The Longest Common Subsequence (LCS) problem is a classic and fundamental problem in computer science and bioinformatics. It involves finding the longest subsequence that two sequences have in common. LCS has applications in various fields, including text comparison, DNA sequence alignment, and version control systems. Dynamic programming is the preferred approach for solving the LCS problem optimally. In this article, we will explore the Longest Common Subsequence (LCS) problem, understanding Dynamic Programming and its Time Complexity.

The Longest Common Subsequence Problem

The Longest Common Subsequence problem can be defined as follows: Given two sequences, X[1..m] and Y[1..n], we want to find the longest sequence Z[1..k] such that Z is a subsequence of both X and Y. In other words, Z may not necessarily be contiguous in X and Y, but its elements must maintain their relative order in both sequences.

For example, consider the sequences:

X: ABCBDAB
Y: BDCAB

The longest common subsequence for these sequences is “BCAB,” which has a length of 4.

Dynamic Programming Solution

Dynamic programming is the preferred approach for solving the LCS problem because it efficiently computes and stores intermediate results, avoiding redundant calculations. The key idea is to break down the problem into smaller subproblems, solve them independently, and then combine their solutions to find the longest common subsequence.

Steps for Dynamic Programming Solution:

  1. Define the Subproblem: To apply dynamic programming, we define a subproblem as finding the longest common subsequence for a prefix of sequences X and Y. We’ll create a table to store the lengths of LCS for different prefixes.
  2. Formulate the Recurrence Relation: The recurrence relation defines the length of the LCS for prefixes of sequences X and Y. For each pair of prefixes (X[1..i] and Y[1..j]), we consider different cases and use previously computed lengths to determine the current length.
  3. Fill in the Table Bottom-Up: Using a bottom-up approach, we start by solving subproblems for shorter prefixes and gradually build up to the full sequences. We fill in the table with the lengths of LCS for various prefixes.
  4. Reconstruct the LCS: After filling in the table, we can reconstruct the longest common subsequence by tracing back the steps from the bottom-right corner of the table to the top-left corner. This step helps us obtain the actual LCS.

Example

Let’s walk through a simple example to illustrate the dynamic programming solution for the LCS problem. Consider the sequences:

X: ABCBDAB
Y: BDCAB

Here are the steps:

  1. Define the Subproblem: We define subproblems for different prefixes of sequences X and Y. For example, we consider the subproblem of finding the LCS for prefixes X[1..i] and Y[1..j].
  2. Formulate the Recurrence Relation: The recurrence relation defines the length of the LCS for prefixes. If X[i] is equal to Y[j], we can extend the LCS by one and consider the prefixes X[1..(i-1)] and Y[1..(j-1)]. If they are not equal, we consider two cases: either we skip an element in X (i.e., consider X[1..(i-1)] and Y[1..j]) or we skip an element in Y (i.e., consider X[1..i] and Y[1..(j-1)]).
  3. Fill in the Table Bottom-Up: We start with subproblems involving prefixes of length 1 and fill in the table with the lengths of LCS for various prefixes. The final result, LCS[m][n], represents the length of the longest common subsequence for the full sequences.
  4. Reconstruct the LCS: To reconstruct the LCS, we start at the bottom-right corner of the table (LCS[m][n]) and trace back to the top-left corner. We follow the rules defined in the recurrence relation to determine which elements belong to the LCS.

Time Complexity

The dynamic programming solution for the LCS problem has a time complexity of O(m * n), where m and n are the lengths of the input sequences X and Y, respectively. This is because we need to compute and store the lengths of LCS for all possible pairs of prefixes of X and Y.

Conclusion

The Longest Common Subsequence (LCS) problem is a fundamental problem with applications in diverse fields. By leveraging dynamic programming, we can efficiently find the longest common subsequence of two sequences, even when they are of substantial length.


more related content onΒ Advanced AlgorithmsΒ (AA)

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