This document discusses the longest common subsequence problem and presents an algorithm for solving it using dynamic programming. Specifically, it introduces the longest common subsequence problem, describes how it is different from the substring problem, and presents a dynamic programming algorithm that uses a matrix to store the lengths of the longest common subsequences between prefixes of the two strings in order to efficiently find the longest common subsequence between the full strings.

1. The Longest Common
Subsequence Problem
CSE 373
Data Structures

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Common Subsequences
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Reading
Goodrich and Tamassia, 3rd ed, Chapter
12, section 11.5, pp.570-574.

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Common Subsequences
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Motivation
• Two Problems and Methods for String
Comparison:
› The substring problem
› The longest common subsequence
problem.
• In both cases, good algorithms do
substantially better than the brute force
methods.

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String Matching Problem
• Given two strings TEXT and PATTERN,
find the first occurrence of PATTERN in
TEXT.
• Useful in text editing, document
analysis, genome analysis, etc.

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String Matching Problem:
Brute-Force Algorithm
For i = 0 to n – m {
For j = 0 to m – 1 {
If TEXT[j]  PATTERN[i] then break
If j = m – 1 then return i
}
return -1;
}
Suppose
TEXT = 0000000000001
PATTERN = 0000001
This type of problem has (n2) behavior.
A more efficient algorithm is the Boyer-Moore algorithm. (We
will not be covering it in this course.)

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Longest Common
Subsequence Problem
• A Longest Common Subsequence LCS
of two strings S1 and S2 is a longest
string the can be obtained from S1 and
from S2 by deleting elements.
• For example, S1 = “thoughtful” and S2
= “shuffle” have an LCS: “hufl”.
• Useful in spelling correction, document
comparison, etc.

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Dynamic Programming
• Analyze the problem in terms of a
number of smaller subproblems.
• Solve the subproblems and keep their
answers in a table.
• Each subproblem’s answer is easily
computed from the answers to its own
subproblems.

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Longest Common Subsequence:
Algorithm using Dynamic Programming
• For every prefix of S1 and prefix of S2
we’ll compute the length L of an LCS.
• In the end, we’ll get the length of an
LCS for S1 and S2 themselves.
• The subsequence can be recovered
from the matrix of L values.
• (see demonstration)