Longest Common Subsequence in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP table with zeros.

DSA Rust

let mut dp = vec![vec![[1]; n + 1]; m + 1];

Drag options to blanks, or click blank then click option'

A0

B1

CNone

Dvec![]

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 initializes with wrong lengths.

Using None causes type errors.

Using vec![] creates empty vectors, not integers.

The DP table must be initialized with zeros to represent empty subsequences.

2fill in blank

medium

Complete the code to compare characters from both strings.

DSA Rust

if s1.chars().nth(i - 1) == s2.chars().nth(j - 1) {
    dp[i][j] = dp[i - 1][j - 1] + [1];
}

Drag options to blanks, or click blank then click option'

A0

Bdp[i - 1][j]

Cdp[i][j]

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Adding 0 does not increase length.

Using dp[i][j] causes self-reference error.

Using dp[i - 1][j] is incorrect for matches.

When characters match, increase the subsequence length by 1.

3fill in blank

hard

Fix the error in choosing the maximum value when characters do not match.

DSA Rust

dp[i][j] = std::cmp::max(dp[i - 1][j], dp[i][[1]]);

Drag options to blanks, or click blank then click option'

Ai - 1

Bj - 1

Cj + 1

Di + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using i - 1 accesses wrong row.

Using j + 1 or i + 1 goes out of bounds.

When characters differ, take the max between dp[i-1][j] and dp[i][j-1].

4fill in blank

hard

Fill both blanks to correctly iterate over the DP table indices.

DSA Rust

for i in 1..=[1] {
    for j in 1..=[2] {
        // DP logic here
    }
}

Drag options to blanks, or click blank then click option'

Am

Bn

Cm + 1

Dn + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using m + 1 or n + 1 causes index out of range.

Using 0 as upper bound causes empty loops.

Loop indices go from 1 to m and 1 to n for strings of length m and n.

5fill in blank

hard

Fill all three blanks to reconstruct the LCS string from the DP table.

DSA Rust

let mut i = m;
let mut j = n;
let mut lcs = String::new();
while i > 0 && j > 0 {
    if s1.chars().nth(i - 1) == s2.chars().nth(j - 1) {
        lcs.push(s1.chars().nth(i - 1).unwrap());
        i -= [1];
        j -= [2];
    } else if dp[i - 1][j] > dp[i][j - 1] {
        i -= [3];
    } else {
        j -= 1;
    }
}
lcs = lcs.chars().rev().collect();

Drag options to blanks, or click blank then click option'

A1

B0

C2

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Decreasing by 0 causes infinite loops.

Decreasing by 2 skips characters incorrectly.

When characters match, move diagonally up-left by decreasing both i and j by 1.

When moving up, decrease i by 1.

Practice

1. What does the Longest Common Subsequence (LCS) between two strings represent?

easy

A. The longest sequence of characters that appear in both strings in the same order, but not necessarily consecutively.

B. The longest substring that appears exactly the same in both strings.

C. The shortest sequence of characters that can be removed to make the strings identical.

D. The total number of characters that differ between the two strings.

Solution

1. Step 1: Understand the definition of LCS

   The LCS is the longest sequence of characters that appear in both strings in the same order, but not necessarily consecutively.

2. Step 2: Differentiate from substring and other concepts

   Unlike substring, LCS does not require characters to be consecutive. It is about order and presence.

3. Final Answer:

   The longest sequence of characters that appear in both strings in the same order, but not necessarily consecutively. -> Option A

4. Quick Check:

   LCS = longest ordered common sequence [OK]

Hint: LCS keeps order, skips gaps, unlike substring [OK]

Common Mistakes:

• Confusing LCS with longest common substring
• Thinking characters must be consecutive
• Mixing LCS with edit distance

2. Which of the following Rust function signatures correctly represents a function to compute LCS length between two string slices?

easy

A. fn lcs_length(s1: &str, s2: &str) -> bool

B. fn lcs_length(s1: String, s2: String) -> &str

C. fn lcs_length(s1: &String, s2: &String) -> String

D. fn lcs_length(s1: &str, s2: &str) -> usize

Solution

1. Step 1: Identify correct parameter types for string inputs

   Using &str is idiomatic for string slices in Rust, allowing borrowing without ownership transfer.

2. Step 2: Identify correct return type for LCS length

   LCS length is a number, so usize is appropriate. Returning &str or String is incorrect for length.

