Longest Common Subsequence in DSA C - Time & Space Complexity

Time Complexity: Longest Common Subsequence

O(2^(m+n))

Understanding Time Complexity

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

How does the work grow when the two strings become bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int lcs(char *X, char *Y, int m, int n) {
    if (m == 0 || n == 0) {
        return 0;
    }
    if (X[m - 1] == Y[n - 1]) {
        return 1 + lcs(X, Y, m - 1, n - 1);
    } else {
        int a = lcs(X, Y, m, n - 1);
        int b = lcs(X, Y, m - 1, n);
        return (a > b) ? a : b;
    }
}
    

This code finds the length of the longest common subsequence between two strings using recursion.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls that explore all combinations of substrings.
• How many times: Each call splits into two more calls unless base case is reached, leading to many repeated calls.

How Execution Grows With Input

As the lengths of the two strings increase, the number of recursive calls grows very fast because the function tries many substring pairs.

Input Size (m, n)	Approx. Operations
10, 10	Millions of calls
20, 20	Trillions of calls
30, 30	Quintillions of calls

Pattern observation: The work roughly doubles with each increase in string length, growing exponentially.

Final Time Complexity

Time Complexity: O(2^(m+n))

This means the time needed grows very fast, doubling with each added character in the strings.

Common Mistake

[X] Wrong: "The recursive calls only happen once per substring pair, so it should be much faster."

[OK] Correct: The recursion repeats many calls for the same substring pairs multiple times, causing exponential growth.

Interview Connect

Understanding this complexity helps you explain why naive recursion is slow and why dynamic programming is needed to solve this problem efficiently.

Self-Check

"What if we use a table to store results of substring pairs (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 in both strings consecutively.

C. The total number of characters in both strings combined.

D. The shortest sequence of characters that can be removed to make the strings identical.

Solution

1. Step 1: Understand the definition of LCS

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

2. Step 2: Differentiate from substring and other concepts

   Unlike substring, LCS does not require characters to be consecutive. It is about order and presence.

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 = longest ordered common sequence [OK]

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

Common Mistakes:

• Confusing LCS with longest common substring
• Thinking characters must be consecutive
• Mixing LCS with string length or edits

2. Which of the following is the correct way to declare a 2D array in C for storing LCS lengths of two strings of length m and n?

easy

A. int lcs[m][n];

B. int lcs[m-1][n-1];

C. int lcs[m+1][n+1];

D. int lcs[m*n];

Solution

1. Step 1: Understand LCS table size requirements

   We need a table of size (m+1) x (n+1) to include empty prefixes for both strings.

2. Step 2: Check array declaration syntax

   int lcs[m+1][n+1]; correctly declares int lcs[m+1][n+1]; which fits the DP table size.

3. Final Answer:

   int lcs[m+1][n+1]; -> Option C

4. Quick Check:

   DP table size = (m+1) x (n+1) [OK]

Hint: Add 1 to lengths for DP table size [OK]

Common Mistakes:

• Using m or n instead of m+1 or n+1
• Declaring 1D array instead of 2D
• Using multiplication instead of 2D array

3. What is the output of the following C code snippet that computes the length of LCS for strings "ABCBDAB" and "BDCABA"?

int main() {
    char X[] = "ABCBDAB";
    char Y[] = "BDCABA";
    int m = 7, n = 6;
    int lcs[8][7];
    for (int i = 0; i <= m; i++) lcs[i][0] = 0;
    for (int j = 0; j <= n; j++) lcs[0][j] = 0;
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (X[i-1] == Y[j-1])
                lcs[i][j] = lcs[i-1][j-1] + 1;
            else
                lcs[i][j] = (lcs[i-1][j] > lcs[i][j-1]) ? lcs[i-1][j] : lcs[i][j-1];
        }
    }
    printf("%d", lcs[m][n]);
    return 0;
}

medium

A. 6

B. 3

C. 5

D. 4

Solution

1. Step 1: Understand the DP table filling

   The code fills the LCS table using standard DP approach comparing characters of X and Y.

