Longest Common Subsequence in DSA C - Execution Trace

Concept Flow - Longest Common Subsequence

Start with two strings

↓

Create DP table with size (m+1)x(n+1)

↓

Fill first row and column with 0

↓

For each character pair (i,j):

↓

i

↓

i

↓

DP table filled

↓

Trace back from dp[m

↓

Output LCS length and sequence

↓

End

Build a table to store lengths of common subsequences for prefixes, then trace back to find the longest sequence.

Execution Sample

DSA C

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

This code calculates the length of the longest common subsequence between two strings X and Y using dynamic programming.

Execution Table

Step	Operation	Indices (i,j)	Characters Compared	DP Table Update	DP Table State (partial)
1	Initialize dp[0][*] and dp[*][0]	-	-	dp[i][0]=0 and dp[0][j]=0	[Row0 and Col0 all zeros]
2	Compare X[0] and Y[0]	(1,1)	X='A', Y='B'	No match, dp[1][1] = max(dp[0][1], dp[1][0]) = 0	[[0,0,0],[0,0,...],...]
3	Compare X[0] and Y[1]	(1,2)	X='A', Y='D'	No match, dp[1][2] = max(dp[0][2], dp[1][1]) = 0	[[0,0,0],[0,0,0],...]
4	Compare X[0] and Y[2]	(1,3)	X='A', Y='C'	No match, dp[1][3] = max(dp[0][3], dp[1][2]) = 0	[[0,0,0,0],[0,0,0,0],...]
5	Compare X[0] and Y[3]	(1,4)	X='A', Y='A'	Match, dp[1][4] = dp[0][3] + 1 = 1	[[0,0,0,0,0],[0,0,0,0,1],...]
6	Compare X[1] and Y[0]	(2,1)	X='B', Y='B'	Match, dp[2][1] = dp[1][0] + 1 = 1	[[0,0,0,0,0],[0,0,0,0,1],[0,1,...],...]
7	Compare X[1] and Y[1]	(2,2)	X='B', Y='D'	No match, dp[2][2] = max(dp[1][2], dp[2][1]) = 1	[... dp[2][2]=1 ...]
8	Compare X[1] and Y[2]	(2,3)	X='B', Y='C'	No match, dp[2][3] = max(dp[1][3], dp[2][2]) = 1	[... dp[2][3]=1 ...]
9	Compare X[1] and Y[3]	(2,4)	X='B', Y='A'	No match, dp[2][4] = max(dp[1][4], dp[2][3]) = 1	[... dp[2][4]=1 ...]
10	Compare X[2] and Y[0]	(3,1)	X='C', Y='B'	No match, dp[3][1] = max(dp[2][1], dp[3][0]) = 1	[... dp[3][1]=1 ...]
11	Compare X[2] and Y[1]	(3,2)	X='C', Y='D'	No match, dp[3][2] = max(dp[2][2], dp[3][1]) = 1	[... dp[3][2]=1 ...]
12	Compare X[2] and Y[2]	(3,3)	X='C', Y='C'	Match, dp[3][3] = dp[2][2] + 1 = 2	[... dp[3][3]=2 ...]
13	Compare X[2] and Y[3]	(3,4)	X='C', Y='A'	No match, dp[3][4] = max(dp[2][4], dp[3][3]) = 2	[... dp[3][4]=2 ...]
14	Compare X[3] and Y[0]	(4,1)	X='D', Y='B'	No match, dp[4][1] = max(dp[3][1], dp[4][0]) = 1	[... dp[4][1]=1 ...]
15	Compare X[3] and Y[1]	(4,2)	X='D', Y='D'	Match, dp[4][2] = dp[3][1] + 1 = 2	[... dp[4][2]=2 ...]
16	Compare X[3] and Y[2]	(4,3)	X='D', Y='C'	No match, dp[4][3] = max(dp[3][3], dp[4][2]) = 2	[... dp[4][3]=2 ...]
17	Compare X[3] and Y[3]	(4,4)	X='D', Y='A'	No match, dp[4][4] = max(dp[3][4], dp[4][3]) = 2	[... dp[4][4]=2 ...]
18	DP table complete	-	-	LCS length = dp[4][4] = 2	[Full dp table filled]
19	Trace back to find LCS	-	-	LCS sequence found: 'BA' or 'CA' depending on path	-
20	End	-	-	Return LCS length 2	-

💡 All characters compared, DP table fully filled, LCS length found at dp[m][n]

Variable Tracker

Variable	Start	After Step 5	After Step 12	After Step 15	Final
i	0	1	3	4	4
j	0	4	3	2	4
dp[1][4]	uninitialized	1	1	1	1
dp[3][3]	uninitialized	0	2	2	2
dp[4][2]	uninitialized	0	0	2	2
LCS length	0	0	0	0	2

Key Moments - 3 Insights

Why do we initialize the first row and column of the dp table with zeros?

Because they represent comparisons with an empty string prefix, so the longest common subsequence length is zero at those points (see Step 1 in execution_table).

Why do we add 1 to dp[i-1][j-1] when characters match?

Because a matching character extends the previous longest subsequence by one (see Steps 5, 6, 12, 15 where dp[i][j] = dp[i-1][j-1] + 1).

Why do we take the max of dp[i-1][j] and dp[i][j-1] when characters don't match?

Because the LCS up to that point is the longest subsequence found by either skipping a character from one string or the other (see Steps 2, 3, 4, 7, 8, 9).

Visual Quiz - 3 Questions

Look at the execution_table at Step 12. What is the value of dp[3][3] after comparing X[2] and Y[2]?

A 2

B 3

C 1

D 0

💡 Hint

Check the DP Table Update column at Step 12 in execution_table.

At which step does the DP table first record a match and increase the subsequence length?

A Step 2

B Step 7

C Step 5

D Step 10

💡 Hint

Look for the first step where 'Match' appears in the Operation column in execution_table.

According to variable_tracker, what is the final LCS length after all steps?

A 3

B 2

C 1

D 4

💡 Hint

Check the 'LCS length' row in variable_tracker under the 'Final' column.

Concept Snapshot

Longest Common Subsequence (LCS):
- Use a 2D dp table of size (m+1)x(n+1) for strings of length m and n.
- Initialize first row and column to 0.
- If characters match: dp[i][j] = dp[i-1][j-1] + 1.
- Else: dp[i][j] = max(dp[i-1][j], dp[i][j-1]).
- Result is dp[m][n], trace back to find the sequence.

Full Transcript

Longest Common Subsequence finds the longest sequence common to two strings. We create a table dp where dp[i][j] stores the length of LCS of prefixes X[0..i-1] and Y[0..j-1]. We fill the first row and column with zeros because an empty string has no common subsequence. For each character pair, if they match, we add 1 to dp[i-1][j-1]. If not, we take the maximum of dp[i-1][j] and dp[i][j-1]. After filling the table, dp[m][n] gives the length of the LCS. We can trace back from dp[m][n] to find the actual subsequence. This method uses dynamic programming to avoid repeated work and efficiently find the LCS length and sequence.

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