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# 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])

## Python3

 `# 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)))`

Output:

`Length of LCS is  4`

Time Complexity: O(2n)

Auxiliary Space: O(n)

Following is a tabulated implementation for the LCS problem.

## Python3

 `# 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` `range``(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) `

Output:

`Length of LCS is  4`

Time Complexity: O(n*m)

Auxiliary Space: O(n*m)

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

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