Longest Common Subsequence in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Longest Common Subsequence (LCS)?

The Longest Common Subsequence is the longest sequence of characters that appear in the same relative order (not necessarily contiguous) in two given sequences.

beginner

How does the LCS problem differ from the Longest Common Substring problem?

LCS allows characters to be non-contiguous but in order, while Longest Common Substring requires characters to be contiguous in both sequences.

intermediate

What is the main approach used to solve the LCS problem efficiently?

Dynamic Programming is used to solve LCS by building a table that stores lengths of LCS for substrings, avoiding repeated calculations.

intermediate

In the LCS dynamic programming table, what does each cell dp[i][j] represent?

dp[i][j] represents the length of the Longest Common Subsequence of the first i characters of the first string and the first j characters of the second string.

advanced

How do you reconstruct the LCS string from the DP table after computing its length?

Start from dp[m][n] and move backwards: if characters match, include that character and move diagonally; else move in the direction of the larger dp value (up or left) until reaching dp[0][0].

What does the Longest Common Subsequence problem find?

A The longest sequence common to both strings in order, but not necessarily contiguous

B The longest substring common to both strings

C The shortest sequence common to both strings

D The longest sequence common to both strings but reversed

LCS finds the longest sequence common to both strings in the same order, but characters do not have to be next to each other.

Which technique is commonly used to solve the LCS problem efficiently?

A Dynamic Programming

B Greedy Algorithm

C Divide and Conquer

D Backtracking only

Dynamic Programming stores results of subproblems to avoid repeated work, making LCS efficient.

If characters at positions i and j in two strings match, how is dp[i][j] computed?

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

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

C 0

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

When characters match, we add 1 to the LCS length found by excluding those characters.

What is the time complexity of the standard LCS dynamic programming solution for strings of length m and n?

A O(m + n)

B O(m^2 + n^2)

C O(m * n)

D O(2^(m+n))

The DP table has m rows and n columns, so the time complexity is O(m * n).

Which of the following is true about the LCS of strings "ABCBDAB" and "BDCABA"?

A One LCS is "BCBA"

B One LCS is "BDAB"

C One LCS is "BAB"

D One LCS is "BDA"

"BDAB" is a valid longest common subsequence for these strings.

Explain how to build the dynamic programming table for the Longest Common Subsequence problem.

Think about comparing characters and filling the table row by row.

Describe the steps to reconstruct the actual Longest Common Subsequence string from the DP table.

Trace back from the end of the table to the start.

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