3. Final Answer:

   fn lcs_length(s1: &str, s2: &str) -> usize -> Option D

4. Quick Check:

   Input: &str, Output: usize for length [OK]

Hint: Use &str for inputs and usize for length output [OK]

Common Mistakes:

• Returning &str or String instead of length
• Using owned String unnecessarily
• Returning bool instead of length

3. What is the output of the following Rust code snippet that computes the LCS length of "AGGTAB" and "GXTXAYB"?

fn main() {
    let s1 = "AGGTAB";
    let s2 = "GXTXAYB";
    let lcs_len = lcs_length(s1, s2);
    println!("{}", lcs_len);
}

fn lcs_length(s1: &str, s2: &str) -> usize {
    let m = s1.len();
    let n = s2.len();
    let mut dp = vec![vec![0; n + 1]; m + 1];
    for i in 1..=m {
        for j in 1..=n {
            if s1.as_bytes()[i - 1] == s2.as_bytes()[j - 1] {
                dp[i][j] = dp[i - 1][j - 1] + 1;
            } else {
                dp[i][j] = dp[i - 1][j].max(dp[i][j - 1]);
            }
        }
    }
    dp[m][n]
}

medium

A. 6

B. 3

C. 4

D. 5

Solution

1. Step 1: Understand the input strings and LCS algorithm

   Strings are "AGGTAB" and "GXTXAYB". The code uses dynamic programming to find LCS length.

2. Step 2: Trace the LCS calculation

   The LCS is "GTAB" with length 4. The code correctly computes this.

3. Step 3: Verify carefully

   Common subsequence: G T A B (positions match in order). Length is 4.

4. Final Answer:

   4 -> Option C

5. Quick Check:

   LCS length of "AGGTAB" and "GXTXAYB" = 4 [OK]

Hint: Trace matching chars in order, ignore gaps [OK]

Common Mistakes:

• Counting common characters not in order
• Confusing substring with subsequence
• Off-by-one errors in indexing

4. Identify the bug in the following Rust function that attempts to compute the LCS length:

fn lcs_length(s1: &str, s2: &str) -> usize {
    let m = s1.len();
    let n = s2.len();
    let mut dp = vec![vec![0; n]; m];
    for i in 1..m {
        for j in 1..n {
            if s1.as_bytes()[i] == s2.as_bytes()[j] {
                dp[i][j] = dp[i - 1][j - 1] + 1;
            } else {
                dp[i][j] = dp[i - 1][j].max(dp[i][j - 1]);
            }
        }
    }
    dp[m - 1][n - 1]
}

medium

A. The dp vector is initialized with size m x n instead of (m+1) x (n+1).

B. The loops start from 0 instead of 1, causing index errors.

C. The loops use 1..=m and 1..=n instead of 1..m and 1..n, causing out of bounds.

D. The function returns dp[m][n] which is out of bounds.

Solution

1. Step 1: Check dp vector initialization

   The dp vector is initialized as m x n, but it should be (m+1) x (n+1) to handle base cases and indexing from 1.

2. Step 2: Understand loop ranges and indexing

   Loops start from 1..

3. Final Answer:

   The dp vector is initialized with size m x n instead of (m+1) x (n+1). -> Option A

4. Quick Check:

   dp size must be (m+1) x (n+1) [OK]

Hint: dp table size must include extra row and column [OK]

Common Mistakes:

• Initializing dp with wrong size
• Incorrect loop ranges causing panic
• Wrong indexing of string bytes

5. Given two strings, write a Rust function that returns the actual Longest Common Subsequence string (not just length). Which approach below correctly reconstructs the LCS from the dp table?

hard

A. Start from dp[0][0], move right or down collecting characters whenever dp[i][j] increases.

B. Start from dp[m][n], move diagonally up-left if characters match, else move to the cell with the larger value, collecting matched characters in reverse order.

C. Traverse dp table row-wise and append characters when dp[i][j] is non-zero.

D. Reverse both strings and find LCS length, then reverse the result to get LCS string.

Solution

1. Step 1: Understand dp table usage for reconstruction

   The dp table stores lengths of LCS for substrings. To reconstruct, start from dp[m][n] and move backwards.

2. Step 2: Trace path to build LCS string

   If characters match at s1[i-1] and s2[j-1], move diagonally up-left and add character. Else move to cell with larger dp value (up or left).

3. Step 3: Reverse collected characters

   Since traversal is from end to start, collected characters must be reversed to get correct LCS.

4. Final Answer:

   Start from dp[m][n], move diagonally up-left if characters match, else move to the cell with the larger value, collecting matched characters in reverse order. -> Option B

5. Quick Check:

   Backtrack dp table from bottom-right to reconstruct LCS [OK]

Hint: Backtrack dp from bottom-right, collect matches, then reverse [OK]

Common Mistakes:

• Starting reconstruction from top-left
• Not reversing collected characters
• Ignoring dp values when moving