Longest Common Subsequence in DSA Typescript - 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.

The question is: how does the work increase when the strings become bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function lcs(X: string, Y: string): number {
  const m = X.length, n = Y.length;
  const dp: number[][] = 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 (X[i - 1] === Y[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 lengths of the two strings grow, 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 lengths of the two input strings.

Common Mistake

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

[OK] Correct: Actually, it uses nested loops, so it compares every character of one string with every character of the other, making the work multiply.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves efficiency and why nested loops matter in real problems.

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 total number of characters in both strings combined

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

C. The shortest sequence of characters to convert one string into another

D. The longest substring that appears in both strings consecutively

Solution

1. Step 1: Understand the definition of LCS

   The LCS is the longest sequence that appears in both strings in the same order but not necessarily next to each other.

2. Step 2: Differentiate from substring and edit distance

   Unlike substring, LCS does not require characters to be consecutive. It is not about shortest edits either.

3. Final Answer:

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

4. Quick Check:

   LCS means longest common sequence, not substring or edits [OK]

Hint: LCS is about order, not consecutive characters [OK]

Common Mistakes:

• Confusing LCS with longest common substring
• Thinking LCS requires consecutive characters
• Mixing LCS with edit distance or shortest transformation

2. Which of the following TypeScript function signatures correctly represents a function to compute LCS length between two strings s1 and s2?

easy

A. function lcsLength(s1: string, s2: string): string[]

B. function lcsLength(s1: number[], s2: number[]): string

C. function lcsLength(s1: string, s2: string): void

D. function lcsLength(s1: string, s2: string): number

Solution

1. Step 1: Identify input and output types for LCS length

   The function takes two strings and returns a number representing the length of their LCS.

2. Step 2: Check each option's signature

   function lcsLength(s1: string, s2: string): number matches input strings and returns a number. Others either use wrong input types or return wrong types.

3. Final Answer:

   function lcsLength(s1: string, s2: string): number -> Option D

4. Quick Check:

   Input strings, output number for LCS length [OK]

Hint: LCS length function returns a number, inputs are strings [OK]

Common Mistakes:

• Using arrays instead of strings as input
• Returning string or void instead of number
• Confusing LCS length with LCS sequence output

3. What is the output of the following TypeScript code snippet?

function lcsLength(s1: string, s2: string): number {
  const m = s1.length, n = s2.length;
  const dp: number[][] = 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'))

medium

A. 4

B. 5

C. 3

D. 6

Solution

1. Step 1: Understand the LCS dynamic programming table

   The code builds a 2D table dp where dp[i][j] stores LCS length of s1[0..i-1] and s2[0..j-1].

2. Step 2: Trace LCS for 'AGGTAB' and 'GXTXAYB'

   The LCS is 'GTAB' with length 4, but careful tracing shows length 4 is correct. However, the known LCS length for these strings is 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   LCS('AGGTAB','GXTXAYB') = 4 [OK]

Hint: Trace dp table or recall classic example LCS length 4 [OK]

Common Mistakes:

• Confusing LCS length with substring length
• Miscounting matching characters
• Off-by-one errors in indexing

4. Identify the bug in the following TypeScript code that attempts to compute the LCS length:

function lcsLength(s1: string, s2: string): number {
  const m = s1.length, n = s2.length;
  const dp: number[][] = 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];
}

medium

A. dp array initialization shares inner arrays causing incorrect updates

B. Loop indices should start from 0 instead of 1

C. Return statement should be dp[m][n] instead of dp[m - 1][n - 1]

D. Comparison should be s1[i] === s2[j] instead of s1[i - 1] === s2[j - 1]

Solution

1. Step 1: Analyze dp array initialization

   Using Array(m).fill(Array(n).fill(0)) creates m references to the same inner array, causing shared rows.

2. Step 2: Understand impact on dp updates

   Updating dp[i][j] affects all rows because they point to the same array, corrupting the dp table.

3. Final Answer:

   dp array initialization shares inner arrays causing incorrect updates -> Option A

4. Quick Check:

   Array fill with same inner array causes shared references [OK]

Hint: Use map to create independent inner arrays for dp [OK]

Common Mistakes:

• Using fill with array causes shared references
• Incorrect dp return indices
• Off-by-one errors in loops

5. Given two strings s1 = 'ABCBDAB' and s2 = 'BDCABA', which of the following is a valid Longest Common Subsequence?

hard

A. 'BDCA'

B. 'BCBD'

C. 'BDAB'

D. 'BDA'

Solution

1. Step 1: Determine length of LCS

   The LCS length for these strings is 4.

2. Step 2: Check each option for validity and length

   'BDAB' is length 4 and appears in order in both strings. Others are either shorter or invalid sequences.

3. Final Answer:

   'BDAB' -> Option C

4. Quick Check:

   Valid LCS must be subsequence and max length 4 [OK]

Hint: Check subsequence order and length equals LCS length [OK]

Common Mistakes:

• Choosing subsequences not in order in both strings
• Picking shorter subsequences ignoring max length
• Confusing substring with subsequence