Longest Common Subsequence in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of LCS length calculation

What is the output of the following Python code that calculates the length of the Longest Common Subsequence (LCS) between two strings?

DSA Python

def lcs_length(X, Y):
    m = len(X)
    n = 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"))

A 3

B 5

C 4

D 6

💡 Hint

Trace the matching characters and count the longest sequence common to both strings.

The longest common subsequence between "AGGTAB" and "GXTXAYB" is "GTAB" which has length 4. The code returns 4.

❓ Predict Output

intermediate

Output of LCS sequence reconstruction

What is the output of the following Python code that reconstructs the Longest Common Subsequence (LCS) string between two strings?

DSA Python

def lcs_sequence(X, Y):
    m = len(X)
    n = 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])
    i, j = m, n
    lcs_str = []
    while i > 0 and j > 0:
        if X[i - 1] == Y[j - 1]:
            lcs_str.append(X[i - 1])
            i -= 1
            j -= 1
        elif dp[i - 1][j] > dp[i][j - 1]:
            i -= 1
        else:
            j -= 1
    return ''.join(reversed(lcs_str))

print(lcs_sequence("ABCBDAB", "BDCABA"))

A "BCBA"

B "BDCB"

C "BCAB"

D "BDAB"

💡 Hint

Follow the path from dp[m][n] back to dp[0][0] to build the LCS string.

The longest common subsequence between "ABCBDAB" and "BDCABA" is "BDAB".

🧠 Conceptual

advanced

Understanding LCS dynamic programming table size

Given two strings of lengths m and n, what is the size of the dynamic programming table used to compute the Longest Common Subsequence (LCS)?

A (m + 1) x (n + 1)

B (m - 1) x (n - 1)

C m x n

D m + n

💡 Hint

The table includes an extra row and column for base cases.

The DP table has (m+1) rows and (n+1) columns to handle empty prefixes of the strings.

🔧 Debug

advanced

Identify the error in LCS length code

What error will the following code produce when trying to compute the LCS length?

DSA Python

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

print(lcs_length("ABC", "AC"))

A IndexError due to negative indices on first row/column

B TypeError due to invalid max arguments

C SyntaxError due to missing colon

D No error, outputs 2

💡 Hint

Check how dp[i-1][j-1] is accessed when i or j is 0.

The code does not handle the base case properly and tries to access dp[-1][-1] which causes IndexError.

🚀 Application

expert

Number of distinct LCS sequences

Given two strings "ABCBDAB" and "BDCABA", how many distinct Longest Common Subsequences (LCS) of maximum length exist?

A 4

B 3

C 2

D 1

💡 Hint

There can be multiple LCS strings of the same maximum length; count them carefully.

For these strings, there are 3 distinct LCS sequences of length 4: "BCAB", "BDAB", and "BCBA".

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