Haga que la string palindrómica no sea palindrómica reorganizando sus letras

Dada la string str que contiene letras en minúsculas (a – z). La tarea es imprimir la string después de reorganizar algunos caracteres de modo que la string se vuelva no palindrómica. Si es imposible hacer que la string no sea palíndromo, imprima -1 .
Ejemplos: 
 

Entrada: str = “abba” 
Salida: aabb
Entrada: str = “zzz” 
Salida: -1 
 

Enfoque: si todos los caracteres de la string son iguales, no importa cómo los reorganice, la string seguirá siendo la misma y será palindrómica. Ahora, si existe un arreglo no palindrómico, la mejor manera de reorganizar los caracteres es ordenar la string que formará un segmento continuo de los mismos caracteres y nunca será palindrómico. Para reducir el tiempo necesario para clasificar la string, podemos almacenar las frecuencias de los 26 caracteres e imprimirlos de forma ordenada.
A continuación se muestra la implementación del enfoque anterior: 
 

// CPP Program to rearrange letters of string
// to find a non-palindromic string if it exists
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the non-palindromic string
// if it exists, otherwise prints -1
void findNonPalinString(string s)
{
    int freq[26] = { 0 }, flag = 0;
 
    for (int i = 0; i < s.size(); i++) {
 
        // If all characters are not
        // same, set flag to 1
        if (s[i] != s[0])
            flag = 1;
 
        // Update frequency of the current character
        freq[s[i] - 'a']++;
    }
 
    // If all characters are same
    if (!flag)
        cout << "-1";
    else {
 
        // Print characters in sorted manner
        for (int i = 0; i < 26; i++)
            for (int j = 0; j < freq[i]; j++)
                cout << char('a' + i);
    }
}
 
// Driver Code
int main()
{
    string s = "abba";
 
    findNonPalinString(s);
 
    return 0;
}

// Java Program to rearrange letters of string
// to find a non-palindromic string if it exists
class GfG
{
 
// Function to print the non-palindromic string
// if it exists, otherwise prints -1
static void findNonPalinString(char s[])
{
    int freq[] = new int[26];
    int flag = 0;
 
    for (int i = 0; i < s.length; i++)
    {
 
        // If all characters are not
        // same, set flag to 1
        if (s[i] != s[0])
            flag = 1;
 
        // Update frequency of
        // the current character
        freq[s[i] - 'a']++;
    }
 
    // If all characters are same
    if (flag == 0)
        System.out.println("-1");
    else
    {
 
        // Print characters in sorted manner
        for (int i = 0; i < 26; i++)
            for (int j = 0; j < freq[i]; j++)
                System.out.print((char)('a' + i));
    }
}
 
// Driver Code
public static void main(String[] args)
{
    String s = "abba";
     
    findNonPalinString(s.toCharArray());
}
}
 
// This code is contributed by
// Prerna Saini.

# Python3 Program to rearrange letters of string
# to find a non-palindromic string if it exists
 
# Function to print the non-palindromic string
# if it exists, otherwise prints -1
def findNonPalinString(s):
 
    freq = [0] * (26)
    flag = 0
 
    for i in range(0, len(s)):
 
        # If all characters are not same,
        # set flag to 1
        if s[i] != s[0]:
            flag = 1
 
        # Update frequency of the current
        # character
        freq[ord(s[i]) - ord('a')] += 1
 
    # If all characters are same
    if not flag:
        print("-1")
     
    else:
         
        # Print characters in sorted manner
        for i in range(0, 26):
            for j in range(0, freq[i]):
                print(chr(ord('a') + i),
                               end = "")
 
# Driver Code
if __name__ == "__main__":
 
    s = "abba"
    findNonPalinString(s)
 
# This code is contributed by
# Rituraj Jain

// C# Program to rearrange letters
// of string to find a non-palindromic
// string if it exists
using System;
 
class GfG
{
 
    // Function to print the
    // non-palindromic string
    // if it exists, otherwise
    // prints -1
    static void findNonPalinString(char []s)
    {
        int []freq = new int[26];
        int flag = 0;
     
        for (int i = 0; i < s.Length; i++)
        {
     
            // If all characters are not
            // same, set flag to 1
            if (s[i] != s[0])
                flag = 1;
     
            // Update frequency of
            // the current character
            freq[s[i] - 'a']++;
        }
     
        // If all characters are same
        if (flag == 0)
            Console.WriteLine("-1");
        else
        {
     
            // Print characters in sorted manner
            for (int i = 0; i < 26; i++)
                for (int j = 0; j < freq[i]; j++)
                        Console.Write((char)('a' + i));
        }
    }
     
    // Driver Code
    public static void Main()
    {
        string s = "abba";
         
        findNonPalinString(s.ToCharArray());
    }
}
 
// This code is contributed by Ryuga

<script>
 
// JavaScript Program to rearrange letters of string
// to find a non-palindromic string if it exists
 
// Function to print the non-palindromic string
// if it exists, otherwise prints -1
function findNonPalinString(s)
{
    var freq = Array.from({length: 26}, (_, i) => 0);
    var flag = 0;
 
    for (var i = 0; i < s.length; i++)
    {
 
        // If all characters are not
        // same, set flag to 1
        if (s[i] != s[0])
            flag = 1;
 
        // Update frequency of
        // the current character
        freq[s[i].charCodeAt(0) -
        'a'.charCodeAt(0)]++;
    }
 
    // If all characters are same
    if (flag == 0)
        document.write("-1");
    else
    {
 
        // Print characters in sorted manner
        for (var i = 0; i < 26; i++)
            for (var j = 0; j < freq[i]; j++)
                document.write(String.fromCharCode
                ('a'.charCodeAt(0) + i));
    }
}
 
// Driver Code
var s = "abba";
 
findNonPalinString(s.split(''));
 
// This code contributed by shikhasingrajput
 
</script>

aabb

Complejidad temporal: O(N)
Espacio auxiliar: O(1)