Longest Palindromic Substring in DSA C++ - Time & Space Complexity

Time Complexity: Longest Palindromic Substring

O(n²)

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.

int expandAroundCenter(const string &s, int left, int right) {
    while (left >= 0 && right < (int)s.size() && s[left] == s[right]) {
        left--;
        right++;
    }
    return right - left - 1;
}

string longestPalindrome(string s) {
    int start = 0, maxLen = 0;
    for (int i = 0; i < (int)s.size(); i++) {
        int len1 = expandAroundCenter(s, i, i);
        int len2 = expandAroundCenter(s, i, i + 1);
        int len = max(len1, len2);
        if (len > maxLen) {
            maxLen = len;
            start = i - (len - 1) / 2;
        }
    }
    return s.substr(start, maxLen);
}
    

This code finds the longest palindromic substring by expanding around each character and its neighbor.

Identify Repeating Operations

• Primary operation: Expanding around each center to check palindrome length.
• How many times: For each character (n times), two expansions happen, each can expand up to n steps in worst case.

How Execution Grows With Input

As the string length grows, the number of expansions grows roughly with the square of the length.

Input Size (n)	Approx. Operations
10	About 100 expansions
100	About 10,000 expansions
1000	About 1,000,000 expansions

Pattern observation: Doubling the input size roughly quadruples the work.

Final Time Complexity

Time Complexity: O(n²)

This means the time needed grows roughly with the square of the string length.

Common Mistake

[X] Wrong: "Checking palindromes around each center is only O(n) because it looks like a single loop."

[OK] Correct: Each center can expand up to n steps, so the inner expansions multiply the work, making it O(n²).

Interview Connect

Understanding this time complexity helps you explain why some palindrome solutions are slower and shows your grasp of nested operations in strings.

Self-Check

"What if we used dynamic programming to store palindrome results? How would the time complexity change?"

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 the problem statement

   The problem asks to find the longest part of a string that is a palindrome, meaning it reads the same forwards and backwards.

2. Step 2: Compare options with the problem goal

   Only Find the longest substring that reads the same forwards and backwards describes finding a substring that reads the same forwards and backwards, which matches the palindrome definition.

3. Final Answer:

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

4. Quick Check:

   Longest Palindromic Substring = Find the longest substring that reads the same forwards and backwards [OK]

Hint: Look for substring that reads same forwards and backwards [OK]

Common Mistakes:

• Confusing palindrome with unique characters
• Thinking about substrings with vowels only
• Mixing substring with subsequence

2. Which of the following C++ code snippets correctly initializes a 2D vector dp of size n x n with false values for the Longest Palindromic Substring problem?

easy

A. std::vector dp(n, std::vector<int>(n, false));

B. std::vector<bool> dp(n, std::vector<bool>(n, false));

C. std::vector dp(n, n, false);

D. std::vector dp(n, std::vector<bool>(n, false));

Solution

1. Step 1: Understand 2D vector initialization syntax

   To create a 2D vector of booleans with size n x n, use vector of vectors: std::vector dp(n, std::vector<bool>(n, false));

2. Step 2: Check each option's correctness

   std::vector dp(n, std::vector<bool>(n, false)); uses correct syntax. std::vector<bool> dp(n, std::vector<bool>(n, false)); tries to assign vector<bool> to vector<bool>, which is invalid. std::vector dp(n, std::vector<int>(n, false)); uses int instead of bool. std::vector dp(n, n, false); uses invalid constructor syntax.

3. Final Answer:

   std::vector dp(n, std::vector<bool>(n, false)); -> Option D

4. Quick Check:

   Correct 2D vector initialization = std::vector dp(n, std::vector<bool>(n, false)); [OK]

Hint: Use vector of vectors with inner vector initialized [OK]

Common Mistakes:

• Using wrong data type for dp vector
• Incorrect constructor syntax for 2D vector
• Confusing vector<bool> with vector

3. What is the output of the following C++ code snippet that finds the longest palindromic substring length?

std::string s = "babad";
int n = s.size();
int maxLength = 1;
for (int i = 0; i < n; i++) {
  for (int j = i; j < n; j++) {
    int left = i, right = j;
    bool isPalindrome = true;
    while (left < right) {
      if (s[left] != s[right]) {
        isPalindrome = false;
        break;
      }
      left++; right--;
    }
    if (isPalindrome && (j - i + 1) > maxLength) {
      maxLength = j - i + 1;
    }
  }
}
std::cout << maxLength << std::endl;

medium

A. 3

B. 2

C. 4

D. 5

Solution

1. Step 1: Understand the code logic

   The code checks all substrings of "babad" and updates maxLength if the substring is a palindrome and longer than current maxLength.

2. Step 2: Identify longest palindromic substrings in "babad"

   "bab" and "aba" are palindromes of length 3. No longer palindrome exists.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Longest palindrome length in "babad" = 3 [OK]

Hint: Check all substrings and update max length if palindrome [OK]

Common Mistakes:

• Assuming entire string is palindrome
• Counting non-palindromic substrings
• Off-by-one errors in substring length

4. Identify the bug in the following C++ code snippet intended to find the longest palindromic substring length:

std::string s = "cbbd";
int n = s.size();
int maxLength = 1;
for (int i = 0; i < n; i++) {
  for (int j = i; j < n; j++) {
    int left = i, right = j;
    while (left < right) {
      if (s[left] != s[right]) {
        break;
      }
      left++; right--;
    }
    if ((j - i + 1) > maxLength) {
      maxLength = j - i + 1;
    }
  }
}
std::cout << maxLength << std::endl;

medium

A. Incorrect loop ranges for i and j

B. Wrong initialization of maxLength

C. Missing check if substring is palindrome before updating maxLength

D. Using break instead of continue in while loop

Solution

1. Step 1: Analyze palindrome check logic

   The code breaks out of while loop if characters don't match but does not track if substring is palindrome.

2. Step 2: Check maxLength update condition

   maxLength is updated without confirming substring is palindrome, causing incorrect results.

3. Final Answer:

   Missing check if substring is palindrome before updating maxLength -> Option C

4. Quick Check:

   Must confirm palindrome before updating maxLength [OK]

Hint: Always confirm palindrome before updating max length [OK]

Common Mistakes:

• Updating maxLength without palindrome check
• Confusing break and continue in loops
• Incorrect loop boundaries

5. Given the string "forgeeksskeegfor", which approach below efficiently finds the longest palindromic substring and what is its length?

hard

A. Dynamic Programming approach, length 7

B. Expand Around Center approach, length 10

C. Brute Force approach, length 12

D. Manacher's Algorithm, length 8

Solution

1. Step 1: Identify longest palindrome in "forgeeksskeegfor"

   The longest palindromic substring is "geeksskeeg" with length 10.

2. Step 2: Choose efficient approach

   Expand Around Center approach efficiently finds palindromes by expanding from each center, suitable for this problem.

3. Final Answer:

   Expand Around Center approach, length 10 -> Option B

4. Quick Check:

   Longest palindrome length = 10 using Expand Around Center [OK]

Hint: Expand from centers to find longest palindrome efficiently [OK]

Common Mistakes:

• Confusing length of palindrome substring
• Choosing less efficient brute force for large strings
• Mixing up algorithm names and their complexities