longest common subsequence

2023_11_09/longest-common-subsequence/Cargo.toml

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+[package]
+name = "longest-common-subsequence"
+version = "0.1.0"
+edition = "2021"
+
+# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
+
+[dependencies]

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

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+## 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(m*n) due to the 2D DP table. Each cell in the table stores an integer, and there are (m+1)*(n+1) cells in total.

2023_11_09/longest-common-subsequence/src/main.rs

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+impl Solution {
+    pub fn longest_common_subsequence(text1: String, text2: String) -> i32 {
+        //  bottom-up dynamic programming approach
+        // we just created a 2D grid of length len(text1)+1 x len(text2) + 1 and initializing it to all zeros
+        let mut dp = vec![vec![0; text2.len() + 1]; text1.len() + 1];
+
+        //  converting the strings to character vectors is to allow for easier indexing. In Rust, strings are not indexable due to their encoding in UTF-8, which means a single character can span more than one byte. When you try to index a string directly, you might end up indexing in the middle of a character, which would result in an error. By converting the string to a vector of characters, you ensure that each index corresponds to a complete character, avoiding this issue.
+        let text1 = text1.chars().collect::<Vec<char>>();
+        let text2 = text2.chars().collect::<Vec<char>>();
+
+        for i in (0..text1.len()).rev() {
+            for j in (0..text2.len()).rev() {
+                if text1[i] == text2[j] {
+                    dp[i][j] = 1 + dp[i+1][j+1];
+                } else {
+                    dp[i][j] = std::cmp::max(dp[i][j+1], dp[i+1][j]);
+                }
+            }
+        }
+
+        dp[0][0]
+    }
+}
+
+struct Solution;
+
+fn main() {
+    assert_eq!(Solution::longest_common_subsequence("abcde".to_string(), "ace".to_string()), 3);
+    assert_eq!(Solution::longest_common_subsequence("abc".to_string(), "abc".to_string()), 3);
+    assert_eq!(Solution::longest_common_subsequence("abc".to_string(), "def".to_string()), 0);
+}