Python Program for Longest Common Subsequence

• Difficulty Level : Medium
• Last Updated : 18 Apr, 2020

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 functionX = "AGGTAB"Y = "GXTXAYB"print ("Length of LCS is ", lcs(X, Y, len(X), len(Y)))
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 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 functionX = "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

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

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