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 implementation of LCS problem */ #include <bits/stdc++.h> int max( int a, int b);
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */ int lcs( char * X, char * Y, int m, int n)
{ if (m == 0 || n == 0)
return 0;
if (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));
} /* Utility function to get max of 2 integers */ int max( int a, int b)
{ return (a > b) ? a : b;
} /* Driver program to test above function */ int main()
{ char X[] = "AGGTAB" ;
char Y[] = "GXTXAYB" ;
int m = strlen (X);
int n = strlen (Y);
printf ( "Length of LCS is %d\n" , lcs(X, Y, m, n));
return 0;
} |
Length of LCS is 4
Following is a tabulated implementation for the LCS problem.
/* Dynamic Programming C/C++ implementation of LCS problem */ #include <bits/stdc++.h> int max( int a, int b);
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */ int lcs( char * X, char * Y, int m, int n)
{ int L[m + 1][n + 1];
int i, j;
/* Following steps build L[m+1][n+1] in bottom up fashion. Note
that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
for (i = 0; i <= m; i++) {
for (j = 0; j <= n; j++) {
if (i == 0 || j == 0)
L[i][j] = 0;
else if (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 length of LCS for X[0..n-1] and Y[0..m-1] */
return L[m][n];
} /* Utility function to get max of 2 integers */ int max( int a, int b)
{ return (a > b) ? a : b;
} /* Driver program to test above function */ int main()
{ char X[] = "AGGTAB" ;
char Y[] = "GXTXAYB" ;
int m = strlen (X);
int n = strlen (Y);
printf ( "Length of LCS is %d\n" , lcs(X, Y, m, n));
return 0;
} |
Length of LCS is 4
Please refer complete article on Dynamic Programming | Set 4 (Longest Common Subsequence) for more details!