To solve this problem, we need to form the longest palindrome using elements from the words array. Here's a structured approach to achieve this:

Steps to Solution: Understanding Palindrome Construction:

A palindrome reads the same forwards and backwards. To maximize its length, we can use characters symmetrically around a potential center. Character Frequency Count:

Count the frequency of each character (or each element in the case of strings) in the words array. Constructing the Palindrome:

For each element in words, add characters in pairs (or even number of times) to ensure they can form a palindrome. If there are any characters left with odd frequencies, we can use one of them as the central character of the palindrome (since a palindrome can have at most one odd-character frequency in its center). Implementation Strategy:

Count frequencies of characters (or strings). Accumulate pairs of characters (or strings). If there are characters left with odd frequencies, use one of them in the center of the palindrome. Calculate the length of the palindrome based on the accumulated pairs and the optional center character.

I think it came in Salesforce OA recently instead of DE Shaw about which you mentioned in comments of the same blog

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define MOD 1000000007

int maxConcatenatedPalindrome(vector<string>& words){
    unordered_map<string, int> mp;
    int len = 0;
    bool hasOddPalindrome = false;

    for(string& word : words) mp[word]++;

    int x = -1;

    for(auto& pair : mp){
        string word = pair.first;
        int wordCount = pair.second;

        int n = word.length();

        string temp = word;
        reverse(temp.begin(), temp.end());

        if(word == temp){
            if(wordCount % 2 == 0){
                len += (wordCount * n);
            } 
            else{
                len += (wordCount - 1) * n;
                x = n;
                hasOddPalindrome = true;
            }
        } 
        else if(mp.find(temp) != mp.end()){
            int tempCount = mp[temp];
            int pairs = min(wordCount, tempCount);
            len += pairs * 2 * n;
            mp[temp] = 0;
        }
    }

    if(hasOddPalindrome) len += x; //middle

    return len;
}

int main() {
    vector<string> words1 = {"y", "xyx", "abc", "cba", "bac", "y", "y"};
    vector<string> words2 = {"y", "xyx", "abc", "cba", "bac", "y"};
    vector<string> words3 = {"ab", "ba", "xyx", "de"};
    vector<string> words4 = {"xy", "abc", "xyx", "bac", "cab", "cba"};

    cout << maxConcatenatedPalindrome(words1) << "\n";
    cout << maxConcatenatedPalindrome(words2) << "\n";
    cout << maxConcatenatedPalindrome(words3) << "\n";
    cout << maxConcatenatedPalindrome(words4) << "\n";

    return 0;
}

Have anyone got the solution?