Longest Common Subsequence in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Longest Common Subsequence (LCS)?

LCS is the longest sequence that appears in the same relative order in two sequences, but not necessarily contiguously.

intermediate

How does dynamic programming help in finding LCS?

Dynamic programming breaks the problem into smaller overlapping subproblems, stores their results, and builds up the solution efficiently.

beginner

What does the DP table represent in LCS algorithm?

Each cell dp[i][j] stores the length of LCS of the first i characters of string1 and first j characters of string2.

intermediate

What is the time complexity of the classic LCS dynamic programming solution?

The time complexity is O(m * n), where m and n are the lengths of the two input strings.

advanced

How do you reconstruct the LCS string from the DP table?

Start from dp[m][n], move backwards: if characters match, add to LCS and move diagonally; else move in the direction of the larger dp value.

What does the Longest Common Subsequence represent?

A Longest sequence appearing in both strings in order

B Longest substring appearing in both strings

C Shortest sequence appearing in both strings

D Longest sequence appearing in either string

LCS is the longest sequence that appears in both strings in the same order, but not necessarily contiguously.

What is the base case for the LCS dynamic programming table?

A dp[0][j] = 0 and dp[i][0] = 0 for all i, j

B dp[0][j] = 1 and dp[i][0] = 1 for all i, j

C dp[i][j] = 1 if characters match

D dp[i][j] = max(i, j)

When either string is empty, LCS length is zero, so dp[0][j] and dp[i][0] are zero.

If characters at position i and j match, how is dp[i][j] computed?

A max(dp[i-1][j], dp[i][j-1])

B dp[i-1][j-1] + 1

C dp[i-1][j-1]

D dp[i][j-1] + 1

When characters match, LCS length increases by 1 from dp[i-1][j-1].

What is the time complexity of the LCS algorithm using dynamic programming?

A O(m^2 + n^2)

B O(m + n)

C O(m * n)

D O(2^(m+n))

The DP table has m rows and n columns, so time complexity is O(m * n).

Which of these is NOT true about LCS?

A LCS length is at most min(m, n)

B LCS characters appear in order

C LCS can be empty if no common characters

D LCS characters must be contiguous

LCS characters do not need to be contiguous, only in the same order.

Explain how to build the DP table for the Longest Common Subsequence problem.

Think about comparing characters and using previous results.

Describe the steps to reconstruct the LCS string from the completed DP table.

Trace back from the bottom-right corner of the table.

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