Largest palindromic string possible from given strings by rearranging the characters

Given two strings S and P, the task is to find the largest palindrome possible string by choosing characters from the given strings S and P after rearranging the characters. 
Note: If many answers are possible, then find the lexicographically smallest T with maximum length. 

Examples:

Input: S = “abad”, T = “eeff”
Output: aefbfea
Explanation: 
From the first string “aba” and from the second string “effe” are palindromes.
Use these two to make a new palindrome T “aefbfea” which is lexicographically smallest.
Also T is of maximum length possible.
Note: “ada”, although this will give T of same length i.e., 7 but that won’t be lexicographically smallest.

Input: S = “aeabb”, T = “dfedf”
Output: abdeffedba
Explanation: 
From the first string “abeba” and from the second string “dfefd”, are palindrome. Combine the two to get T: “abdeffedba” is lexicographically smallest.

Approach: The idea is to take all the characters from the strings S and P which are present even a number of times respectively. As the substrings, A and B of both the string have all characters which are present even number of times in S and P respectively to become the largest palindrome from parent string. Below are the steps:



  1. Take a single character from strings S and P, such that they will be present in the mid of palindromes taken from each string respectively. Place that unique element in mid of T.
  2. The possible cases to take a unique element from substring A and B are:
    • If a unique element is present in both A and B, then take both of them because it will increase the length of T by 2.
    • If unique elements are present in both S and P but not the same element, then take that which is small because that will make lexicographically the smallest palindrome T.
    • If no unique element is present in both, then make the string T just with elements from A and B.
  3. Use the unordered map to take count of characters occurring even number of times in parent string. Also, use another map to have count of characters to be used in making T.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find largest palindrome
// possible from S and P after rearranging
// characters to make palindromic string T
string mergePalindromes(string S, string P)
{
 
    // Using unordered_map to store
    // frequency of elements
    // mapT will have freq of elements of T
    unordered_map<char, int> mapS, mapP, mapT;
 
    // Size of both the strings
    int n = S.size(), m = P.size();
 
    for (int i = 0; i < n; i++) {
        mapS[S[i]]++;
    }
    for (int i = 0; i < m; i++) {
        mapP[P[i]]++;
    }
 
    // Take all character in mapT which
    // occur even number of times in
    // respective strings & simultaneously
    // update number of characters left in map
    for (char i = 'a'; i <= 'z'; i++) {
 
        if (mapS[i] % 2 == 0) {
            mapT[i] += mapS[i];
            mapS[i] = 0;
        }
        else {
            mapT[i] += mapS[i] - 1;
            mapS[i] = 1;
        }
 
        if (mapP[i] % 2 == 0) {
            mapT[i] += mapP[i];
            mapP[i] = 0;
        }
        else {
            mapT[i] += mapP[i] - 1;
            mapP[i] = 1;
        }
    }
 
    // Check if a unique character is
    // present in both string S and P
    int check = 0;
 
    for (char i = 'a'; i <= 'z'; i++) {
        if (mapS[i] > 0 && mapP[i] > 0) {
            mapT[i] += 2;
            check = 1;
            break;
        }
    }
 
    // Making string T in two halves
    // half1 - first half
    // half2 - second half
    string half1 = "", half2 = "";
 
    for (char i = 'a'; i <= 'z'; i++) {
        for (int j = 0; (2 * j) < mapT[i]; j++) {
            half1 += i;
            half2 += i;
        }
    }
 
    // Reverse the half2 to attach with half1
    // to make palindrome T
    reverse(half2.begin(), half2.end());
 
    // If same unique element is present
    // in both S and P, then taking that only
    // which is already taken through mapT
    if (check) {
        return half1 + half2;
    }
 
    // If same unique element is not
    // present in S and P, then take
    // characters that make string T
    // lexicographically smallest
    for (char i = 'a'; i <= 'z'; i++) {
        if (mapS[i] > 0 || mapP[i] > 0) {
            half1 += i;
            return half1 + half2;
        }
    }
 
    // If no unique element is
    // present in both string S and P
    return half1 + half2;
}
 
// Driver Code
int main()
{
    // Given two strings S and P
    string S = "aeabb";
    string P = "dfedf";
 
    // Function Call
    cout << mergePalindromes(S, P);
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
import java.lang.*;
 
class GFG{
 
// Function to find largest palindrome
// possible from S and P after rearranging
// characters to make palindromic string T
static String mergePalindromes(StringBuilder S,
                               StringBuilder P)
{
    // Using unordered_map to store
    // frequency of elements
    // mapT will have freq of elements of T
    Map<Character, Integer> mapS = new HashMap<>(),
                            mapP = new HashMap<>(),
                            mapT = new HashMap<>();
 
