Longest Common Substring in DSA C - Execution Trace

Concept Flow - Longest Common Substring

Start with two strings

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Create 2D table of size (len1+1) x (len2+1)

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Initialize all cells to 0

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For each character pair (i,j)

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If chars match: table[i

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Update max length and position if needed

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After filling table, extract substring from max position

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Return longest common substring

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Done

We build a table to track matching characters between two strings, updating lengths of common substrings and remembering the longest found.

Execution Sample

DSA C

char* longestCommonSubstring(char* s1, char* s2) {
  int len1 = strlen(s1), len2 = strlen(s2);
  int table[len1+1][len2+1];
  int maxLen = 0, endPos = 0;
  for (int i = 0; i <= len1; i++)
    for (int j = 0; j <= len2; j++)
      table[i][j] = 0;
  for (int i = 1; i <= len1; i++) {
    for (int j = 1; j <= len2; j++) {
      if (s1[i-1] == s2[j-1]) {
        table[i][j] = table[i-1][j-1] + 1;
        if (table[i][j] > maxLen) {
          maxLen = table[i][j];
          endPos = i;
        }
      } else {
        table[i][j] = 0;
      }
    }
  }
  char* result = malloc(maxLen + 1);
  strncpy(result, s1 + endPos - maxLen, maxLen);
  result[maxLen] = '\0';
  return result;
}

This code finds the longest substring common to both input strings by building a table of matching character lengths.

Execution Table

Step	Operation	Indices (i,j)	Chars Compared	Table[i][j]	Max Length	End Position	Visual State (Table snippet)
1	Initialize table	-	-	-	0	0	All zeros
2	Compare chars	(1,1)	'a' vs 'b'	0	0	0	table[1][1]=0
3	Compare chars	(1,2)	'a' vs 'a'	1	1	1	table[1][2]=1
4	Compare chars	(1,3)	'a' vs 'c'	0	1	1	table[1][3]=0
5	Compare chars	(2,1)	'b' vs 'b'	1	1	1	table[2][1]=1
6	Compare chars	(2,2)	'b' vs 'a'	0	1	1	table[2][2]=0
7	Compare chars	(2,3)	'b' vs 'c'	0	1	1	table[2][3]=0
8	Compare chars	(3,1)	'c' vs 'b'	0	1	1	table[3][1]=0
9	Compare chars	(3,2)	'c' vs 'a'	0	1	1	table[3][2]=0
10	Compare chars	(3,3)	'c' vs 'c'	1	1	1	table[3][3]=1
11	Compare chars	(4,1)	'd' vs 'b'	0	1	1	table[4][1]=0
12	Compare chars	(4,2)	'd' vs 'a'	0	1	1	table[4][2]=0
13	Compare chars	(4,3)	'd' vs 'c'	0	1	1	table[4][3]=0
14	Compare chars	(5,1)	'e' vs 'b'	0	1	1	table[5][1]=0
15	Compare chars	(5,2)	'e' vs 'a'	0	1	1	table[5][2]=0
16	Compare chars	(5,3)	'e' vs 'c'	0	1	1	table[5][3]=0
17	End of loops	-	-	-	1	1	Longest substring length=1, ends at s1 index 1
18	Extract substring	-	-	-	-	-	Substring: 'a'

💡 All character pairs compared; longest common substring length found and extracted.

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 10	After Step 17	Final
maxLen	0	1	1	1	1	1
endPos	0	1	1	1	1	1
table[1][2]	0	1	1	1	1	1
table[2][1]	0	0	1	1	1	1
table[3][3]	0	0	0	1	1	1

Key Moments - 3 Insights

Why do we start filling the table from index 1 instead of 0?

Because table[0][*] and table[*][0] represent empty prefixes, initialized to zero to handle boundary cases. See execution_table rows 2-3 where i and j start at 1.

Why do we add 1 to table[i-1][j-1] when characters match?

Because a matching character extends the previous common substring by one. This is shown in execution_table row 3 where table[1][2] = table[0][1] + 1 = 1.

How do we know which substring is the longest after filling the table?

We track maxLen and endPos during filling. The longest substring ends at endPos in s1 with length maxLen, as shown in execution_table row 17.

Visual Quiz - 3 Questions

Look at the execution_table at step 3. What is the value of table[1][2] and what does it represent?

A 2, sum of previous matches

B 0, no match at these indices

C 1, length of common substring ending at s1[0] and s2[1]

D Undefined, not calculated yet

💡 Hint

Refer to execution_table row 3 under 'Table[i][j]' and 'Chars Compared'

At which step does the algorithm update maxLen for the first time?

A Step 2

B Step 3

C Step 5

D Step 10

💡 Hint

Check execution_table rows where maxLen changes from 0 to 1

If the input strings were 'abc' and 'abc', what would be the maxLen after filling the table?

A 3

B 1

C 2

D 0

💡 Hint

Longest common substring length equals the full string length if both strings are identical

Concept Snapshot

Longest Common Substring:
- Use 2D table of size (len1+1) x (len2+1)
- Initialize all cells to 0
- For each i,j: if s1[i-1]==s2[j-1], table[i][j] = table[i-1][j-1] + 1
- Track max length and end position
- Extract substring from s1 using end position and max length

Full Transcript

The Longest Common Substring algorithm compares two strings by building a 2D table. Each cell table[i][j] stores the length of the longest common substring ending at s1[i-1] and s2[j-1]. We initialize the table with zeros. For each character pair, if they match, we add 1 to the value from the previous diagonal cell. We keep track of the maximum length found and the position where it ends in s1. After filling the table, we extract the substring from s1 using the recorded position and length. This method efficiently finds the longest substring shared by both strings.