Longest Common Substring in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the 2D array for dynamic programming.

DSA Typescript

const dp: number[][] = Array([1]).fill(0).map(() => Array(m + 1).fill(0));

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

Am + 1

Bn

Cn + 1

Dm

Attempts:

3 left

💡 Hint

Common Mistakes

Using m+1 instead of n+1 for rows.

Not adding 1 to the length causing index errors.

We create a 2D array with (n+1) rows to cover all prefixes of the first string including empty prefix.

2fill in blank

medium

Complete the code to update the dp table when characters match.

DSA Typescript

if (s1[i - 1] === s2[j - 1]) {
  dp[i][j] = dp[i - 1][[1]] + 1;
} else {
  dp[i][j] = 0;
}

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

Aj - 1

Bj

Cj + 1

Dj - 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using dp[i-1][j] or dp[i][j-1] instead of dp[i-1][j-1].

Adding 1 to the wrong index.

When characters match, we extend the substring by 1 from the previous diagonal cell dp[i-1][j-1].

3fill in blank

hard

Fix the error in updating the maximum length of the longest common substring.

DSA Typescript

if (dp[i][j] > [1]) {
  maxLen = dp[i][j];
  endIndex = i;
}

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

AendIndex

Bi

Cj

DmaxLen

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with endIndex instead of maxLen.

Using i or j as comparison variable.

We compare dp[i][j] with maxLen to update the maximum length found so far.

4fill in blank

hard

Fill both blanks to extract the longest common substring from s1.

DSA Typescript

const longestSubstring = s1.slice([1], [2]);

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

AendIndex - maxLen

BmaxLen

CendIndex

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxLen as start index.

Using 0 as start index instead of calculated start.

We slice s1 from (endIndex - maxLen) to endIndex to get the longest common substring.

5fill in blank

hard

Fill all three blanks to complete the nested loops for dp table computation.

DSA Typescript

for (let i = 1; i <= [1]; i++) {
  for (let j = 1; j <= [2]; j++) {
    if (s1[i - 1] === s2[j - 1]) {
      dp[i][j] = dp[i - 1][j - 1] + [3];
    } else {
      dp[i][j] = 0;
    }
  }
}

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

An

Bm

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 as increment instead of 1.

Swapping n and m in loops.

The outer loop runs from 1 to n, inner loop from 1 to m, and we add 1 when characters match.

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 in order but not necessarily contiguously

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

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

D. The shortest substring that contains all characters from both strings

Solution

1. Step 1: Understand substring vs subsequence

   A substring is a continuous part of a string, while a subsequence can skip characters.

2. Step 2: Define Longest Common Substring

   It finds the longest continuous sequence of characters present in both strings.

3. Final Answer:

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

4. Quick Check:

   Longest Common Substring = continuous match [OK]

Hint: Look for continuous matching characters, not scattered ones [OK]

Common Mistakes:

• Confusing substring with subsequence
• Thinking order doesn't matter
• Counting total common characters instead

2. Which of the following TypeScript function signatures correctly represents a function to find the longest common substring length between two strings s1 and s2?

easy

A. function longestCommonSubstring(s1: string, s2: string): number

B. function longestCommonSubstring(s1: number[], s2: number[]): string

C. function longestCommonSubstring(s1: string, s2: string): string[]

D. function longestCommonSubstring(s1: string, s2: string): void

Solution

1. Step 1: Identify input types

   The inputs are two strings, so parameters should be string.

2. Step 2: Identify output type

   The function returns the length of the longest common substring, which is a number.

3. Final Answer:

   function longestCommonSubstring(s1: string, s2: string): number -> Option A

4. Quick Check:

   Input strings, output number length [OK]

Hint: Match input/output types to problem description [OK]

Common Mistakes:

• Using arrays instead of strings as input
• Returning string or array instead of number
• Using void return type mistakenly

3. What is the output of the following TypeScript code?

function longestCommonSubstring(s1: string, s2: string): number {
  const m = s1.length, n = s2.length;
  let maxLen = 0;
  const dp: number[][] = Array(m + 1).fill(0).map(() => Array(n + 1).fill(0));
  for (let i = 1; i <= m; i++) {
    for (let j = 1; j <= n; j++) {
      if (s1[i - 1] === s2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1] + 1;
        maxLen = Math.max(maxLen, dp[i][j]);
      }
    }
  }
  return maxLen;
}

console.log(longestCommonSubstring('abcde', 'abfde'));

medium

A. 2

B. 3

C. 4

D. 1

Solution

1. Step 1: Compare characters of 'abcde' and 'abfde'

   Matching continuous substrings are 'ab' at start and 'de' at end, both length 2.

2. Step 2: Find longest continuous match

   Longest common substring length is 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Longest common substring length = 2 [OK]

Hint: Check continuous matching parts, not scattered matches [OK]

Common Mistakes:

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

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

function longestCommonSubstring(s1: string, s2: string): number {
  const m = s1.length, n = s2.length;
  let maxLen = 0;
  const dp: number[][] = Array(m).fill(0).map(() => Array(n).fill(0));
  for (let i = 1; i <= m; i++) {
    for (let j = 1; j <= n; j++) {
      if (s1[i - 1] === s2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1] + 1;
        maxLen = Math.max(maxLen, dp[i][j]);
      }
    }
  }
  return maxLen;
}

medium

A. Loop indices should start from 0 instead of 1

B. dp array is initialized with wrong size causing index errors

C. maxLen is not updated inside the loop

D. Comparison uses wrong characters from strings

Solution

1. Step 1: Check dp array initialization

   dp is created with size m x n but accessed with indices up to m and n (1-based), causing out-of-bound errors.

2. Step 2: Correct dp size

   dp should be (m+1) x (n+1) to safely use 1-based indexing in loops.

3. Final Answer:

   dp array is initialized with wrong size causing index errors -> Option B

4. Quick Check:

   dp size must match loop indices [OK]

Hint: Match dp size with loop ranges to avoid index errors [OK]

Common Mistakes:

• Not adding +1 to dp dimensions
• Starting loops at 1 but dp size too small
• Ignoring index out-of-bound runtime errors

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

hard

A. 3

B. 5

C. 6

D. 4

Solution

1. Step 1: Identify common substrings

   Common substrings include 'abx' and 'abxa'. 'abxa' is length 4 and appears in both strings.

2. Step 2: Confirm longest substring length

   No longer continuous substring than length 4 exists in both strings.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Longest common substring length = 4 [OK]

Hint: Look for longest continuous matching sequence visually [OK]

Common Mistakes:

• Counting non-continuous matches
• Choosing shorter common substrings
• Confusing substring with subsequence