competitive-programming/2023_11_09/longest-common-subsequence/README.md

Technique

The technique used here is a bottom-up dynamic programming approach. The main steps are as follows:

1. Initialization: A 2D DP table dp is created with dimensions (text1.len() + 1) x (text2.len() + 1). The extra row and column are for the base case when one of the strings is empty.

2. Conversion to Character Vectors: The strings text1 and text2 are converted to vectors of characters. This is done because strings in Rust are not indexable due to their UTF-8 encoding. Converting them to character vectors allows for easier indexing.

3. Filling the DP Table: The outer loop iterates over text1 in reverse order, and the inner loop iterates over text2 in reverse order. For each pair of characters text1[i] and text2[j], it checks if they are equal. If they are equal, it increments the length of the current longest common subsequence, which is stored in dp[i+1][j+1], by 1 and stores it in dp[i][j]. If they are not equal, it takes the maximum length of the common subsequence found so far, which is the maximum of dp[i][j+1] and dp[i+1][j], and stores it in dp[i][j].

4. Result: After all iterations, dp[0][0] will have the length of the longest common subsequence of text1 and text2.

Time and Space Complexity Analysis

Time Complexity: The time complexity of this approach is O(m*n), where m and n are the lengths of text1 and text2, respectively. This is because each cell in the DP table is filled exactly once.

Space Complexity: The space complexity is also O(mn) due to the 2D DP table. Each cell in the table stores an integer, and there are (m+1)(n+1) cells in total.