Given two strings x=x1 x2 ⋯xn and y=y1 y2 ⋯ym, we wish to...

Given two strings $x=x_{1} x_{2} \cdots x_{n}$ and $y=y_{1} y_{2} \cdots y_{m},$ we wish to find the length of their longest common subsequence, that is, the largest $k$ for which there are indices $i_{1}<i_{2}<\cdots<i_{k}$ and $j_{1}<j_{2}<\cdots<j_{k}$ with $x_{i_{1}} x_{i_{2}} \cdots x_{i_{k}}=y_{j_{1}} y_{j_{2}} \cdots y_{j_{k}},$ Show how to do this in time $O(m n)$

Key Concepts

Longest Common Subsequence (LCS)

The longest common subsequence is a sequence that appears in the same relative order (but not necessarily contiguously) in both of the given sequences. It represents the largest sequence that can be derived from both original sequences without reordering the elements, and is fundamental in fields like bioinformatics and text comparison.

Dynamic Programming

Dynamic Programming is an algorithmic technique used to solve complex problems by breaking them down into simpler subproblems and storing the results of these subproblems to avoid repeated computations. In the context of finding the LCS, it involves constructing a table that represents the solutions to subproblems, facilitating an efficient computation of the overall problem.

DP Table Construction

The DP table (or matrix) is built where each cell at position (i, j) represents the length of the LCS for the prefixes of the two strings up to positions i and j. The construction uses a recurrence relation that compares characters and determines whether to extend the subsequence or take the maximum from adjacent cells, leading to a time complexity of O(mn).

Time Complexity Analysis

Time complexity analysis involves determining the amount of time an algorithm takes relative to the input size. In the case of the LCS dynamic programming approach, the algorithm processes an m x n table, resulting in a time complexity of O(mn), which ensures the solution is efficient for the problem size given.

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