# Python Program for Longest Common Subsequence

*LCS Problem Statement:* Given two sequences, find the length of longest subsequence present in both of them. A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous. For example, “abc”, “abg”, “bdf”, “aeg”, ‘”acefg”, .. etc are subsequences of “abcdefg”. So a string of length n has 2^n different possible subsequences.

It is a classic computer science problem, the basis of diff (a file comparison program that outputs the differences between two files), and has applications in bioinformatics.

**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.

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])

`# A Naive recursive Python implementation of LCS problem ` ` ` `def` `lcs(X, Y, m, n): ` ` ` ` ` `if` `m ` `=` `=` `0` `or` `n ` `=` `=` `0` `: ` ` ` `return` `0` `; ` ` ` `elif` `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 program to test the above function ` `X ` `=` `"AGGTAB"` `Y ` `=` `"GXTXAYB"` `print` `"Length of LCS is "` `, lcs(X, Y, ` `len` `(X), ` `len` `(Y)) ` |

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**Output:**

Length of LCS is 4

Following is a tabulated implementation for the LCS problem.

`# Dynamic Programming implementation of LCS problem ` ` ` `def` `lcs(X, Y): ` ` ` `# find the length of the strings ` ` ` `m ` `=` `len` `(X) ` ` ` `n ` `=` `len` `(Y) ` ` ` ` ` `# declaring the array for storing the dp values ` ` ` `L ` `=` `[[` `None` `]` `*` `(n ` `+` `1` `) ` `for` `i ` `in` `xrange` `(m ` `+` `1` `)] ` ` ` ` ` `"""Following steps build L[m + 1][n + 1] in bottom up fashion ` ` ` `Note: 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` `]) ` ` ` ` ` `# L[m][n] contains the length of LCS of X[0..n-1] & Y[0..m-1] ` ` ` `return` `L[m][n] ` `# end of function lcs ` ` ` ` ` `# Driver program to test the above function ` `X ` `=` `"AGGTAB"` `Y ` `=` `"GXTXAYB"` `print` `"Length of LCS is "` `, lcs(X, Y) ` ` ` `# This code is contributed by Nikhil Kumar Singh(nickzuck_007) ` |

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**Output:**

Length of LCS is 4

Please refer complete article on Dynamic Programming | Set 4 (Longest Common Subsequence) for more details!

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