Longest Common Substring in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the maximum length variable for the longest common substring.

DSA C++

int maxLength = [1];

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

A0

B1

C-1

DINT_MAX

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing maxLength to 1 or a negative number.

We start with maxLength as 0 because initially, we haven't found any common substring.

2fill in blank

medium

Complete the code to create a 2D vector dp with dimensions (m+1) x (n+1) initialized to zero.

DSA C++

std::vector<std::vector<int>> dp([1], std::vector<int>([2], 0));

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

Am+1

Bn+1

Cm-1

Dn-1

Em

Fn

Attempts:

3 left

💡 Hint

Common Mistakes

Using m and n instead of m+1 and n+1, causing index errors.

We use m+1 and n+1 to include the base case row and column for empty substrings.

3fill in blank

hard

Fix the error in the condition to check if characters at positions i-1 and j-1 in strings s1 and s2 match.

DSA C++

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 '!=' instead of '==' causing logic to fail.

We check if characters are equal to extend the common substring.

4fill in blank

hard

Fill both blanks to update dp[i][j] and maxLength when characters match.

DSA C++

dp[i][j] = dp[i-1][j-1] [1] 1;
if (dp[i][j] [2] maxLength) maxLength = dp[i][j];

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

A+

B>

C<

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' causing wrong substring length.

Using '<' instead of '>' in comparison.

We add 1 to the previous diagonal dp value to extend substring length and update maxLength if current is greater.

5fill in blank

hard

Fill all three blanks to complete the nested loops iterating over strings s1 and s2.

DSA C++

for (int i = 1; i [1] m+1; i[2]) {
    for (int j = 1; j [3] n+1; j[2]) {

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

A<

B++

C<=

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causing out-of-bound errors.

Using '--' causing infinite loops.

Loops run from 1 to m and n (exclusive of m+1 and n+1) with increment ++ to cover all characters.

Practice

1. What does the Longest Common Substring problem find between two strings s1 and s2?

easy

A. The longest sequence of characters that appear contiguously in both strings

B. The longest sequence of characters that appear in both strings but not necessarily contiguously

C. The total number of common characters between the two strings

D. The longest string formed by concatenating both strings

Solution

1. Step 1: Understand the problem definition

   The Longest Common Substring problem looks for the longest continuous sequence of characters that appears in both strings without any breaks.

2. Step 2: Differentiate from similar problems

   Unlike Longest Common Subsequence, which allows gaps, Longest Common Substring requires characters to be contiguous in both strings.

3. Final Answer:

   The longest sequence of characters that appear contiguously in both strings -> Option A

4. Quick Check:

   Longest Common Substring = contiguous matching characters [OK]

Hint: Look for continuous matching characters in both strings [OK]

Common Mistakes:

• Confusing substring with subsequence
• Thinking characters can be non-contiguous
• Counting total common characters instead of longest continuous

2. Which of the following C++ code snippets correctly initializes a 2D vector dp of size m+1 by n+1 with all zeros for Longest Common Substring?

easy

A. std::vector dp(m+1, std::vector<int>(n+1, 0));

B. std::vector<int> dp(m+1, std::vector<int>(n+1, 0));

C. std::vector dp(m, std::vector<int>(n));

D. std::vector dp(m+1, n+1, 0);

Solution

1. Step 1: Understand 2D vector initialization syntax

   To create a 2D vector with m+1 rows and n+1 columns filled with zeros, use nested vector initialization.

2. Step 2: Check each option

   std::vector dp(m+1, std::vector<int>(n+1, 0)); correctly creates a vector of vectors with m+1 rows, each row a vector of n+1 zeros. Others have syntax errors or wrong types.

