# LCS (Longest Common Subsequence) of three strings

• Difficulty Level : Medium
• Last Updated : 06 Jul, 2022

Given 3 strings of all having length < 100,the task is to find the longest common sub-sequence in all three given sequences.

Examples:

```Input : str1 = "geeks"
str2 = "geeksfor"
str3 = "geeksforgeeks"
Output : 5
Longest common subsequence is "geeks"
i.e., length = 5

Input : str1 = "abcd1e2"
str2 = "bc12ea"
str3 = "bd1ea"
Output : 3
Longest common subsequence is "b1e"
i.e. length = 3.```
Recommended Practice

This problem is simply an extension of LCS
Let the input sequences be X[0..m-1], Y[0..n-1] and Z[0..o-1] of lengths m, n and o respectively. And let L(X[0..m-1], Y[0..n-1], Z[0..o-1]) be the lengths of LCS of the three sequences X, Y and Z.

Following is the implementation:

```The idea is to take a 3D array to store the
length of common subsequence in all 3 given
sequences i. e., L[m + 1][n + 1][o + 1]

1- If any of the string is empty then there
is no common subsequence at all then
L[i][j][k] = 0

2- If the characters of all sequences match
(or X[i] == Y[j] ==Z[k]) then
L[i][j][k] = 1 + L[i-1][j-1][k-1]

3- If the characters of both sequences do
not match (or X[i] != Y[j] || X[i] != Z[k]
|| Y[j] !=Z[k]) then
L[i][j][k] = max(L[i-1][j][k],
L[i][j-1][k],
L[i][j][k-1])```

Below is implementation of above idea.

## C++

 `// C++ program to find LCS of three strings``#include``using` `namespace` `std;` `/* Returns length of LCS for X[0..m-1], Y[0..n-1]``   ``and Z[0..o-1] */``int` `lcsOf3( string X, string Y, string Z, ``int` `m,``                               ``int` `n, ``int` `o)``{``    ``int` `L[m+1][n+1][o+1];` `    ``/* Following steps build L[m+1][n+1][o+1] in``       ``bottom up fashion. Note that L[i][j][k]``       ``contains length of LCS of X[0..i-1] and``       ``Y[0..j-1]  and Z[0.....k-1]*/``    ``for` `(``int` `i=0; i<=m; i++)``    ``{``        ``for` `(``int` `j=0; j<=n; j++)``        ``{``            ``for` `(``int` `k=0; k<=o; k++)``            ``{``                ``if` `(i == 0 || j == 0||k==0)``                    ``L[i][j][k] = 0;` `                ``else` `if` `(X[i-1] == Y[j-1] && X[i-1]==Z[k-1])``                    ``L[i][j][k] = L[i-1][j-1][k-1] + 1;` `                ``else``                    ``L[i][j][k] = max(max(L[i-1][j][k],``                                         ``L[i][j-1][k]),``                                     ``L[i][j][k-1]);``            ``}``        ``}``    ``}` `    ``/* L[m][n][o] contains length of LCS for``      ``X[0..n-1] and Y[0..m-1] and Z[0..o-1]*/``    ``return` `L[m][n][o];``}` `/* Driver program to test above function */``int` `main()``{``    ``string X = ``"AGGT12"``;``    ``string Y = ``"12TXAYB"``;``    ``string Z = ``"12XBA"``;` `    ``int` `m = X.length();``    ``int` `n = Y.length();``    ``int` `o = Z.length();` `    ``cout << ``"Length of LCS is "` `<< lcsOf3(X, Y,``                                    ``Z, m, n, o);` `    ``return` `0;``}`

