Java Program for Longest Common Subsequence
Last Updated :
08 Mar, 2023
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])
Java
public class LongestCommonSubsequence {
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));
}
int max( int a, int b)
{
return (a > b) ? a : b;
}
public static void main(String[] args)
{
LongestCommonSubsequence lcs = new LongestCommonSubsequence();
String s1 = "AGGTAB";
String s2 = "GXTXAYB";
char [] X = s1.toCharArray();
char [] Y = s2.toCharArray();
int m = X.length;
int n = Y.length;
System.out.println("Length of LCS is"
+ " " + lcs.lcs(X, Y, m, n));
}
}
|
Output:
Length of LCS is 4
Time Complexity: O(2^(N+M)) where N and M are the lengths of two input strings.
Space Complexity: O(N+M) for recursion stack used.
Following is a tabulated implementation for the LCS problem.
Java
public class LongestCommonSubsequence {
int lcs( char [] X, char [] Y, int m, int n)
{
int L[][] = new int [m + 1 ][n + 1 ];
for ( int i = 0 ; i <= m; i++) {
for ( int 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 ]);
}
}
return L[m][n];
}
int max( int a, int b)
{
return (a > b) ? a : b;
}
public static void main(String[] args)
{
LongestCommonSubsequence lcs = new LongestCommonSubsequence();
String s1 = "AGGTAB";
String s2 = "GXTXAYB";
char [] X = s1.toCharArray();
char [] Y = s2.toCharArray();
int m = X.length;
int n = Y.length;
System.out.println("Length of LCS is"
+ " " + lcs.lcs(X, Y, m, n));
}
}
|
Output:
Length of LCS is 4
Time Complexity: O(m*n) where m and n are lengths of two input strings.
Space Complexity: O(m*n) as 2d array has been created.
Please refer complete article on Dynamic Programming | Set 4 (Longest Common Subsequence) for more details!
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