Longest Common Substring in DSA C - Time & Space Complexity

Time Complexity: Longest Common Substring

O(m * n)

Understanding Time Complexity

We want to understand how the time needed to find the longest common substring grows as the input strings get longer.

Specifically, how does the work change when the strings increase in size?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int longestCommonSubstring(char *s1, char *s2) {
    int m = strlen(s1);
    int n = strlen(s2);
    int maxLen = 0;
    int dp[m+1][n+1];

    for (int i = 0; i <= m; i++) {
        for (int j = 0; j <= n; j++) {
            if (i == 0 || j == 0) dp[i][j] = 0;
            else 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 common to both input strings using a 2D table.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops over both strings filling the dp table.
• How many times: The inner code runs (m+1) * (n+1) times, where m and n are string lengths.

How Execution Grows With Input

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

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

Pattern observation: Doubling the length roughly quadruples the work, showing a multiplication effect.

Final Time Complexity

Time Complexity: O(m * n)

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

Common Mistake

[X] Wrong: "The algorithm only depends on the length of one string, so it is O(n)."

[OK] Correct: The algorithm compares every character of one string with every character of the other, so both lengths matter and multiply the work.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming solves problems efficiently by breaking them into smaller parts.

Self-Check

"What if we only compared substrings starting at the same index in both strings? 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 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 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 sequence of characters that appear in both strings without any breaks (consecutively).

2. Step 2: Differentiate from similar problems

   Unlike Longest Common Subsequence, which allows characters to be non-consecutive, Longest Common Substring requires consecutive matching characters.

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 consecutive matching characters in both strings [OK]

Common Mistakes:

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

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

easy

A. int dp[m+1][n+1] = {0};

B. int dp[m+1][n+1]; memset(dp, 0, sizeof(dp));

C. int dp[m][n]; for(int i=0;i

D. int dp[m+1][n+1]; for(int i=0;i<=m;i++) for(int j=0;j<=n;j++) dp[i][j] = -1;

Solution

1. Step 1: Understand array initialization in C

   To initialize a 2D array with zeros, we can declare it and then use memset to set all bytes to zero.

2. Step 2: Analyze each option

   int dp[m+1][n+1] = {0}; tries to initialize with {0} but this only works for fixed size arrays, not variable sizes. int dp[m+1][n+1]; memset(dp, 0, sizeof(dp)); declares and uses memset correctly. int dp[m][n]; for(int i=0;i

3. Final Answer:

   int dp[m+1][n+1]; memset(dp, 0, sizeof(dp)); -> Option B

4. Quick Check:

   Use memset to zero initialize arrays in C [OK]

Hint: Use memset to zero initialize arrays in C [OK]

Common Mistakes:

• Using {0} for variable size arrays
• Initializing with wrong values like 1 or -1
• Not initializing dp array before use

3. What is the output of the following C code snippet that finds the length of the longest common substring between s1 = "abcde" and s2 = "abfde"?

int longestCommonSubstring(char* s1, char* s2) {
    int m = strlen(s1), n = strlen(s2);
    int dp[m+1][n+1];
    memset(dp, 0, sizeof(dp));
    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;
            }
        }
    }
    return maxLen;
}

int main() {
    printf("%d", longestCommonSubstring("abcde", "abfde"));
    return 0;
}

medium

A. 2

B. 3

C. 4

D. 5

Solution

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

   Strings: s1 = "abcde", s2 = "abfde". Matching consecutive characters are "ab" at start and "de" at end. "ab" length = 2, "de" length = 2. The longest common substring length is 2.

2. Step 2: Check dp updates for longest substring

   At i=3, j=3, characters differ ('c' vs 'f'), so dp resets. The longest substring is "ab" or "de" with length 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Longest Common Substring length = 2 [OK]

Hint: Check consecutive matching chars, reset on mismatch [OK]

Common Mistakes:

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

4. The following code attempts to find the longest common substring length but has a bug. What is the bug?

int longestCommonSubstring(char* s1, char* s2) {
    int m = strlen(s1), n = strlen(s2);
    int dp[m][n];
    int maxLen = 0;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (s1[i] == s2[j]) {
                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;
}

medium

A. dp array is not initialized before use

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

C. The maxLen variable is not updated correctly

D. dp array size should be (m+1) x (n+1) to avoid out-of-bound access

Solution

1. Step 1: Analyze dp array size and indexing

   The dp array is declared as dp[m][n], but the code accesses dp[i-1][j-1] which can cause out-of-bound access when i=0 or j=0.

2. Step 2: Correct dp array size and indexing

   To safely access dp[i-1][j-1], dp should be sized (m+1) x (n+1) and loops should run from 1 to m and 1 to n, with dp[0][*] and dp[*][0] initialized to 0.

3. Final Answer:

   dp array size should be (m+1) x (n+1) to avoid out-of-bound access -> Option D

4. Quick Check:

   Use dp size (m+1)x(n+1) for safe indexing [OK]

Hint: Use dp size (m+1)x(n+1) to avoid negative indices [OK]

Common Mistakes:

• Using dp size m x n causing index errors
• Starting loops at 0 but accessing dp[i-1][j-1]
• Not initializing dp array elements

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

hard

A. 5

B. 4

C. 3

D. 6

Solution

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

   Check substrings of s1 and s2. Common substrings include "ana", "nan", "ana" again, and the longest is "ana" with length 3.

2. Step 2: Confirm no longer substring exists

   No substring longer than 3 exists in both strings consecutively. "anana" is not a substring of "banana" starting at index 1 (it's "anana" in s2 but not in s1 consecutively).

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Longest common substring between "banana" and "ananas" is "ana" length 3 [OK]

Hint: Look for "ana" substring in both strings [OK]

Common Mistakes:

• Confusing substring with subsequence
• Assuming "anana" is common substring
• Stopping at shorter matches like "an"