Longest Common Subsequence in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of LCS length calculation

What is the output of the following JavaScript code that calculates the length of the Longest Common Subsequence (LCS) between two strings?

DSA Javascript

function lcsLength(s1, s2) {
  const m = s1.length, n = s2.length;
  const dp = Array(m + 1).fill(null).map(() => 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("AGGTAB", "GXTXAYB"));

A 4

B 7

C 6

D 5

💡 Hint

Trace the matching characters between the two strings and count the longest sequence in order.

The longest common subsequence between "AGGTAB" and "GXTXAYB" is "GTAB" which has length 4.

The code uses a standard dynamic programming table to compute the LCS length and returns 4.

❓ Predict Output

intermediate

Output of LCS sequence reconstruction

What is the output of the following JavaScript code that reconstructs the Longest Common Subsequence (LCS) string between two strings?

DSA Javascript

function lcsSequence(s1, s2) {
  const m = s1.length, n = s2.length;
  const dp = Array(m + 1).fill(null).map(() => 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]);
      }
    }
  }
  let i = m, j = n;
  let lcs = [];
  while (i > 0 && j > 0) {
    if (s1[i - 1] === s2[j - 1]) {
      lcs.push(s1[i - 1]);
      i--;
      j--;
    } else if (dp[i - 1][j] > dp[i][j - 1]) {
      i--;
    } else {
      j--;
    }
  }
  return lcs.reverse().join('');
}

console.log(lcsSequence("ABCBDAB", "BDCABA"));

A "BCBA"

B "BDCB"

C "BDAB"

D "BCAB"

💡 Hint

Follow the path from dp[m][n] backwards to find matching characters.

The longest common subsequence between "ABCBDAB" and "BDCABA" is "BDAB".

The code reconstructs this sequence by moving backwards through the dp table.

🧠 Conceptual

advanced

Time complexity of LCS dynamic programming solution

What is the time complexity of the classic dynamic programming solution to find the length of the Longest Common Subsequence (LCS) of two strings of lengths m and n?

A O(m^2 + n^2)

B O(m * n)

C O(m + n)

D O(2^(m + n))

💡 Hint

Consider the nested loops iterating over both strings.

The dynamic programming solution uses a 2D table of size m by n and fills each cell once.

Thus, the time complexity is O(m * n).

🔧 Debug

advanced

Identify the error in LCS length code

What error will the following JavaScript code produce when trying to compute the LCS length?

DSA Javascript

function lcsLength(s1, s2) {
  const m = s1.length, n = s2.length;
  const dp = Array(m).fill(Array(n).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 - 1][n - 1];
}

console.log(lcsLength("ABC", "AC"));

A SyntaxError: Unexpected token

B ReferenceError: dp is not defined

C Returns incorrect LCS length

D TypeError: Cannot set property of undefined

💡 Hint

Check how the dp array is initialized and accessed.

The dp array is initialized with Array(m).fill(Array(n).fill(0)) which creates multiple references to the same inner array.

When the code tries to assign dp[i][j], it tries to modify a shared inner array, leading to unexpected behavior or TypeError when i or j is out of bounds.

🚀 Application

expert

Number of distinct LCS sequences

Given two strings s1 = "ABCBDAB" and s2 = "BDCABA", how many distinct Longest Common Subsequences (LCS) of maximum length exist?

A 3

B 2

C 4

D 5

💡 Hint

Consider all LCS strings of length 4 and count unique sequences.

The maximum LCS length is 4 for these strings.

The distinct LCS sequences are "BCAB", "BDAB", and "BCBA".

Thus, there are 3 distinct LCS sequences.

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