Longest Common Subsequence in DSA Javascript - Time & Space Complexity

Time Complexity: Longest Common Subsequence

O(m x n)

Understanding Time Complexity

We want to understand how the time needed to find the longest common subsequence grows as the input strings get longer.

How does the work increase when the strings become bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function lcs(str1, str2) {
  const m = str1.length, n = str2.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 (str1[i - 1] === str2[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];
}
    

This code finds the length of the longest common subsequence between two strings using a table to store results.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops over both strings to fill the dp table.
• How many times: Outer loop runs m times, inner loop runs n times, so total m x n times.

How Execution Grows With Input

As the strings get longer, the number of operations grows by multiplying their lengths.

Input Size (m, n)	Approx. Operations
10, 10	100
100, 100	10,000
1000, 1000	1,000,000

Pattern observation: Doubling the length of both strings roughly quadruples the work.

Final Time Complexity

Time Complexity: O(m x n)

This means the time needed grows proportionally to the product of the two string lengths.

Common Mistake

[X] Wrong: "The algorithm runs in linear time because it just loops through the strings once."

[OK] Correct: The algorithm uses nested loops, so it compares every character of one string with every character of the other, making it slower than just one loop.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves efficiency by avoiding repeated work, a key skill in problem solving.

Self-Check

"What if we used recursion without memoization? How would the time complexity change?"

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