## Java

 `// Java program to find LCS of three strings``public` `class` `LCS_3Strings {``     ` `    ``/* Returns length of LCS for X[0..m-1], Y[0..n-1]``       ``and Z[0..o-1] */``    ``static` `int` `lcsOf3(String X, String Y, String Z, ``int` `m,``                                   ``int` `n, ``int` `o)``    ``{``        ``int``[][][] L = ``new` `int``[m+``1``][n+``1``][o+``1``];``     ` `        ``/* Following steps build L[m+1][n+1][o+1] in``           ``bottom up fashion. Note that L[i][j][k]``           ``contains length of LCS of X[0..i-1] and``           ``Y[0..j-1]  and Z[0.....k-1]*/``        ``for` `(``int` `i=``0``; i<=m; i++)``        ``{``            ``for` `(``int` `j=``0``; j<=n; j++)``            ``{``                ``for` `(``int` `k=``0``; k<=o; k++)``                ``{``                    ``if` `(i == ``0` `|| j == ``0``||k==``0``)``                        ``L[i][j][k] = ``0``;``     ` `                    ``else` `if` `(X.charAt(i - ``1``) == Y.charAt(j - ``1``)``                                ``&& X.charAt(i - ``1``)==Z.charAt(k - ``1``))``                        ``L[i][j][k] = L[i-``1``][j-``1``][k-``1``] + ``1``;``     ` `                    ``else``                        ``L[i][j][k] = Math.max(Math.max(L[i-``1``][j][k],``                                             ``L[i][j-``1``][k]),``                                         ``L[i][j][k-``1``]);``                ``}``            ``}``        ``}``     ` `        ``/* L[m][n][o] contains length of LCS for``          ``X[0..n-1] and Y[0..m-1] and Z[0..o-1]*/``        ``return` `L[m][n][o];``    ``}``     ` `    ``/* Driver program to test above function */``    ``public` `static` `void` `main(String args[])``    ``{``        ``String X = ``"AGGT12"``;``        ``String Y = ``"12TXAYB"``;``        ``String Z = ``"12XBA"``;``     ` `        ``int` `m = X.length();``        ``int` `n = Y.length();``        ``int` `o = Z.length();``     ` `        ``System.out.println(``"Length of LCS is "` `+``                                ``lcsOf3(X, Y,Z, m, n, o));``     ` `    ``}``}``// This code is contributed by Sumit Ghosh`

## Python3

 `# Python program to find``# LCS of three strings` `# Returns length of LCS``# for X[0..m-1], Y[0..n-1]``# and Z[0..o-1]``def` `lcsOf3(X, Y, Z, m, n, o):``    ` `    ``L ``=` `[[[``0` `for` `i ``in` `range``(o``+``1``)] ``for` `j ``in` `range``(n``+``1``)]``         ``for` `k ``in` `range``(m``+``1``)]` `    ``''' Following steps build L[m+1][n+1][o+1] in``    ``bottom up fashion. Note that L[i][j][k]``    ``contains length of LCS of X[0..i-1] and``    ``Y[0..j-1] and Z[0.....k-1] '''``    ``for` `i ``in` `range``(m``+``1``):``        ``for` `j ``in` `range``(n``+``1``):``            ``for` `k ``in` `range``(o``+``1``):``                ``if` `(i ``=``=` `0` `or` `j ``=``=` `0` `or` `k ``=``=` `0``):``                    ``L[i][j][k] ``=` `0``                    ` `                ``elif` `(X[i``-``1``] ``=``=` `Y[j``-``1``] ``and``                      ``X[i``-``1``] ``=``=` `Z[k``-``1``]):``                    ``L[i][j][k] ``=` `L[i``-``1``][j``-``1``][k``-``1``] ``+` `1` `                ``else``:``                    ``L[i][j][k] ``=` `max``(``max``(L[i``-``1``][j][k],``                    ``L[i][j``-``1``][k]),``                                    ``L[i][j][k``-``1``])` `    ``# L[m][n][o] contains length of LCS for``    ``# X[0..n-1] and Y[0..m-1] and Z[0..o-1]``    ``return` `L[m][n][o]` `# Driver program to test above function` `X ``=` `'AGGT12'``Y ``=` `'12TXAYB'``Z ``=` `'12XBA'` `m ``=` `len``(X)``n ``=` `len``(Y)``o ``=` `len``(Z)` `print``(``'Length of LCS is'``, lcsOf3(X, Y, Z, m, n, o))` `# This code is contributed by Soumen Ghosh.                   `

