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 grows as the input strings get longer.

How does the work increase when the strings become bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int LongestCommonSubstring(string s1, string s2) {
    int maxLen = 0;
    int[,] dp = new int[s1.Length + 1, s2.Length + 1];
    for (int i = 1; i <= s1.Length; i++) {
        for (int j = 1; j <= s2.Length; j++) {
            if (s1[i - 1] == s2[j - 1]) {
                dp[i, j] = dp[i - 1, j - 1] + 1;
                if (dp[i, j] > maxLen) maxLen = dp[i, j];
            } else {
                dp[i, j] = 0;
            }
        }
    }
    return maxLen;
}
    

This code finds the length of the longest substring that appears in both input strings by checking all pairs of characters and building up a table.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops comparing characters of both strings and updating the table.
• How many times: The outer loop runs for each character in the first string (n times), and the inner loop runs for each character in the second string (m times), so total n x m times.

How Execution Grows With Input

As the strings get longer, the number of comparisons grows by multiplying their lengths.

Input Size (n = length of s1, m = length of s2)	Approx. Operations
10 and 10	100
100 and 100	10,000
1000 and 1000	1,000,000

Pattern observation: The work grows roughly by the product of the lengths, so doubling both strings makes the work about four times bigger.

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 strings.

Common Mistake

[X] Wrong: "Since we only check characters one by one, the time is O(n) or O(m)."

[OK] Correct: We actually compare every character of the first string with every character of the second string, so the total checks multiply, not just add.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming solves problems by breaking them into smaller parts and why nested loops matter for performance.

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?

easy

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

B. The longest sequence of characters that appear anywhere in both strings, not necessarily consecutively

C. The total number of common characters between two strings

D. The longest string formed by combining both strings

Solution

1. Step 1: Understand the problem definition

   The Longest Common Substring problem looks for the longest sequence of characters that appear in both strings without any gaps.

2. Step 2: Differentiate from similar problems

   Unlike Longest Common Subsequence, substrings must be continuous in both strings.

3. Final Answer:

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

4. Quick Check:

   Longest Common Substring = consecutive matching characters [OK]

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

Common Mistakes:

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

2. Which of the following C# code snippets correctly initializes a 2D array for dynamic programming in Longest Common Substring?

easy

A. int[][] dp = new int[m][n];

B. int[,] dp = new int[m][n];

C. int[,] dp = new int[m + 1, n + 1];

D. int[][] dp = new int[m + 1, n + 1];

Solution

1. Step 1: Recall C# syntax for 2D arrays

   In C#, a rectangular 2D array is declared as int[,] with sizes specified in new int[m + 1, n + 1].

2. Step 2: Check each option

   int[,] dp = new int[m + 1, n + 1]; uses correct syntax. Options A and D mix jagged and rectangular array syntax incorrectly. int[,] dp = new int[m][n]; uses wrong initialization syntax.

3. Final Answer:

   int[,] dp = new int[m + 1, n + 1]; -> Option C

4. Quick Check:

   Rectangular 2D array syntax = int[,] dp = new int[m + 1, n + 1]; [OK]

Hint: Use int[,] for 2D arrays, not jagged arrays for DP [OK]

Common Mistakes:

• Using jagged array syntax incorrectly
• Missing +1 for boundary conditions
• Mixing rectangular and jagged array syntax

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

string s1 = "abcde";
string s2 = "abfde";
int m = s1.Length;
int n = s2.Length;
int[,] dp = new int[m + 1, n + 1];
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;
      if (dp[i, j] > maxLen) maxLen = dp[i, j];
    } else {
      dp[i, j] = 0;
    }
  }
}
Console.WriteLine(maxLen);

medium

A. 2

B. 3

C. 4

D. 1

Solution

1. Step 1: Trace matching characters between s1 and s2

   Strings: s1 = "abcde", s2 = "abfde". Matching substrings are "ab" at start and "de" at end, both length 2.

2. Step 2: Identify longest common substring length

   Longest common substring length is 2, either "ab" or "de".

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Longest common substring length = 2 [OK]

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

Common Mistakes:

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

4. Identify the bug in this Longest Common Substring C# code snippet:

string s1 = "hello";
string s2 = "yellow";
int m = s1.Length;
int n = s2.Length;
int[,] dp = new int[m, n];
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;
      if (dp[i, j] > maxLen) maxLen = dp[i, j];
    } else {
      dp[i, j] = 0;
    }
  }
}
Console.WriteLine(maxLen);

medium

A. The comparison should be s1[i] == s2[j] without -1

B. The dp array size should be (m+1, n+1) to avoid index errors

C. The loops should start from 0 instead of 1

D. maxLen should be initialized to -1

Solution

1. Step 1: Check dp array size and indexing

   dp is declared as int[m, n], but loops access dp[i, j] with i and j up to m and n, causing index out of range.

2. Step 2: Correct dp size for safe indexing

   dp should be int[m+1, n+1] to allow indices from 0 to m and 0 to n safely.

3. Final Answer:

   The dp array size should be (m+1, n+1) to avoid index errors -> Option B

4. Quick Check:

   dp size = (m+1, n+1) for safe indexing [OK]

Hint: Add +1 to dp array size for boundary indexing [OK]

Common Mistakes:

• Using dp size without +1 causes runtime errors
• Starting loops at 0 causes negative indexing
• Incorrect character indexing without -1

5. Given two strings s1 = "xabxac" and s2 = "abcabxabcd", what is the length of their longest common substring?

hard

A. 3

B. 6

C. 5

D. 4

Solution

1. Step 1: Identify common substrings between s1 and s2

   Check substrings like "abx", "bxa", "xab", "abxa". The longest common substring is "abxa" with length 4.

2. Step 2: Confirm no longer substring exists

   "abxac" is in s1 but not fully in s2 consecutively. So length 4 is maximum.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Longest common substring length = 4 [OK]

Hint: Look for longest continuous matching sequence [OK]

Common Mistakes:

• Counting subsequence instead of substring
• Overestimating length by including non-continuous chars
• Ignoring overlapping substrings