2. Step 2: Calculate LCS length for given strings

   The LCS of "ABCBDAB" and "BDCABA" is "BCBA" or "BDAB" with length 4, but careful tracing shows length 4 is correct. However, the code prints lcs[m][n], which is 4.

3. Step 3: Re-evaluate LCS length

   Actually, the longest common subsequence length is 4 or 5? Let's check carefully:
   - "BCBA" length 4
   - "BDAB" length 4
   - But "BCABA" is not a subsequence.
   Rechecking the classic example, the LCS length is 4.
   But the code output is 4, so 4 is correct.

4. Final Answer:

   4 -> Option D

5. Quick Check:

   LCS length for given strings = 4 [OK]

Hint: Trace DP table or recall classic LCS example [OK]

Common Mistakes:

• Confusing LCS length with substring length
• Off-by-one errors in indexing
• Misreading the strings or code logic

4. Identify the error in the following C code snippet intended to compute the LCS length of two strings X and Y:

int lcs_length(char *X, char *Y, int m, int n) {
    int lcs[m+1][n+1];
    for (int i = 0; i <= m; i++) lcs[i][0] = 0;
    for (int j = 0; j <= n; j++) lcs[0][j] = 0;
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (X[i] == Y[j])
                lcs[i][j] = lcs[i-1][j-1] + 1;
            else
                lcs[i][j] = (lcs[i-1][j] > lcs[i][j-1]) ? lcs[i-1][j] : lcs[i][j-1];
        }
    }
    return lcs[m][n];
}

medium

A. Indexing X[i] and Y[j] should be X[i-1] and Y[j-1].

B. Array size should be lcs[m+1][n+1] instead of lcs[m][n].

C. Initialization loops should start from 1, not 0.

D. Return statement should be lcs[m-1][n-1] instead of lcs[m][n].

Solution

1. Step 1: Check array size and initialization

   The array size (m+1) x (n+1) and initialization loops are correct for the DP table.

2. Step 2: Check indexing of characters in strings

   Strings in C are 0-indexed, but loops run from 1 to m/n, so characters should be accessed as X[i-1] and Y[j-1].

3. Final Answer:

   Indexing X[i] and Y[j] should be X[i-1] and Y[j-1]. -> Option A

4. Quick Check:

   String indexing off by one error [OK]

Hint: Remember C strings start at index 0 [OK]

Common Mistakes:

• Using wrong indices for string characters
• Ignoring array size for DP table
• Incorrect initialization loop ranges

5. Given two strings X = "AGGTAB" and Y = "GXTXAYB", write a C function to print the actual Longest Common Subsequence (LCS) string, not just its length. Which approach correctly reconstructs the LCS from the DP table?

hard

A. Start from lcs[0][0], move right or down to find matching characters and build LCS forward.

B. Start from lcs[m][n], move diagonally up-left if characters match, else move in direction of larger value until reaching lcs[0][0].

C. Use recursion without memoization to print LCS directly from strings.

D. Print characters from X that appear anywhere in Y without order consideration.

Solution

1. Step 1: Understand LCS reconstruction from DP table

   After filling the DP table, start from bottom-right cell lcs[m][n] and move backwards.

2. Step 2: Move diagonally if characters match, else move towards larger neighbor

   If characters match, include that character and move diagonally up-left; else move to the cell with larger value (up or left).

3. Step 3: Continue until reaching lcs[0][0]

   Collect characters in reverse order and then reverse them to get the LCS string.

4. Final Answer:

   Start from lcs[m][n], move diagonally up-left if characters match, else move in direction of larger value until reaching lcs[0][0]. -> Option B

5. Quick Check:

   Backtrack DP table from bottom-right to reconstruct LCS [OK]

Hint: Backtrack DP table from bottom-right to get LCS [OK]

Common Mistakes:

• Trying to build LCS forward from top-left
• Ignoring order of characters in reconstruction
• Printing characters without checking DP table values