## C#

 `// C# program to find``// LCS of three strings``using` `System;` `class` `GFG``{``    ` `    ``/* Returns length of LCS``    ``for X[0..m-1], Y[0..n-1]``    ``and Z[0..o-1] */``    ``static` `int` `lcsOf3(String X, String Y,``                      ``String Z, ``int` `m,``                      ``int` `n, ``int` `o)``    ``{``        ``int` `[,,]L = ``new` `int``[m + 1,``                            ``n + 1, o + 1];``    ` `        ``/* Following steps build``        ``L[m+1][n+1][o+1] in bottom``        ``up fashion. Note that``        ``L[i][j][k] contains length``        ``of LCS of X[0..i-1] and``        ``Y[0..j-1] and Z[0.....k-1]*/``        ``for` `(``int` `i = 0; i <= m; i++)``        ``{``            ``for` `(``int` `j = 0; j <= n; j++)``            ``{``                ``for` `(``int` `k = 0; k <= o; k++)``                ``{``                    ``if` `(i == 0 ||``                        ``j == 0 || k == 0)``                        ``L[i, j, k] = 0;``    ` `                    ``else` `if` `(X[i - 1] == Y[j - 1] &&``                             ``X[i - 1] == Z[k - 1])``                        ``L[i, j, k] = L[i - 1,``                                       ``j - 1,``                                       ``k - 1] + 1;``    ` `                    ``else``                        ``L[i, j, k] = Math.Max(Math.Max(L[i - 1, j, k],``                                                       ``L[i, j - 1, k]),``                                                       ``L[i, j, k - 1]);``                ``}``            ``}``        ``}``    ` `        ``/* L[m][n][o] contains length``        ``of LCS for X[0..n-1] and``        ``Y[0..m-1] and Z[0..o-1]*/``        ``return` `L[m, n, o];``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``string` `X = ``"AGGT12"``;``        ``string` `Y = ``"12TXAYB"``;``        ``string` `Z = ``"12XBA"``;``    ` `        ``int` `m = X.Length;``        ``int` `n = Y.Length;``        ``int` `o = Z.Length;``    ` `        ``Console.Write(``"Length of LCS is "` `+``                       ``lcsOf3(X, Y, Z, m, n, o));``    ``}``}` `// This code is contributed``// by shiv_bhakt.`

## PHP

 ``

## Javascript

 ``

Output

`Length of LCS is 2`

Another approach: (Using recursion)

## C++

 `// C++ program to find LCS of three strings``#include``using` `namespace` `std;` `    ``string X = ``"AGGT12"``;``    ``string Y = ``"12TXAYB"``;``    ``string Z = ``"12XBA"``;` `int` `dp;` `/* Returns length of LCS for X[0..m-1], Y[0..n-1]``and Z[0..o-1] */``int` `lcsOf3(``int` `i, ``int` `j,``int` `k)``{``    ``if``(i==-1||j==-1||k==-1)``        ``return` `0;``    ``if``(dp[i][j][k]!=-1)``        ``return` `dp[i][j][k];``    ` `    ``if``(X[i]==Y[j] && Y[j]==Z[k])``        ``return` `dp[i][j][k] = 1+lcsOf3(i-1,j-1,k-1);``    ``else``        ``return` `dp[i][j][k] = max(max(lcsOf3(i-1,j,k),``                            ``lcsOf3(i,j-1,k)),lcsOf3(i,j,k-1));``}` `// Driver code``int` `main()``{``    ``memset``(dp, -1,``sizeof``(dp));``    ``int` `m = X.length();``    ``int` `n = Y.length();``    ``int` `o = Z.length();` `    ``cout << ``"Length of LCS is "` `<< lcsOf3(m-1,n-1,o-1);``// this code is contributed by Kushdeep Mittal``}`

## Java

 `// Java program to find LCS of three strings``class` `GFG``{` `    ``static` `String X = ``"AGGT12"``;``    ``static` `String Y = ``"12TXAYB"``;``    ``static` `String Z = ``"12XBA"``;` `    ``static` `int``[][][] dp = ``new` `int``[``100``][``100``][``100``];` `    ``/* Returns length of LCS for X[0..m-1],``    ``Y[0..n-1] and Z[0..o-1] */``    ``static` `int` `lcsOf3(``int` `i, ``int` `j, ``int` `k)``    ``{``        ``if` `(i == -``1` `|| j == -``1` `|| k == -``1``)``        ``{``            ``return` `0``;``        ``}``        ``if` `(dp[i][j][k] != -``1``)``        ``{``            ``return` `dp[i][j][k];``        ``}` `        ``if` `(X.charAt(i) == Y.charAt(j) &&``            ``Y.charAt(j) == Z.charAt(k))``        ``{``            ``return` `dp[i][j][k] = ``1` `+ lcsOf3(i - ``1``, j - ``1``, k - ``1``);``        ``} ``else` `{``            ``return` `dp[i][j][k] = Math.max(Math.max(lcsOf3(i - ``1``, j, k),``                                ``lcsOf3(i, j - ``1``, k)),``                                ``lcsOf3(i, j, k - ``1``));``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``for` `(``int` `i = ``0``; i < ``100``; i++)``        ``{``            ``for` `(``int` `j = ``0``; j < ``100``; j++)``            ``{``                ``for` `(``int` `k = ``0``; k < ``100``; k++)``                ``{``                    ``dp[i][j][k] = -``1``;``                ``}``            ``}``        ``}``        ``int` `m = X.length();``        ``int` `n = Y.length();``        ``int` `o = Z.length();` `        ``System.out.print(``"Length of LCS is "``                ``+ lcsOf3(m - ``1``, n - ``1``, o - ``1``));``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Python3

