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Longest palindrome formed by concatenating and reordering strings of equal length

  • Difficulty Level : Medium
  • Last Updated : 27 Apr, 2020

Given an array arr[] consisting of N strings of equal length M, the task is to create the longest palindrome by concatenating the strings. Reordering and discarding some strings from the given set of strings can also be done.

Examples:

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Input: N = 3, arr[] = { “tab”, “one”, “bat” }, M = 3
Output: tabbat
Explanation:
On combining “tab” and “bat” from the given set of input strings, the longest palindromic string is formed.



Input: N = 4, arr[] = { “oo”, “ox”, “xo”, “xx” }, M = 2
Output: oxxxxo

Observation: Let us assume that we chose some K strings from the given array and the string which we get after choosing those K strings is a palindrome. Then the observation which needs to be made for the possible K values are:

  • K is even: Then for every integer x (1 <= x <= K/2):
    Sx = rev(SK - x + 1)
    

    For example, if the strings which form the longest palindrome are S1 = “tab”, S2 = “ab”, S3 = “ba”, S4 = “bat”. Here, K = 4. Therefore, S1 = rev(S4) and S2 = rev(S3).

  • K is odd: Apart from the above observation, the middle string S(K+1)/2 is a palindrome.

Approach: Clearly, from the above observation, to get the largest palindrome, the given array must also contain the reverse of the string. Therefore, for every string, we find if there is another string that is its reverse. If there exists, then we add this pair of strings to the left and right end respectively. If there are one or more strings that are palindrome themselves, pick any one of them and put it in the middle.

Below is the implementation of the above approach:

C++




// C++ program to find the
// Longest palindrome that can be formed
// by concatenating and reordering
// given N strings of equal length
#include <bits/stdc++.h>
using namespace std;
  
// Function to print the longest palindrome
void printPalindrome(vector<string> left,
                     string mid, vector<string> right)
{
    // Printing every string in left vector
    for (string x : left)
        cout << x;
  
    // Printing the palindromic string
    // in the middle
    cout << mid;
  
    // Printing the reverse of the right vector
    // to make the final output palindromic
    reverse(right.begin(), right.end());
    for (string x : right)
        cout << x;
  
    cout << endl;
}
  
// Function to find and print the
// longest palindrome that can be formed
void findPalindrome(vector<string>& S, int N, int M)
{
    set<string> dict;
    for (int i = 0; i < M; i++) {
        cin >> S[i];
        // Inserting each string in the set
        dict.insert(S[i]);
    }
  
    // Vectors to add the strings
    // in the left and right side
    vector<string> left, right;
  
    // To add the already present palindrome
    // string in the middle of the solution
    string mid;
  
    // Iterating through all the given strings
    for (int i = 0; i < N; i++) {
        string t = S[i];
        reverse(t.begin(), t.end());
  
        // If the string is a palindrome
        // it is added in the middle
        if (t == S[i])
            mid = t;
  
        // Checking if the reverse
        // of the string is already
        // present in the set
        else if (dict.find(t) != dict.end()) {
            left.push_back(S[i]);
            right.push_back(t);
            dict.erase(S[i]);
            dict.erase(t);
        }
    }
  
    printPalindrome(left, mid, right);
}
  
// Driver code
int main()
{
    vector<string> S{ "tab", "one", "bat" };
    int M = 3;
    int N = S.size();
    findPalindrome(S, N, M);
}

Java




// Java program to find the
// Longest palindrome that can be formed
// by concatenating and reordering
// given N Strings of equal length
import java.util.*;
  
class GFG{
   
// Function to print the longest palindrome
static void printPalindrome(Vector<String> left,
                     String mid, Vector<String> right)
{
    // Printing every String in left vector
    for (String x : left)
        System.out.print(x);
   
    // Printing the palindromic String
    // in the middle
    System.out.print(mid);
   
    // Printing the reverse of the right vector
    // to make the final output palindromic
    Collections.reverse(right);
    for (String x : right)
        System.out.print(x);
   
    System.out.println();
}
   
// Function to find and print the
// longest palindrome that can be formed
static void findPalindrome(Vector<String> S, int N, int M)
{
    HashSet<String> dict = new HashSet<String>();
    for (int i = 0; i < M; i++) {
        // Inserting each String in the set
        dict.add(S.get(i));
    }
   
    // Vectors to add the Strings
    // in the left and right side
    Vector<String> left = new Vector<String>(), right = new Vector<String>();
   
