
        // base case
        if (m == 0 || n == 0) {
            return 0;
        }

        // if the same state has already been
        // computed
        if (dp[m - 1][n - 1] != -1) {
            return dp[m - 1][n - 1];
        }

        // if equal, then we store the value of the
        // function call
        if (X.charAt(m - 1) == Y.charAt(n - 1)) {

            // store it in arr to avoid further repetitive
            // work in future function calls
            dp[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1, dp);

            return dp[m - 1][n - 1];
        } else {

            // store it in arr to avoid further repetitive
            // work in future function calls
            dp[m - 1][n - 1] = Math.max(lcs(X, Y, m, n - 1, dp),
                    lcs(X, Y, m - 1, n, dp));

            return dp[m - 1][n - 1];
        }
    }

// Driver Code
    public static void main(String[] args) {
        String X = "AGGTAB";
        String Y = "GXTXAYB";
        int m = X.length();
        int n = Y.length();

        int dp[][] = new int[m][maximum];

        // assign -1 to all positions
        for (int[] row : dp) {
            Arrays.fill(row, -1);
        }

        System.out.println("Length of LCS: " + lcs(X, Y, m, n, dp));
    }
}
/* This Java code is contributed by 29AjayKumar*/

Python3

# Python3 program to memoize
# recursive implementation of LCS problem
maximum = 1000

# Returns length of LCS for X[0..m-1], Y[0..n-1] */
# memoization applied in recursive solution
def lcs(X, Y, m, n, dp):

    # base case
    if (m == 0 or n == 0):
        return 0

    # if the same state has already been
    # computed
    if (dp[m - 1][n - 1] != -1):
        return dp[m - 1][n - 1]

    # if equal, then we store the value of the
    # function call
    if (X[m - 1] == Y[n - 1]):

        # store it in arr to avoid further repetitive
        # work in future function calls
        dp[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1, dp)

        return dp[m - 1][n - 1]

    else :

        # store it in arr to avoid further repetitive
        # work in future function calls
        dp[m - 1][n - 1] = max(lcs(X, Y, m, n - 1, dp),
                               lcs(X, Y, m - 1, n, dp))

        return dp[m - 1][n - 1]

# Driver Code
X = "AGGTAB"
Y = "GXTXAYB"
m = len(X)
n = len(Y)

dp = [[-1 for i in range(maximum)]
          for i in range(m)]

print("Length of LCS:", lcs(X, Y, m, n, dp))

# This code is contributed by Mohit Kumar

C#

// C# program to memoize
// recursive implementation of LCS problem
using System;

class GFG
{
static readonly int maximum = 1000;

// Returns length of LCS for
// X[0..m-1], Y[0..n-1]
// memoization applied in
// recursive solution
static int lcs(String X, String Y,
               int m, int n, int [,]dp)
{
    // base case
    if (m == 0 || n == 0)
    {
        return 0;
    }

    // if the same state has already been
    // computed
    if (dp[m - 1, n - 1] != -1)
    {
        return dp[m - 1, n - 1];
    }

    // if equal, then we store the value
    // of the function call
    if (X[m - 1] == Y[n - 1])
    {

        // store it in arr to avoid
        // further repetitive work
        // in future function calls
        dp[m - 1,
           n - 1] = 1 + lcs(X, Y, m - 1,
                                  n - 1, dp);

        return dp[m - 1, n - 1];
    }
    else
    {

        // store it in arr to avoid
        // further repetitive work
        // in future function calls
        dp[m - 1,
           n - 1] = Math.Max(lcs(X, Y, m, n - 1, dp),
                             lcs(X, Y, m - 1, n, dp));

        return dp[m - 1, n - 1];
    }
}

// Driver Code
public static void Main(String[] args)
{
    String X = "AGGTAB";
    String Y = "GXTXAYB";
    int m = X.Length;
    int n = Y.Length;

    int [,]dp = new int[m, maximum];

    // assign -1 to all positions
    for(int i = 0; i < m; i++)
    {
        for(int j = 0; j < maximum; j++)
        {
            dp[i, j] = -1;
        }
    }
    Console.WriteLine("Length of LCS: " +
                    lcs(X, Y, m, n, dp));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript program to memoize
// recursive implementation of LCS problem

let maximum = 1000;

// Returns length of LCS for X[0..m-1], Y[0..n-1] */
// memoization applied in recursive solution
function lcs(X,Y,m,n,dp)
{
    // base case
        if (m == 0 || n == 0) {
            return 0;
        }

        // if the same state has already been
        // computed
        if (dp[m - 1][n - 1] != -1) {
            return dp[m - 1][n - 1];
        }

        // if equal, then we store the value of the
        // function call
        if (X[m-1] == Y[n-1]) {

            // store it in arr to avoid further repetitive
            // work in future function calls
            dp[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1, dp);

            return dp[m - 1][n - 1];
        } else {

            // store it in arr to avoid further repetitive
            // work in future function calls
            dp[m - 1][n - 1] = Math.max(lcs(X, Y, m, n - 1, dp),
                    lcs(X, Y, m - 1, n, dp));

            return dp[m - 1][n - 1];
        }
}

// Driver Code
let X = "AGGTAB";
let Y = "GXTXAYB";
let m = X.length;
let n = Y.length;

let dp=new Array(m);
for(let i=0;i<dp.length;i++)
{
    dp[i]=new Array(maximum);
    for(let j=0;j<dp[i].length;j++)
    {
        dp[i][j]=-1;
    }
}

document.write("Length of LCS: " + lcs(X, Y, m, n, dp));



// This code is contributed by avanitrachhadiya2155
</script>

Output

Length of LCS: 4

complexity Analysis:

• Time Complexity: O(N * M), where N and M are lengths of the first and second string respectively. 
• Auxiliary Space: (N * M)