 `# Python3 program to find LCS of``# three strings``X ``=` `"AGGT12"``Y ``=` `"12TXAYB"``Z ``=` `"12XBA"` `dp ``=` `[[[``-``1` `for` `i ``in` `range``(``100``)]``           ``for` `j ``in` `range``(``100``)]``           ``for` `k ``in` `range``(``100``)]``        ` `# Returns length of LCS for``# X[0..m-1], Y[0..n-1] and Z[0..o-1]``def` `lcsOf3(i, j, k) :` `    ``if``(i ``=``=` `-``1` `or` `j ``=``=` `-``1` `or` `k ``=``=` `-``1``) :``        ``return` `0``        ` `    ``if``(dp[i][j][k] !``=` `-``1``) :``        ``return` `dp[i][j][k]``    ` `    ``if``(X[i] ``=``=` `Y[j] ``and` `Y[j] ``=``=` `Z[k]) :``        ``dp[i][j][k] ``=` `1` `+` `lcsOf3(i ``-` `1``,``                                 ``j ``-` `1``, k ``-` `1``)``        ``return` `dp[i][j][k]``        ` `    ``else` `:``        ``dp[i][j][k] ``=` `max``(``max``(lcsOf3(i ``-` `1``, j, k),``                              ``lcsOf3(i, j ``-` `1``, k)),``                              ``lcsOf3(i, j, k ``-` `1``))``        ` `        ``return` `dp[i][j][k]` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:``    ``m ``=` `len``(X)``    ``n ``=` `len``(Y)``    ``o ``=` `len``(Z)` `    ``print``(``"Length of LCS is"``,``           ``lcsOf3(m ``-` `1``, n ``-` `1``, o ``-` `1``))``    ` `# This code is contributed by Ryuga`

## C#

 `// C# program to find LCS of three strings``using` `System;` `class` `GFG``{``static` `string` `X = ``"AGGT12"``;``static` `string` `Y = ``"12TXAYB"``;``static` `string` `Z = ``"12XBA"``;` `static` `int``[,,] dp = ``new` `int``[100, 100, 100];` `/* Returns length of LCS for X[0..m-1],``Y[0..n-1] and Z[0..o-1] */``static` `int` `lcsOf3(``int` `i, ``int` `j, ``int` `k)``{``    ``if``(i == -1 || j == -1 || k == -1)``        ``return` `0;``    ``if``(dp[i, j, k] != -1)``        ``return` `dp[i, j, k];``    ` `    ``if``(X[i] == Y[j] && Y[j] == Z[k])``        ``return` `dp[i, j, k] = 1 + lcsOf3(i - 1, j - 1, k - 1);``    ``else``        ``return` `dp[i, j, k] = Math.Max(Math.Max(lcsOf3(i - 1, j, k),``                                               ``lcsOf3(i, j - 1, k)),``                                               ``lcsOf3(i, j, k - 1));``}` `// Driver code``static` `void` `Main()``{``    ``for``(``int` `i = 0; i < 100; i++)``        ``for``(``int` `j = 0; j < 100; j++)``            ``for``(``int` `k = 0; k < 100; k++)``                ``dp[i, j, k] = -1;``    ``int` `m = X.Length;``    ``int` `n = Y.Length;``    ``int` `o = Z.Length;` `    ``Console.Write(``"Length of LCS is "` `+``                   ``lcsOf3(m - 1, n - 1, o - 1));``}``}` `// This code is contributed by DrRoot_`

## PHP

 ``

## Javascript

 ``

Output

`Length of LCS is 2`

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

My Personal Notes arrow_drop_up