Longest Common Subsequence in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the 2D list for dynamic programming.

DSA Python

dp = [[0 for _ in range(len(s2) + 1)] for _ in range([1] + 1)]

Drag options to blanks, or click blank then click option'

Alen(s1) - 1

Blen(s1)

Clen(s2) - 1

Dlen(s2)

Attempts:

3 left

💡 Hint

Common Mistakes

Using length of s2 instead of s1

Not adding 1 to length

We create a 2D list with rows equal to length of s1 plus one, to include the empty prefix.

2fill in blank

medium

Complete the code to check if characters match at current indices.

DSA Python

if s1[i - 1] [1] s2[j - 1]:

Drag options to blanks, or click blank then click option'

A!=

B>

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using inequality instead of equality

Using comparison operators like > or <

We check if characters at positions i-1 and j-1 are equal to extend the subsequence.

3fill in blank

hard

Fix the error in updating dp when characters do not match.

DSA Python

dp[i][j] = max(dp[i - 1][j], dp[i][[1]])

Drag options to blanks, or click blank then click option'

Aj - 1

Bj + 1

Ci - 1

Di + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using j+1 which is out of range

Using i-1 instead of j-1

When characters don't match, we take the max of dp[i-1][j] and dp[i][j-1].

4fill in blank

hard

Fill both blanks to complete the loop headers for building dp table.

DSA Python

for i in range(1, [1] + 1):
    for j in range(1, [2] + 1):

Drag options to blanks, or click blank then click option'

Alen(s1)

Blen(s2)

Clen(s1) - 1

Dlen(s2) - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using lengths minus one

Swapping s1 and s2 lengths

We loop from 1 to length of s1 and s2 to fill dp table.

5fill in blank

hard

Fill all three blanks to reconstruct the LCS string from dp table.

DSA Python

while i > 0 and j > 0:
    if s1[i - 1] == s2[j - 1]:
        lcs = s1[[1]] + lcs
        i -= [2]
        j -= [3]
    elif dp[i - 1][j] > dp[i][j - 1]:
        i -= 1
    else:
        j -= 1

Drag options to blanks, or click blank then click option'

Ai - 1

B1

Cj - 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using j-1 for character index

Decreasing i or j by 0 instead of 1

We add character at i-1, then move diagonally up-left by decreasing i and j by 1.

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 total number of characters in both strings combined.

D. The shortest sequence of characters that can be formed from both strings.

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 consecutively.

2. Step 2: Differentiate from substring

   Unlike substring, LCS does not require characters to be contiguous.

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 sequence [OK]

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

Common Mistakes:

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

2. Which of the following Python code snippets correctly initializes a 2D list for LCS dynamic programming of size (m+1) x (n+1)?

easy

A. dp = [[0] * (m+1)] * (n+1)

B. dp = [[0] * m for _ in range(n)]

C. dp = [0] * (m+1) * (n+1)

D. dp = [[0] * (n+1) for _ in range(m+1)]

Solution

1. Step 1: Understand 2D list initialization

   We need a list of lists with (m+1) rows and (n+1) columns filled with zeros.

2. Step 2: Check each option

   dp = [[0] * (n+1) for _ in range(m+1)] uses list comprehension correctly to create independent rows. dp = [[0] * (m+1)] * (n+1) uses multiplication which creates references to the same list, causing bugs.

3. Final Answer:

   dp = [[0] * (n+1) for _ in range(m+1)] -> Option D

4. Quick Check:

   Use list comprehension for 2D lists [OK]

Hint: Use list comprehension to avoid shared references [OK]

Common Mistakes:

• Using multiplication for 2D lists causing shared rows
• Mixing up m and n dimensions
• Initializing with 1D list instead of 2D

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

def lcs_length(X, Y):
    m, n = len(X), len(Y)
    dp = [[0] * (n+1) for _ in range(m+1)]
    for i in range(1, m+1):
        for j in range(1, n+1):
            if X[i-1] == Y[j-1]:
                dp[i][j] = dp[i-1][j-1] + 1
            else:
                dp[i][j] = max(dp[i-1][j], dp[i][j-1])
    return dp[m][n]

print(lcs_length('AGGTAB', 'GXTXAYB'))

medium

A. 5

B. 4

C. 6

D. 7

Solution

1. Step 1: Understand the LCS calculation

   The function calculates the length of the longest common subsequence between 'AGGTAB' and 'GXTXAYB'.

2. Step 2: Trace the LCS

   A common LCS is 'GTAB' (G at pos 1 or 2 in X, T at 3, A at 4, B at 5; matching positions in Y: G0, T2, A4, B6), length 4. The standard DP code correctly computes dp[6][7] = 4.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   LCS length = 4 [OK]

Hint: Trace characters matching in order for LCS length [OK]

Common Mistakes:

• Counting longest common substring instead of subsequence
• Miscounting characters due to off-by-one errors
• Confusing length with actual subsequence

4. The following code attempts to compute the LCS length but has a bug. What is the error?

def lcs_bug(X, Y):
    m, n = len(X), len(Y)
    dp = [[0] * (m+1) for _ in range(n+1)]
    for i in range(1, m+1):
        for j in range(1, n+1):
            if X[i-1] == Y[j-1]:
                dp[i][j] = dp[i-1][j-1] + 1
            else:
                dp[i][j] = max(dp[i-1][j], dp[i][j-1])
    return dp[m][n]

print(lcs_bug('ABC', 'AC'))

medium

A. The dp array dimensions are swapped; should be (m+1) x (n+1).

B. The loops iterate over wrong ranges; should start from 0.

C. The comparison uses wrong indices; should be X[i] and Y[j].

D. The return statement returns dp[0][0] instead of dp[m][n].

Solution

1. Step 1: Check dp array initialization

   The dp array is initialized with dimensions (n+1) x (m+1), which is swapped. It should be (m+1) x (n+1) to match loops.

2. Step 2: Consequences of swapped dimensions

   This causes index errors or incorrect results because dp[i][j] expects i in [0..m], j in [0..n].

3. Final Answer:

   The dp array dimensions are swapped; should be (m+1) x (n+1). -> Option A

4. Quick Check:

   dp size matches string lengths [OK]

Hint: Match dp size with string lengths order [OK]

Common Mistakes:

• Swapping m and n in dp initialization
• Starting loops at 0 instead of 1
• Using wrong indices for string characters

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

hard

A. 'BBCA'

B. 'BCBD'

C. 'BDAB'

D. 'BDA'

Solution

1. Step 1: Identify LCS length and candidates

   The LCS length for these strings is 4. Verify which option of length 4 appears in order in both X='ABCBDAB' and Y='BDCABA'.

2. Step 2: Verify each option

   A: 'BDAB' - In X: B(1), D(4), A(5), B(6); in Y: B(0), D(1), A(3), B(4). Order preserved.
   B: 'BCBD' - Requires B then C(2) then B(4) then D after 4, but no D after pos 4 in Y (only D at 1).
   C: 'BBCA' - Requires two B's before C(2) in Y, but only one B at 0 before C(2).
   D: 'BDA' length 3 < 4.

3. Final Answer:

   'BDAB' -> Option C

4. Quick Check:

   Check subsequence order in both strings [OK]

Hint: Check character order in both strings carefully [OK]

Common Mistakes:

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