Longest Common Subsequence in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the 2D array for dynamic programming.

DSA Javascript

const dp = Array(m + 1).fill(null).map(() => Array([1]).fill(0));

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

Am + 1

Bm

Cn

Dn + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using m instead of n for the second dimension.

Not adding 1 to the lengths.

We create a 2D array with dimensions (m+1) x (n+1) to store LCS lengths.

2fill in blank

medium

Complete the condition to check if characters match in the LCS algorithm.

DSA Javascript

if (text1[i - 1] [1] text2[j - 1]) {

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

A<

B===

C!=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of ===.

Using < or > which are not correct for equality check.

We check if characters at positions i-1 and j-1 are equal to extend the subsequence.

3fill in blank

hard

Fix the error in updating dp when characters do not match.

DSA Javascript

dp[i][j] = Math.max(dp[i - 1][j], dp[i][[1]]);

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

Aj - 1

Bi - 1

Cj + 1

Di + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using i - 1 instead of j - 1.

Using j + 1 or i + 1 which go out of bounds.

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

4fill in blank

hard

Fill both blanks to complete the nested loops iterating over strings.

DSA Javascript

for (let i = 1; i [1] m + 1; i++) {
  for (let j = 1; j [2] n + 1; j++) {

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

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= which causes index out of range.

Using > or >= which are incorrect for loop conditions.

Loops run from 1 up to but not including m+1 and n+1 using < operator.

5fill in blank

hard

Fill all three blanks to return the final LCS length from dp array.

DSA Javascript

return dp[[1]][[2]]; // length of LCS for text1 and text2

// Variables: m = [3]

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

Am

Bn

Ctext1.length

Dtext2.length

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up m and n indices.

Using incorrect variables for lengths.

Return dp[m][n] where m and n are lengths of text1 and text2 respectively.

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 that can be formed from both strings

D. The total number of characters in both strings combined

Solution

1. Step 1: Understand the definition of LCS

   The LCS is about finding a sequence that appears in both strings in the same order, but characters do not need to be next to each other.

2. Step 2: Differentiate LCS from substring

   A substring requires characters to be consecutive, but LCS allows gaps, so it is not necessarily a substring.

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 means longest common sequence, not substring [OK]

Hint: LCS allows gaps; substring does not [OK]

Common Mistakes:

• Confusing LCS with longest common substring
• Thinking characters must be consecutive
• Assuming LCS is the combined length

2. Which of the following is the correct way to initialize a 2D array dp of size (m+1) x (n+1) with zeros in JavaScript for LCS dynamic programming?

easy

A. const dp = Array(m+1).fill(Array(n+1).fill(0));

B. const dp = Array.from({length: m+1}, () => Array(n+1).fill(0));

C. const dp = new Array(m+1).map(new Array(n+1).fill(0));

D. const dp = Array(m+1).map(Array(n+1).fill(0));

Solution

1. Step 1: Understand how to create independent inner arrays

   Using Array.from with a function creates new inner arrays for each row, avoiding shared references.

2. Step 2: Analyze each option

   Array(m+1).fill(Array(n+1).fill(0)) shares the same inner array for all rows, causing bugs. Options C and D pass an array to map() instead of a function, causing a TypeError. Array.from({length: m+1}, () => Array(n+1).fill(0)) correctly creates independent arrays.

3. Final Answer:

   const dp = Array.from({length: m+1}, () => Array(n+1).fill(0)); -> Option B

4. Quick Check:

   Use Array.from with function for 2D arrays [OK]

Hint: Use Array.from with function to avoid shared inner arrays [OK]

Common Mistakes:

• Using fill with array directly causing shared references
• Passing an array to map() instead of a function
• Not initializing inner arrays properly

3. What is the output of the following JavaScript code that calculates the length of the LCS for strings s1 = 'abcde' and s2 = 'ace'?

function lcsLength(s1, s2) {
  const m = s1.length, n = s2.length;
  const dp = Array.from({length: m+1}, () => Array(n+1).fill(0));
  for (let i = 1; i <= m; i++) {
    for (let j = 1; j <= n; j++) {
      if (s1[i-1] === s2[j-1]) {
        dp[i][j] = dp[i-1][j-1] + 1;
      } else {
        dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
      }
    }
  }
  return dp[m][n];
}
console.log(lcsLength('abcde', 'ace'));

medium

A. 4

B. 2

C. 3

D. 5

Solution

1. Step 1: Trace matching characters between 'abcde' and 'ace'

   The common subsequence is 'ace' with length 3.

2. Step 2: Confirm DP calculation

   The code correctly fills dp table and returns dp[5][3] = 3 as the LCS length.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   LCS('abcde','ace') = 'ace' length 3 [OK]

Hint: Count matching chars in order, skipping mismatches [OK]

Common Mistakes:

• Counting only consecutive matches (substring)
• Confusing length with total characters
• Off-by-one errors in indexing

4. The following code attempts to compute the LCS length but has a bug. What is the error?

function lcsLength(s1, s2) {
  const m = s1.length, n = s2.length;
  const dp = Array(m+1).fill(Array(n+1).fill(0));
  for (let i = 1; i <= m; i++) {
    for (let j = 1; j <= n; j++) {
      if (s1[i-1] === s2[j-1]) {
        dp[i][j] = dp[i-1][j-1] + 1;
      } else {
        dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
      }
    }
  }
  return dp[m][n];
}

medium

A. The inner arrays in dp share the same reference causing incorrect updates

B. The loops start from 0 instead of 1

C. The dp array is not initialized with zeros

D. The comparison s1[i-1] === s2[j-1] is incorrect

Solution

1. Step 1: Analyze dp initialization

   Using fill with an array fills dp with references to the same inner array, so all rows share one array.

2. Step 2: Understand impact on updates

   Updating dp[i][j] changes all rows at column j, corrupting the DP table and causing wrong results.

3. Final Answer:

   The inner arrays in dp share the same reference causing incorrect updates -> Option A

4. Quick Check:

   Array fill shares references, use Array.from instead [OK]

Hint: Avoid fill with arrays; use Array.from for 2D arrays [OK]

Common Mistakes:

• Using fill with arrays causing shared references
• Starting loops at 0 causing index errors
• Incorrect character comparison

5. Given two strings s1 = 'AGGTAB' and s2 = 'GXTXAYB', which of the following is a valid Longest Common Subsequence (LCS)?

hard

A. "GXAYB"

B. "GXTAB"

C. "AGTXB"

D. "GTAB"

Solution

1. Step 1: Identify common subsequences in order

   Check which sequences appear in both strings in order without rearranging characters.

2. Step 2: Validate each option

   "GTAB" appears in both strings in order. Options A, B, and C contain characters out of order or extra characters not matching subsequence rules.

3. Final Answer:

   "GTAB" -> Option D

4. Quick Check:

   Valid LCS must appear in both strings in order [OK]

Hint: Check order and presence of characters in both strings [OK]

Common Mistakes:

• Choosing sequences with characters out of order
• Including characters not in both strings
• Confusing subsequence with substring