    // To add the already present palindrome
    // String in the middle of the solution
    String mid="";
   
    // Iterating through all the given Strings
    for (int i = 0; i < N; i++) {
        String t = S.get(i);
        t = reverse(t);
   
        // If the String is a palindrome
        // it is added in the middle
        if (t == S.get(i))
            mid = t;
   
        // Checking if the reverse
        // of the String is already
        // present in the set
        else if (dict.contains(t)) {
            left.add(S.get(i));
            right.add(t);
            dict.remove(S.get(i));
            dict.remove(t);
        }
    }
   
    printPalindrome(left, mid, right);
}
static String reverse(String input) {
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.valueOf(a);
// Driver code
public static void main(String[] args)
{
    String arr[] = { "tab", "one", "bat" };
    Vector<String> S = new Vector<String>(Arrays.asList(arr));
    int M = 3;
    int N = S.size();
    findPalindrome(S, N, M);
}
}
  
// This code contributed by PrinciRaj1992

Python3




# Function to print the longest palindrome
def printPalindrome(left,mid,right):
  
    # Printing every string in left vector
    for x in left:
        print(x, end="")
  
    # Printing the palindromic string
    # in the middle
    print(mid, end="")
  
    # Printing the reverse of the right vector
    # to make the final output palindromic
    right = right[::-1]
    for x in right:
        print(x, end = "")
  
    print('\n', end = "")
  
# Function to find and print the
# longest palindrome that can be formed
def findPalindrome(S, N, M):
    d = set()
    for i in range(M):
        # Inserting each string in the set
        d.add(S[i])
  
    # Vectors to add the strings
    # in the left and right side
    left = []
    right = []
  
    # To add the already present palindrome
    # string in the middle of the solution
    mid = ""
  
    # Iterating through all the given strings
    for i in range(N):
        t = S[i]
        t = t[::-1]
  
        # If the string is a palindrome
        # it is added in the middle
        if (t == S[i]):
            mid = t
  
        # Checking if the reverse
        # of the string is already
        # present in the set
        elif (t in d):
            left.append(S[i])
            right.append(t)
            d.remove(S[i])
            d.remove(t)
  
    printPalindrome(left, mid, right)
  
# Driver code
if __name__ == '__main__':
    S = ["tab", "one", "bat"]
    M = 3
    N = len(S)
    findPalindrome(S, N, M)
  
# This code is contributed by Surendra_Gangwar

C#




// C# program to find the
// longest palindrome that can be formed
// by concatenating and reordering
// given N Strings of equal length
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to print the longest palindrome
static void printPalindrome(List<String> left,
                    String mid, List<String> right)
{
    // Printing every String in left vector
    foreach (String x in left)
        Console.Write(x);
  
    // Printing the palindromic String
    // in the middle
    Console.Write(mid);
  
    // Printing the reverse of the right vector
    // to make the readonly output palindromic
    right.Reverse();
    foreach (String x in right)
        Console.Write(x);
  
    Console.WriteLine();
}
  
// Function to find and print the
// longest palindrome that can be formed
static void findPalindrome(List<String> S, int N, int M)
{
    HashSet<String> dict = new HashSet<String>();
    for (int i = 0; i < M; i++) {
          
        // Inserting each String in the set
        dict.Add(S[i]);
    }
  
    // Lists to add the Strings
    // in the left and right side
    List<String> left = new List<String>(), 
    right = new List<String>();
  
    // To add the already present palindrome
    // String in the middle of the solution
    String mid="";
  
    // Iterating through all the given Strings
    for (int i = 0; i < N; i++) {
        String t = S[i];
        t = reverse(t);
  
        // If the String is a palindrome
        // it is added in the middle
        if (t == S[i])
            mid = t;
  
        // Checking if the reverse
        // of the String is already
        // present in the set
        else if (dict.Contains(t)) {
            left.Add(S[i]);
            right.Add(t);
            dict.Remove(S[i]);
            dict.Remove(t);
        }
    }
  
    printPalindrome(left, mid, right);
}
static String reverse(String input) {
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.Join("",a);
  
// Driver code
public static void Main(String[] args)
{
    String []arr = { "tab", "one", "bat" };
    List<String> S = new List<String>(arr);
    int M = 3;
    int N = S.Count;
    findPalindrome(S, N, M);
}
}
  
// This code is contributed by Princi Singh
Output:
tabbat



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