15. Longest Palindrome in a String

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

My Approach

To find the longest palindrome in a given string, I have used the following approach:

• I implemented a straightforward approach using two pointers to identify substrings and determine whether they are palindromic.

• To optimize efficiency and avoid unnecessary checks, I started with larger substrings and gradually reduced their size.

• I initialized an output string out with the first character of the input string s.

• I also initialized a variable ms to store the maximum length of the palindromic substring found so far, which is initially set to 1 (length of the first character).

• I iterated over the string s using two nested loops. The outer loop starts from the first character and goes up to the second-to-last character, while the inner loop starts from the last character and goes down to the character next to the current outer loop index.

• For each combination of indices, I checked if the substring from i to j (inclusive) is a palindrome by comparing the characters at the corresponding indices using the ispalin helper function.

• If the current substring is a palindrome and its length is greater than ms, I updated ms with the new maximum length and stored the substring in out using the substr function.

• Finally, I returned the out string containing the longest palindromic substring.

Time and Auxiliary Space Complexity

• Time Complexity: O(N^3), where N is the length of the input string. This is because we have nested loops that iterate over all possible substrings of s, resulting in a cubic time complexity.

• Auxiliary Space Complexity: O(1) since we are not using any extra space that scales with the input.

Code (C++)

class Solution {
  public:
    bool ispalin(string& s, int i ,int j){
        while(i<j){
            if(s[i]!=s[j])
                return false;

            ++i; --j;
        }
        return true;
    }

    string longestPalin (string s) {
        int n = s.size();
        string out;
        out.push_back(s[0]);
        int ms = 1;
        for(int i = 0; i<n-1 ;++i){
            for(int j = n-1; j>i; --j){
                if((j-i+1)>ms && ispalin(s,i,j)){
                    ms = j-i+1;
                    out = s.substr(i,j-i+1);
                }
            }
        }
        return out;
        
    }
};

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