Longest Palindromic Substring in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of longest palindromic substring length

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

DSA C++

int longestPalindromeLength(const std::string& s) {
    int n = s.size();
    if (n == 0) return 0;
    int maxLen = 1;
    for (int i = 0; i < n; ++i) {
        int left = i, right = i;
        while (left >= 0 && right < n && s[left] == s[right]) {
            maxLen = std::max(maxLen, right - left + 1);
            --left;
            ++right;
        }
        left = i; right = i + 1;
        while (left >= 0 && right < n && s[left] == s[right]) {
            maxLen = std::max(maxLen, right - left + 1);
            --left;
            ++right;
        }
    }
    return maxLen;
}

#include <iostream>
int main() {
    std::string s = "babad";
    std::cout << longestPalindromeLength(s) << std::endl;
    return 0;
}

A 4

B 5

C 2

D 3

💡 Hint

Check palindromes centered at each character and between characters.

The longest palindromic substrings in "babad" are "bab" and "aba", both length 3.

❓ Predict Output

intermediate

Output of longest palindromic substring

What is the output of this C++ code that returns the longest palindromic substring itself?

DSA C++

std::string longestPalindrome(const std::string& s) {
    int n = s.size();
    if (n == 0) return "";
    int start = 0, maxLen = 1;
    for (int i = 0; i < n; ++i) {
        int left = i, right = i;
        while (left >= 0 && right < n && s[left] == s[right]) {
            if (right - left + 1 > maxLen) {
                start = left;
                maxLen = right - left + 1;
            }
            --left;
            ++right;
        }
        left = i; right = i + 1;
        while (left >= 0 && right < n && s[left] == s[right]) {
            if (right - left + 1 > maxLen) {
                start = left;
                maxLen = right - left + 1;
            }
            --left;
            ++right;
        }
    }
    return s.substr(start, maxLen);
}

#include <iostream>
int main() {
    std::string s = "cbbd";
    std::cout << longestPalindrome(s) << std::endl;
    return 0;
}

A b

B cbbd

C bb

D bd

💡 Hint

Look for the longest substring that reads the same forward and backward.

The longest palindromic substring in "cbbd" is "bb".

🔧 Debug

advanced

Identify the runtime error in palindrome code

What runtime error will this C++ code produce when run with input "abc"?

DSA C++

int longestPalindromeLength(const std::string& s) {
    int n = s.size();
    int maxLen = 1;
    for (int i = 0; i <= n; ++i) {
        int left = i, right = i;
        while (left >= 0 && right < n && s[left] == s[right]) {
            maxLen = std::max(maxLen, right - left + 1);
            --left;
            ++right;
        }
    }
    return maxLen;
}

#include <iostream>
int main() {
    std::string s = "abc";
    std::cout << longestPalindromeLength(s) << std::endl;
    return 0;
}

A No error, output is 1

B Out of range error (accessing s[n])

C Segmentation fault due to null pointer

D Infinite loop

💡 Hint

Check the loop boundary conditions carefully.

The for loop runs with i <= n, so when i == n, s[i] accesses out of range causing runtime error.

🧠 Conceptual

advanced

Time complexity of longest palindromic substring algorithm

What is the time complexity of the expand-around-center approach to find the longest palindromic substring in a string of length n?

A O(n^2)

B O(n log n)

C O(n^3)

D O(n)

💡 Hint

Consider how many centers and expansions happen.

There are 2n-1 centers and each expansion can take O(n) in worst case, so total O(n^2).

🚀 Application

expert

Longest palindromic substring after one character removal

Given a string, which approach best finds the longest palindromic substring after removing at most one character?

A Use dynamic programming to check all substrings and skip one character if needed

B Use Manacher's algorithm directly without modification

C Check all substrings of length n-1 only

D Use a greedy approach removing the first mismatched character and check palindrome

💡 Hint

Consider how to handle the removal of one character in palindrome checks.

Dynamic programming can be adapted to allow skipping one character while checking substrings, ensuring all cases are covered.

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