# Longest Common Subsequence | DP using Memoization

Given two strings s1 and s2, the task is to find the length of 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 term 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. Total possible combinations will be 2n. Hence recursive solution will take O(2n).

Optimal Substructure:

• Let the input sequences be 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:

## C++

 `// A Naive C++ recursive implementation ` `// of LCS of two strings ` `#include ` `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

 ` `

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”) is being solved twice. On drawing the complete recursion tree, it has been observed that there are many subproblems which 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 do 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 then once.

Below is the implementation of the above approach:

## C++

 `// C++ program to memoize ` `// recursive implementation of LCS problem ` `#include ` `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 `

Output:

```Length of LCS: 4
```

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

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