    // Size of both the strings
    int n = S.length(), m = P.length();
 
    for(int i = 0; i < n; i++)
    {
        mapS.put(S.charAt(i),
        mapS.getOrDefault(S.charAt(i), 0) + 1);
    }
    for(int i = 0; i < m; i++)
    {
        mapP.put(P.charAt(i),
        mapP.getOrDefault(P.charAt(i), 0) + 1);
    }
 
    // Take all character in mapT which
    // occur even number of times in
    // respective strings & simultaneously
    // update number of characters left in map
    for(char i = 'a'; i <= 'z'; i++)
    {
        if (mapS.getOrDefault(i, 0) % 2 == 0)
        {
           mapT.put(i,mapT.getOrDefault(i, 0) +
           mapS.getOrDefault(i, 0));
           mapS.put(i, 0);
        }
        else
        {
            mapT.put(i,mapT.getOrDefault(i, 0) +
            mapS.getOrDefault(i, 0) - 1);
            mapS.put(i, 1);
        }
 
        if (mapP.getOrDefault(i, 0) % 2 == 0)
        {
            mapT.put(i, mapT.getOrDefault(i, 0) +
            mapP.getOrDefault(i, 0));
            mapP.put(i, 0);
        }
        else
        {
            mapT.put(i, mapT.getOrDefault(i, 0) +
            mapP.getOrDefault(i, 0) - 1);
            mapP.put(i, 1);
        }
    }
 
    // Check if a unique character is
    // present in both string S and P
    int check = 0;
 
    for(char i = 'a'; i <= 'z'; i++)
    {
        if (mapS.getOrDefault(i, 0) > 0 &&
            mapP.getOrDefault(i, 0) > 0)
        {
            mapT.put(i, mapT.getOrDefault(i, 0) + 2);
            check = 1;
            break;
        }
    }
 
    // Making string T in two halves
    // half1 - first half
    // half2 - second half
    StringBuilder half1 = new StringBuilder(),
                  half2 = new StringBuilder();
 
    for(char i = 'a'; i <= 'z'; i++)
    {
        for(int j = 0;
           (2 * j) < mapT.getOrDefault(i, 0);
                j++)
        {
            half1.append(i);
            half2.append(i);
        }
    }
 
    // Reverse the half2 to attach with half1
    // to make palindrome T
    StringBuilder tmp = half2.reverse();
 
    // If same unique element is present
    // in both S and P, then taking that only
    // which is already taken through mapT
    if (check == 1)
    {
        return half1.append(tmp).toString();
    }
 
    // If same unique element is not
    // present in S and P, then take
    // characters that make string T
    // lexicographically smallest
    for(char i = 'a'; i <= 'z'; i++)
    {
        if (mapS.getOrDefault(i, 0) > 0 ||
            mapP.getOrDefault(i, 0) > 0)
        {
            half1.append(i);
            return half1.append(tmp).toString();
        }
    }
 
    // If no unique element is
    // present in both string S and P
    return half1.append(tmp).toString();
}
 
// Driver code
public static void main (String[] args)
{
     
    // Given two strings S and P
    StringBuilder S = new StringBuilder("aeabb");
    StringBuilder P = new StringBuilder("dfedf");
     
    // Function call
    System.out.println(mergePalindromes(S, P));
}
}
 
// This code is contributed by offbeat

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Python3

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# Python3 program for the above approach
from collections import defaultdict
 
# Function to find largest palindrome
# possible from S and P after rearranging
# characters to make palindromic string T
def mergePalindromes(S, P):
 
    # Using unordered_map to store
    # frequency of elements
    # mapT will have freq of elements of T
    mapS = defaultdict(lambda : 0)
    mapP = defaultdict(lambda : 0)
    mapT = defaultdict(lambda : 0)
 
    # Size of both the strings
    n = len(S)
    m = len(P)
 
    for i in range(n):
        mapS[ord(S[i])] += 1
 
    for i in range(m):
        mapP[ord(P[i])] += 1
 
    # Take all character in mapT which
    # occur even number of times in
    # respective strings & simultaneously
    # update number of characters left in map
    for i in range(ord('a'), ord('z') + 1):
        if(mapS[i] % 2 == 0):
            mapT[i] += mapS[i]
            mapS[i] = 0
        else:
            mapT[i] += mapS[i] - 1
            mapS[i] = 1
 
        if (mapP[i] % 2 == 0):
            mapT[i] += mapP[i]
            mapP[i] = 0
        else:
            mapT[i] += mapP[i] - 1
            mapP[i] = 1
 
