Why Longest Palindromic Substring in DSA Javascript?

The Big Idea

Discover how a clever trick turns a huge search into a quick find!

The Scenario

Imagine you have a long word or sentence, and you want to find the longest part of it that reads the same forwards and backwards, like 'racecar' or 'level'. Doing this by checking every possible part by hand is like looking for a needle in a haystack.

The Problem

Manually checking each part is slow and tiring. You might miss some parts or make mistakes because there are so many possible pieces to check. It takes too long and is easy to get confused.

The Solution

The Longest Palindromic Substring method uses a smart way to check parts of the word quickly. It expands around centers and finds the longest matching palindrome without checking every piece one by one. This saves time and effort.

Before vs After

✗ Before

for (let i = 0; i < s.length; i++) {
  for (let j = i; j < s.length; j++) {
    let substring = s.substring(i, j + 1);
    if (isPalindrome(substring)) {
      // check and update longest
    }
  }
}

✓ After

for (let center = 0; center < s.length; center++) {
  expandAroundCenter(s, center, center);
  expandAroundCenter(s, center, center + 1);
}

What It Enables

This lets you quickly find the longest palindrome inside any string, making text analysis and pattern finding easy and fast.

Real Life Example

In DNA research, finding palindromic sequences helps identify important genetic markers that read the same forwards and backwards, which can be crucial for understanding diseases.

Key Takeaways

Manually checking all substrings is slow and error-prone.

Expanding around centers finds palindromes efficiently.

This method speeds up finding the longest palindromic substring.

Practice

1. What is the main goal of the Longest Palindromic Substring problem?

easy

A. Find the longest substring that reads the same forwards and backwards

B. Find the longest substring with all unique characters

C. Find the longest substring with increasing characters

D. Find the longest substring that contains only vowels

Solution

1. Step 1: Understand palindrome definition

   A palindrome reads the same forwards and backwards, like 'madam' or 'racecar'.

2. Step 2: Identify problem goal

   The problem asks to find the longest substring inside a string that is a palindrome.

3. Final Answer:

   Find the longest substring that reads the same forwards and backwards -> Option A

4. Quick Check:

   Longest Palindromic Substring = Find palindrome substring [OK]

Hint: Look for substrings that mirror around a center [OK]

Common Mistakes:

• Confusing palindrome with unique characters
• Thinking about subsequences instead of substrings
• Ignoring the order of characters

2. Which of the following JavaScript code snippets correctly initializes a 2D array dp for dynamic programming in the Longest Palindromic Substring problem?

easy

A. const dp = Array.from({ length: n }, () => Array(n).fill(false));

B. const dp = Array(n).fill(Array(n).fill(false));

C. const dp = new Array(n).fill(new Array(n).fill(false));

D. const dp = Array(n).map(Array(n).fill(false));

Solution

1. Step 1: Understand 2D array initialization in JavaScript

   Using Array.from with a mapping function creates independent inner arrays.

2. Step 2: Analyze each option

   B and C fill with the same inner array reference, causing bugs. D uses map incorrectly by passing an array instead of a function, which won't work. A correctly creates independent arrays.

3. Final Answer:

   const dp = Array.from({ length: n }, () => Array(n).fill(false)); -> Option A

4. Quick Check:

   Correct 2D array init = const dp = Array.from({ length: n }, () => Array(n).fill(false)); [OK]

Hint: Use Array.from with arrow function for independent inner arrays [OK]

Common Mistakes:

• Using fill with object references causing shared inner arrays
• Using map on uninitialized arrays
• Not initializing inner arrays separately

3. What will be the output of the following JavaScript code snippet?

function longestPalindrome(s) {
  let maxLen = 1, start = 0;
  for (let i = 0; i < s.length; i++) {
    for (let j = i; j < s.length; j++) {
      let substr = s.slice(i, j + 1);
      if (substr === substr.split('').reverse().join('') && substr.length > maxLen) {
        maxLen = substr.length;
        start = i;
      }
    }
  }
  return s.slice(start, start + maxLen);
}
console.log(longestPalindrome('babad'));

medium

A. "aba"

B. "babad"

C. "bad"

D. "bab"

Solution

1. Step 1: Understand the nested loops and palindrome check

   The code checks all substrings and updates the longest palindrome found.

