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 problemdef 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))) |
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) |
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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