Longest Common Substring in DSA C++ - Time & Space Complexity

Time Complexity: Longest Common Substring

O(n x m)

Understanding Time Complexity

We want to understand how the time needed to find the longest common substring between two strings changes as the strings get longer.

How does the work grow when input strings increase in size?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int longestCommonSubstring(const string &s1, const string &s2) {
    int n = s1.size(), m = s2.size();
    vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
    int maxLen = 0;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            if (s1[i - 1] == s2[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
                maxLen = max(maxLen, dp[i][j]);
            } else {
                dp[i][j] = 0;
            }
        }
    }
    return maxLen;
}
    

This code finds the length of the longest substring that appears in both strings by building a table of matching substring lengths.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops comparing characters of both strings and updating a table.
• How many times: Outer loop runs n times, inner loop runs m times, so total n x m times.

How Execution Grows With Input

As the lengths of the two strings grow, the number of operations grows by multiplying their sizes.

Input Size (n = m)	Approx. Operations
10	100
100	10,000
1000	1,000,000

Pattern observation: Doubling the string length roughly quadruples the work because both loops multiply.

Final Time Complexity

Time Complexity: O(n x m)

This means the time needed grows proportionally to the product of the lengths of the two input strings.

Common Mistake

[X] Wrong: "The algorithm only needs to scan one string once, so it is O(n)."

[OK] Correct: The algorithm compares every character of the first string with every character of the second, so it must do work proportional to n times m.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming solves problems by breaking them into smaller parts and why nested loops often lead to multiplied work.

Self-Check

"What if we used a suffix tree or suffix array instead? How would the time complexity change?"

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