2. Step 2: Trace substrings in 'babad'

   Palindromes found include 'bab' and 'aba', both length 3. The code updates maxLen and start when it finds 'aba' last (i=1,j=3), because the condition uses >, so it updates only if strictly longer. 'bab' is found first but 'aba' is found later and has the same length, so it is not updated. Actually, the code updates only if substr.length > maxLen, so the first palindrome of length 3 is 'bab' (i=0,j=2), maxLen becomes 3. Then 'aba' (i=1,j=3) is also length 3, so substr.length > maxLen is false (3 > 3 is false), so maxLen and start are not updated. Therefore, the returned palindrome is 'bab'.

3. Final Answer:

   "bab" -> Option D

4. Quick Check:

   First longest palindrome found = "bab" [OK]

Hint: Check all substrings; first longest palindrome returned [OK]

Common Mistakes:

• Assuming 'aba' is returned instead of 'bab'
• Confusing substring with subsequence
• Ignoring order of substring checks

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

function longestPalindrome(s) {
  let maxLen = 1, start = 0;
  for (let i = 0; i < s.length; i++) {
    for (let j = i; j < s.length; j++) {
      let substr = s.slice(i, j + 1);
      if (substr === substr.split('').reverse().join('') && substr.length >= maxLen) {
        maxLen = substr.length;
        start = i;
      }
    }
  }
  return s.slice(start, start + maxLen);
}
console.log(longestPalindrome('cbbd'));

medium

A. The reverse method does not work on strings

B. The slice method is used incorrectly

C. Using >= causes the last palindrome to be chosen instead of the first

D. The loop ranges are off by one

Solution

1. Step 1: Analyze the condition with >=

   The condition uses >=, so if two palindromes have the same length, the later one overwrites the earlier.

2. Step 2: Understand effect on output

   For input 'cbbd', palindromes 'bb' and 'b' exist. Using >= picks the last palindrome of max length, which may not be desired.

3. Final Answer:

   Using >= causes the last palindrome to be chosen instead of the first -> Option C

4. Quick Check:

   Condition >= picks last palindrome [OK]

Hint: Use > to pick first longest palindrome found [OK]

Common Mistakes:

• Thinking slice is incorrect
• Believing reverse doesn't work on strings
• Ignoring >= vs > difference

5. Given the string s = 'forgeeksskeegfor', which approach below efficiently finds the longest palindromic substring and what is that substring?

hard

A. Expand around center approach; longest palindrome is 'eeksskee'

B. Dynamic programming with 2D table; longest palindrome is 'geeksskeeg'

C. Sorting characters and checking; longest palindrome is 'ss'

D. Brute force checking all substrings; longest palindrome is 'forgeeksskeegfor'

Solution

1. Step 1: Understand problem and input

   The string 'forgeeksskeegfor' contains a long palindrome 'geeksskeeg'.

2. Step 2: Evaluate approaches

   Brute force checking all substrings is inefficient and won't find the full string palindrome. Sorting characters loses order, so 'ss' is incorrect. Dynamic programming with 2D table finds 'geeksskeeg' correctly, unlike other options.

3. Final Answer:

   Dynamic programming with 2D table; longest palindrome is 'geeksskeeg' -> Option B

4. Quick Check:

   DP approach finds 'geeksskeeg' efficiently [OK]

Hint: Use DP table to track palindromes efficiently [OK]

Common Mistakes:

• Assuming whole string is palindrome
• Using sorting which breaks character order
• Confusing expand around center with brute force