8. Longest Palindrome in a String

The problem can be found at the following link: Question Link

Problem Description

Given a string s, your task is to find the longest palindromic substring within s.

• A substring is a contiguous sequence of characters within the string (i.e., s[i...j] for 0 ≤ i ≤ j < len(s)).

• A palindrome reads the same forwards and backwards, i.e., reverse(s) == s.

• If there are multiple palindromic substrings of the same maximum length, return the first occurrence from left to right.

Examples

Example 1:

Input:

s = "forgeeksskeegfor"

Output:

"geeksskeeg"

Explanation:

Possible palindromic substrings include "kssk", "ss", "eeksskee", etc. However, "geeksskeeg" is the longest among them.

Example 2:

Input:

s = "Geeks"

Output:

"ee"

Explanation:

The substring "ee" is the longest palindrome in "Geeks".

Example 3:

Input:

s = "abc"

Output:

"a"

Explanation:

All substrings "a", "b", and "c" have length 1, which is the maximum. Returning the first occurrence results in "a".

Constraints:

• $(1 \leq s.\text{size()} \leq 10^3)$

• s consists of only lowercase English letters.

My Approach

Expand Around Center

1. A palindrome can be expanded from its center.

2. For each index i in the string:

   • Consider two scenarios for the center:

     1. Odd-length palindromes (center at i).

     2. Even-length palindromes (center between i and i+1).

3. Expand outwards while left and right characters match.

4. Track the maximum length palindrome found and its start position.

This approach checks each possible center in O(N) and potentially expands in O(N), leading to O(N²) overall time.

Time and Auxiliary Space Complexity

• Expected Time Complexity: O(N²), because for each index we can expand in both directions.

• Expected Auxiliary Space Complexity: O(1), since we use only a few extra variables for tracking indices and lengths.

Code (C++)

class Solution {
  public:
    string longestPalindrome(string &s) {
        int n = s.size(), start = 0, maxLen = 0;
        for (int i = 0; i < n; i++) {
            for (int l : {i, i + 1}) {
                int j = i;
                while (j >= 0 && l < n && s[j] == s[l]) j--, l++;
                if (l - j - 1 > maxLen) start = j + 1, maxLen = l - j - 1;
            }
        }
        return s.substr(start, maxLen);
    }
};

⚡ Alternative Approaches

2️⃣ Dynamic Programming Approach (O(N²) Time, O(N²) Space)

Algorithm Steps:

1. Create a 2D DP table dp[i][j], where dp[i][j] is true if s[i:j] is a palindrome.

2. Set dp[i][i] = true (single-character substrings are palindromes).

3. If two adjacent characters are equal (s[i] == s[i+1]), set dp[i][i+1] = true.

4. For substrings of length 3 or more, use the formula:

   • dp[i][j] = (s[i] == s[j] && dp[i+1][j-1])

5. Keep track of the longest palindrome found and return it.

class Solution {
public:
    string longestPalindrome(string &s) {
        string t = "#";
        for (char c : s) t += c, t += "#";
        int n = t.size(), center = 0, right = 0, maxLen = 0, start = 0;
        vector<int> p(n, 0);

        for (int i = 0; i < n; i++) {
            int mirror = 2 * center - i;
            if (i < right) p[i] = min(right - i, p[mirror]);
            while (i - p[i] - 1 >= 0 && i + p[i] + 1 < n && t[i - p[i] - 1] == t[i + p[i] + 1])
                p[i]++;
            if (i + p[i] > right) center = i, right = i + p[i];
            if (p[i] > maxLen) maxLen = p[i], start = (i - maxLen) / 2;
        }
        return s.substr(start, maxLen);
    }
};

🔹 Easy to understand 🔹 Uses O(N²) space for the DP table

3️⃣ Manacher’s Algorithm (O(N) Time, O(N) Space)

Algorithm Steps:

1. Transform the original string by inserting special characters (#) between characters to handle even-length palindromes.

   • Example: "abc" → "#a#b#c#"

2. Use a palindrome radius array p[i] to store the length of the longest palindrome centered at i.

3. Maintain a center (C) and right boundary (R), representing the rightmost palindrome found.

4. If i is within R, mirror the value of p[i] from the symmetric point across C.

5. Expand the palindrome at i while characters match.

6. If the palindrome at i expands beyond R, update C and R.

7. Extract the longest palindrome from p[i].

class Solution {
public:
    string longestPalindrome(string &s) {
        if (s.empty()) return "";
        string t = "#";
        for (char c : s) t += c, t += "#";

        int n = t.size(), C = 0, R = 0, maxLen = 0, center = 0;
        vector<int> p(n, 0);

        for (int i = 0; i < n; i++) {
            int mirror = 2 * C - i;
            if (i < R) p[i] = min(R - i, p[mirror]);

            while (i + p[i] + 1 < n && i - p[i] - 1 >= 0 && t[i + p[i] + 1] == t[i - p[i] - 1])
                p[i]++;

            if (i + p[i] > R) C = i, R = i + p[i];

            if (p[i] > maxLen) maxLen = p[i], center = i;
        }

        int start = (center - maxLen) / 2;
        return s.substr(start, maxLen);
    }
};

🔹 Fastest solution (O(N)) 🔹 Requires string transformation

📊 Comparison of Approaches

Approach	⏱️ Time Complexity	🗂️ Space Complexity	✅ Pros	⚠️ Cons
Expand Around Center	🟡 O(N²)	🟢 O(1)	Simple and uses constant extra space	Slower for larger strings
Dynamic Programming	🟡 O(N²)	🟡 O(N²)	Straight-forward to implement	High space usage (DP table)
Manacher’s Algorithm	🟢 O(N)	🟢 O(N)	Fastest known approach	String transformation can be tricky to code

💡 Best Choice?

• ✅ For best runtime performance: Use Manacher’s Algorithm (O(N)).

• ✅ For simplicity and minimal space usage: Use Expand Around Center (O(N²)).

• ✅ For detailed table-based logic understanding: Use Dynamic Programming (O(N²)).

Code (Java)

class Solution {
    static String longestPalindrome(String s) {
        int n = s.length(), start = 0, maxLen = 0;
        for (int i = 0; i < n; i++)
            for (int l : new int[]{i, i + 1}) {
                int j = i;
                while (j >= 0 && l < n && s.charAt(j) == s.charAt(l)) {
                    j--;
                    l++;
                }
                if (l - j - 1 > maxLen) {
                    start = j + 1;
                    maxLen = l - j - 1;
                }
            }
        return s.substring(start, start + maxLen);
    }
}

Code (Python)

class Solution:
    def longestPalindrome(self, s):
        start, max_len = 0, 0
        for i in range(len(s)):
            for l in [i, i + 1]:
                j = i
                while j >= 0 and l < len(s) and s[j] == s[l]: j, l = j - 1, l + 1
                if l - j - 1 > max_len: start, max_len = j + 1, l - j - 1
        return s[start:start + max_len]

Contribution and Support

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