    # Check if a unique character is
    # present in both string S and P
    check = 0
 
    for i in range(ord('a'), ord('z') + 1):
        if (mapS[i] > 0 and mapP[i] > 0):
            mapT[i] += 2
            check = 1
            break
 
    # Making string T in two halves
    # half1 - first half
    # half2 - second half
    half1, half2 = "", ""
 
    for i in range(ord('a'), ord('z') + 1):
        j = 0
        while((2 * j) < mapT[i]):
            half1 += chr(i)
            half2 += chr(i)
            j += 1
 
    # Reverse the half2 to attach with half1
    # to make palindrome
    half2 = half2[::-1]
 
    # If same unique element is present
    # in both S and P, then taking that only
    # which is already taken through mapT
    if(check):
        return half1 + half2
 
    # If same unique element is not
    # present in S and P, then take
    # characters that make string T
    # lexicographically smallest
    for i in range(ord('a'), ord('z') + 1):
        if(mapS[i] > 0 or mapP[i] > 0):
            half1 += chr(i)
            return half1 + half2
 
    # If no unique element is
    # present in both string S and P
    return half1 + half2
 
# Driver Code
 
# Given string S and P
S = "aeabb"
P = "dfedf"
 
# Function call
print(mergePalindromes(S, P))
 
# This code is contributed by Shivam Singh

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C#

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// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Text;
 
class GFG{
 
// Function to find largest palindrome
// possible from S and P after rearranging
// characters to make palindromic string T
static string mergePalindromes(string S,
                               string P)
{
 
    // Using unordered_map to store
    // frequency of elements mapT will
    // have freq of elements of T
    Dictionary<char,
               int> mapS = new Dictionary<char,
                                          int>();
    Dictionary<char,
               int> mapP = new Dictionary<char,
                                          int>();
    Dictionary<char,
               int> mapT = new Dictionary<char,
                                          int>();
 
    // Size of both the strings
    int n = S.Length, m = P.Length;
     
    for(char i = 'a'; i <= 'z'; i++)
    {
        mapS[i] = 0;
        mapP[i] = 0;
        mapT[i] = 0;
    }
     
    for(int i = 0; i < n; i++)
    {
        mapS[S[i]]++;
    }
     
    for(int i = 0; i < m; i++)
    {
        mapP[P[i]]++;
    }
 
    // Take all character in mapT which
    // occur even number of times in
    // respective strings & simultaneously
    // update number of characters left in map
    for(char i = 'a'; i <= 'z'; i++)
    {
        if (mapS[i] % 2 == 0)
        {
            mapT[i] += mapS[i];
            mapS[i] = 0;
        }
        else
        {
            mapT[i] += mapS[i] - 1;
            mapS[i] = 1;
        }
 
        if (mapP[i] % 2 == 0)
        {
            mapT[i] += mapP[i];
            mapP[i] = 0;
        }
        else
        {
            mapT[i] += mapP[i] - 1;
            mapP[i] = 1;
        }
    }
 
    // Check if a unique character is
    // present in both string S and P
    int check = 0;
 
    for(char i = 'a'; i <= 'z'; i++)
    {
        if (mapS[i] > 0 && mapP[i] > 0)
        {
            mapT[i] += 2;
            check = 1;
            break;
        }
    }
 
    // Making string T in two halves
    // half1 - first half
    // half2 - second half
    string half1 = "", half2 = "";
 
    for(char i = 'a'; i <= 'z'; i++)
    {
        for(int j = 0; (2 * j) < mapT[i]; j++)
        {
            half1 += i;
            half2 += i;
        }
    }
 
    // Reverse the half2 to attach with half1
    // to make palindrome T
    char[] tmp = half2.ToCharArray();
    Array.Reverse(tmp);
    half2 = new string(tmp);
 
    // If same unique element is present
    // in both S and P, then taking that only
    // which is already taken through mapT
    if (check != 0)
    {
        return half1 + half2;
    }
 
    // If same unique element is not
    // present in S and P, then take
    // characters that make string T
    // lexicographically smallest
    for(char i = 'a'; i <= 'z'; i++)
    {
        if (mapS[i] > 0 || mapP[i] > 0)
        {
            half1 += i;
            return half1 + half2;
        }
    }
 
    // If no unique element is
    // present in both string S and P
    return half1 + half2;
}
 
// Driver Code
public static void Main(string []args)
{
     
    // Given two strings S and P
    string S = "aeabb";
    string P = "dfedf";
     
    Console.Write(mergePalindromes(S, P));
}
}
 
// This code is contributed by rutvik_56

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Output: 

abdeffedba



 

Time Complexity: O(N), where N is the length of String S or P.
Auxiliary Space: O(N)
 

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