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DSA Javascriptprogramming~5 mins

Longest Palindromic Substring in DSA Javascript - Time & Space Complexity

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Time Complexity: Longest Palindromic Substring
O(n³)
Understanding Time Complexity

We want to understand how the time needed to find the longest palindromic substring changes as the input string gets longer.

How does the number of steps grow when the string size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


function longestPalindrome(s) {
  let maxLen = 1;
  let start = 0;
  for (let i = 0; i < s.length; i++) {
    for (let j = i; j < s.length; j++) {
      let flag = true;
      for (let k = 0; k < Math.floor((j - i + 1) / 2); k++) {
        if (s[i + k] !== s[j - k]) {
          flag = false;
          break;
        }
      }
      if (flag && (j - i + 1) > maxLen) {
        start = i;
        maxLen = j - i + 1;
      }
    }
  }
  return s.substring(start, start + maxLen);
}
    

This code checks every possible substring and tests if it is a palindrome, keeping track of the longest one found.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Checking all substrings and verifying if each is a palindrome.
  • How many times: Outer loop runs n times, inner loop runs up to n times, and palindrome check runs up to substring length (up to n/2).
How Execution Grows With Input

As the string length grows, the number of substrings grows roughly with the square of the length, and checking each substring takes time proportional to its length.

Input Size (n)Approx. Operations
10About 10 * 10 * 5 = 500
100About 100 * 100 * 50 = 500,000
1000About 1000 * 1000 * 500 = 500,000,000

Pattern observation: The operations grow roughly with the cube of the input size, so time increases very quickly as input grows.

Final Time Complexity

Time Complexity: O(n³)

This means the time needed grows very fast, roughly proportional to the cube of the string length.

Common Mistake

[X] Wrong: "Checking all substrings is only O(n²) because there are n² substrings."

[OK] Correct: Each substring check itself can take up to O(n) time, so total time is more than just counting substrings.

Interview Connect

Understanding this helps you explain why naive solutions can be slow and motivates learning faster methods, showing your grasp of algorithm efficiency.

Self-Check

"What if we used a method that expands around centers instead of checking all substrings? How would the time complexity change?"

Practice

(1/5)
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
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