3. Final Answer:

   std::vector dp(m+1, std::vector<int>(n+1, 0)); -> Option A

4. Quick Check:

   Correct 2D vector initialization = std::vector dp(m+1, std::vector<int>(n+1, 0)); [OK]

Hint: Use nested vector constructor for 2D zero initialization [OK]

Common Mistakes:

• Using wrong vector types in initialization
• Missing inner vector initialization
• Trying to pass three arguments to vector constructor

3. What is the output of the following C++ code snippet?

std::string s1 = "abcdxyz";
std::string s2 = "xyzabcd";
int m = s1.size(), n = s2.size();
std::vector dp(m+1, std::vector<int>(n+1, 0));
int maxLen = 0;
for (int i = 1; i <= m; i++) {
  for (int j = 1; j <= n; j++) {
    if (s1[i-1] == s2[j-1]) {
      dp[i][j] = dp[i-1][j-1] + 1;
      maxLen = std::max(maxLen, dp[i][j]);
    }
  }
}
std::cout << maxLen << std::endl;

medium

A. 3

B. 4

C. 1

D. 0

Solution

1. Step 1: Trace matching substrings

   The longest common substring is "abcd" (length 4), which appears contiguously at the start of s1 and end of s2. "xyz" is length 3.

2. Step 2: Check dp updates and maxLen

   dp updates length for matching characters. The longest continuous match found is length 4 for "abcd".

3. Final Answer:

   4 -> Option B

4. Quick Check:

   Longest common substring length = 4 [OK]

Hint: Check continuous matching segments, not total matches [OK]

Common Mistakes:

• Confusing substring with subsequence length
• Counting non-contiguous matches
• Off-by-one errors in indexing

4. Identify the bug in the following C++ code snippet for Longest Common Substring:

std::string s1 = "abcde";
std::string s2 = "abfde";
int m = s1.size(), n = s2.size();
std::vector dp(m+1, std::vector<int>(n+1, 0));
int maxLen = 0;
for (int i = 1; i <= m; i++) {
  for (int j = 1; j <= n; j++) {
    if (s1[i] == s2[j]) {
      dp[i][j] = dp[i-1][j-1] + 1;
      maxLen = std::max(maxLen, dp[i][j]);
    }
  }
}
std::cout << maxLen << std::endl;

medium

A. dp vector size should be (m+1) x (n+1) to avoid out-of-bound errors

B. maxLen should be initialized to -1 instead of 0

C. The loop indices should start from 0 instead of 1

D. The comparison should be s1[i-1] == s2[j-1] instead of s1[i] == s2[j]

Solution

1. Step 1: Check string indexing inside loops

   Loop variables i and j start from 1, so to access string characters correctly, indices should be i-1 and j-1.

2. Step 2: Identify the incorrect comparison

   The code compares s1[i] and s2[j], which causes out-of-bound access and wrong comparisons. It should be s1[i-1] == s2[j-1].

3. Final Answer:

   The comparison should be s1[i-1] == s2[j-1] instead of s1[i] == s2[j] -> Option D

4. Quick Check:

   Correct string indexing = i-1 and j-1 [OK]

Hint: Use i-1 and j-1 to index strings when loops start at 1 [OK]

Common Mistakes:

• Accessing string with i or j directly causing out-of-bounds
• Not initializing dp with correct size
• Starting loops at wrong indices

5. Given two strings s1 = "banana" and s2 = "ananas", write a C++ function to find the longest common substring and return it. What is the correct longest common substring?

hard

A. "anana"

B. "nana"

C. "ana"

D. "ban"

Solution

1. Step 1: Understand the longest common substring candidates

   Check substrings common to both strings. "ana" (length 3) appears multiple times in both strings. "anana" is not a substring of both strings contiguously.

2. Step 2: Verify other options

   "nana" length 4 is not a substring of s2; "ban" only in s1; "anana" is not a substring of s2.

3. Final Answer:

   "ana" -> Option C

4. Quick Check:

   Longest common substring between "banana" and "ananas" is "ana" [OK]

Hint: Check all continuous matches, pick longest common substring [OK]

Common Mistakes:

• Choosing longest common subsequence instead of substring
• Ignoring continuity requirement
• Assuming longer substrings without verifying presence