Longest Common Subsequence in DSA C# - 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 running time change when the strings become bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int LCS(string s1, string s2) {
  int m = s1.Length, n = s2.Length;
  int[,] dp = new int[m + 1, n + 1];
  for (int i = 1; i <= m; i++) {
    for (int 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];
}
    

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: The inner code runs once for each pair of characters, so m x n times.

How Execution Grows With Input

As the strings get longer, the number of operations grows with the product of 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 strings.

Common Mistake

[X] Wrong: "The algorithm runs in linear time because it just compares characters."

[OK] Correct: It compares pairs of characters from both strings, so the work grows with both lengths multiplied, not just one.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves efficiency over simple recursion, 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 appearing in both strings in the same order, not necessarily contiguous

B. The longest sequence of consecutive characters shared by both strings

C. The string formed by concatenating both inputs

D. The shortest string that contains both inputs as subsequences

Solution

1. Step 1: Define subsequence

   A subsequence maintains relative order of characters but does not require them to be contiguous. 'ace' is a subsequence of 'abcde'.

2. Step 2: LCS definition

   LCS is the longest such subsequence common to both strings. For 'ABCBDAB' and 'BDCAB', LCS is 'BCAB' or 'BDAB', length 4.

3. Final Answer:

   Longest common subsequence - same order, not necessarily contiguous. -> Option A

4. Quick Check:

   Subsequence = order preserved, not contiguous. LCS = longest shared one. [OK]

Hint: Subsequence = order preserved, gaps allowed. LCS = longest one common to both strings. [OK]

Common Mistakes:

• Confusing subsequence with substring (substring must be contiguous)
• Thinking LCS is the shortest common supersequence
• Assuming characters must be adjacent

2. For strings X = "ABCD" and Y = "ACDF", what is the length of the LCS?

medium

A. 2

B. 3

C. 4

D. 1

Solution

1. Step 1: Find common subsequences

   Characters in both: A, C, D. Check order: A appears before C in both. C appears before D in both. So 'ACD' is a common subsequence of length 3.

2. Step 2: Can we do better?

   'ABCD' and 'ACDF' share A, C, D in order. B is not in Y. F is not in X. No common subsequence of length 4 exists.

3. Final Answer:

   LCS = "ACD", length 3. -> Option B

4. Quick Check:

   A→C→D in both strings: length 3. [OK]

Hint: Trace common characters in order: A(pos 0,0)→C(pos 2,1)→D(pos 3,2) = 3. [OK]

Common Mistakes:

• Counting all common characters without checking order
• Including B which is not in Y
• Stopping at 2 by missing D

3. In the LCS DP recurrence, when X[i] == Y[j], what is dp[i][j]?

medium

A. dp[i-1][j-1]

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

C. max(dp[i-1][j], dp[i][j-1])

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

Solution

1. Step 1: When characters match

   If X[i] == Y[j], the current characters extend the LCS found without them. We add 1 to the LCS of X[0..i-1] and Y[0..j-1].

2. Step 2: Recurrence

   dp[i][j] = dp[i-1][j-1] + 1. We extend the diagonal neighbor by 1.

3. Step 3: When they don't match

   dp[i][j] = max(dp[i-1][j], dp[i][j-1]) - take the best from skipping either character.

4. Final Answer:

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

5. Quick Check:

   Match → diagonal + 1. No match → max of left and above. [OK]

Hint: Match: diagonal + 1. No match: max(left, above). [OK]

Common Mistakes:

• Using dp[i-1][j-1] without adding 1
• Using max formula when characters match
• Adding 1 to dp[i][j-1] instead of diagonal

4. What is the time and space complexity of the standard 2D DP LCS solution for strings of length m and n?

medium

A. Time O(m+n), Space O(m+n)

B. Time O(m×n), Space O(m×n)

C. Time O(m×n), Space O(min(m,n))

D. Time O(2^n), Space O(n)

Solution

1. Step 1: Time complexity

   We fill a 2D table of size (m+1)×(n+1), computing each cell in O(1). Total: O(m×n).

2. Step 2: Space complexity

   The full 2D DP table requires O(m×n) space. Space can be optimized to O(min(m,n)) using two rows, but standard solution uses O(m×n).

3. Final Answer:

   Time O(m×n), Space O(m×n) -> Option B

4. Quick Check:

   Standard 2D table = O(m×n) both time and space. [OK]

Hint: Standard LCS: O(m×n) time and space (2D table). [OK]

Common Mistakes:

• Saying O(m+n) which is linear - LCS needs quadratic
• Reporting space-optimized O(min(m,n)) for the standard solution
• Confusing with exponential recursion without memoization

5. In C#, after filling the LCS DP table for X="AGGTAB" and Y="GXTXAYB", what value does dp[6][7] hold?

hard

A. 3

B. 4

C. 5

D. 6

Solution

1. Step 1: Identify LCS

   X = 'AGGTAB', Y = 'GXTXAYB'. Common subsequences include 'GTAB' (length 4): G(1,0)→T(2,3)→A(4,4)→B(5,6).

2. Step 2: Verify length 4

   G: appears in both at valid positions. T: X[2]='G',X[2]='G'... Let's check: X='A,G,G,T,A,B'. Y='G,X,T,X,A,Y,B'. Common: G(X[1],Y[0]), T(X[3],Y[2]), A(X[4],Y[4]), B(X[5],Y[6]) → GTAB length 4.

3. Final Answer:

   dp[6][7] = 4 -> Option B

4. Quick Check:

   LCS of 'AGGTAB' and 'GXTXAYB' is 'GTAB' = length 4. [OK]

Hint: LCS('AGGTAB','GXTXAYB') = 'GTAB' = 4. This is a classic example - memorize it. [OK]

Common Mistakes:

• Confusing LCS length with string length
• Incorrectly tracing the DP table
• Getting 3 by missing one character in the subsequence