Ordenar una string sin alterar la posición de las vocales

Dada una string S de tamaño N , la tarea es ordenar la string sin cambiar la posición de las vocales.

Ejemplos:

Entrada: S = “geeksforgeeks”
Salida: feeggkokreess
Explicación:
Las consonantes presentes en la string son gksfrgks . Ordenar las consonantes modifica su secuencia a fggkkrss .
Ahora, actualice la string colocando las consonantes ordenadas en esas posiciones.

Entrada: S = “manzana”
Salida: alppe

Enfoque: siga los pasos a continuación para resolver el problema:

• Inicialice una string , digamos temp .
• Iterar sobre los caracteres de la string S.
• Si el carácter actual es una consonante, inserte el carácter en temp .
• Ordene la string temporal en orden lexicográfico.
• Inicialice un puntero, digamos ptr = 0 , para apuntar al carácter actual en la string temp .
• Ahora, recorra la string S y reemplace cada consonante de la string S con temp[ptr] . Incrementar ptr .

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to sort the string
// leaving the vowels unchanged
void sortStr(string S)
{
    // Length of string S
    int N = S.size();
 
    string temp = "";
 
    // Traverse the string S
    for (int i = 0; i < N; i++) {
        if (S[i] != 'a' && S[i] != 'e' && S[i] != 'i'
            && S[i] != 'o' && S[i] != 'u')
            temp += S[i];
    }
 
    // Sort the string temp
    if (temp.size())
        sort(temp.begin(), temp.end());
 
    // Pointer to traverse the
    // sorted string of consonants
    int ptr = 0;
 
    // Traverse the string S
    for (int i = 0; i < N; i++) {
        if (S[i] != 'a' && S[i] != 'e' && S[i] != 'i'
            && S[i] != 'o' && S[i] != 'u')
            S[i] = temp[ptr++];
    }
 
    cout << S;
}
 
// Driver Code
int main()
{
    string S = "geeksforgeeks";
    sortStr(S);
    return 0;
}

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to sort the string
// leaving the vowels unchanged
static void sortStr(String str)
{
    char S[] = str.toCharArray();
 
    // Length of string S
    int N = S.length;
 
    ArrayList<Character> temp = new ArrayList<>();
 
    // Traverse the string S
    for(int i = 0; i < N; i++)
    {
        if (S[i] != 'a' && S[i] != 'e' &&
            S[i] != 'i' && S[i] != 'o' &&
            S[i] != 'u')
            temp.add(S[i]);
    }
 
    // Sort the string temp
    if (temp.size() != 0)
        Collections.sort(temp);
 
    // Pointer to traverse the
    // sorted string of consonants
    int ptr = 0;
 
    // Traverse the string S
    for(int i = 0; i < N; i++)
    {
        if (S[i] != 'a' && S[i] != 'e' &&
            S[i] != 'i' && S[i] != 'o' &&
            S[i] != 'u')
            S[i] = temp.get(ptr++);
    }
    System.out.println(new String(S));
}
 
// Driver Code
public static void main(String[] args)
{
    String S = "geeksforgeeks";
     
    sortStr(S);
}
}
 
// This code is contributed by Kingash

# Python3 program for the above approach
 
# Function to sort the string
# leaving the vowels unchanged
def sortStr(S):
 
    # Length of string S
    N = len(S)
    temp = ""
 
    # Traverse the string S
    for i in range(N):
        if (S[i] != 'a' and S[i] != 'e' and S[i] != 'i'
                and S[i] != 'o'and S[i] != 'u'):
            temp += S[i]
 
    # Sort the string temp
    if (len(temp)):
        p = list(temp)
        p.sort()
        temp=''.join(p)
         
    # Pointer to traverse the
    # sorted string of consonants
    ptr = 0
 
    # Traverse the string S
    for i in range(N):
      S = list(S)
      if (S[i] != 'a' and S[i] != 'e' and S[i] != 'i'
                and S[i] != 'o' and S[i] != 'u'):
            S[i] = temp[ptr]
            ptr += 1
 
    print(''.join(S))
 
# Driver Code
if __name__ == "__main__":
 
    S = "geeksforgeeks"
    sortStr(S)
 
    # This code is contributed by ukasp.

// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG{
 
// Function to sort the string
// leaving the vowels unchanged
static void sortStr(String str)
{
    char []S = str.ToCharArray();
 
    // Length of string S
    int N = S.Length;
 
    List<char> temp = new List<char>();
 
    // Traverse the string S
    for(int i = 0; i < N; i++)
    {
        if (S[i] != 'a' && S[i] != 'e' &&
            S[i] != 'i' && S[i] != 'o' &&
            S[i] != 'u')
            temp.Add(S[i]);
    }
 
    // Sort the string temp
    if (temp.Count != 0)
        temp.Sort();
 
    // Pointer to traverse the
    // sorted string of consonants
    int ptr = 0;
 
    // Traverse the string S
    for(int i = 0; i < N; i++)
    {
        if (S[i] != 'a' && S[i] != 'e' &&
            S[i] != 'i' && S[i] != 'o' &&
            S[i] != 'u')
            S[i] = temp[ptr++];
    }
    Console.WriteLine(new String(S));
}
 
// Driver Code
public static void Main(String[] args)
{
    String S = "geeksforgeeks";
     
    sortStr(S);
}
}
 
// This code is contributed by Amit Katiyar

<script>
 
// JavaScript program for the above approach
 
// Function to sort the string
// leaving the vowels unchanged
function sortStr(str)
{
    var S = str.split('');
 
    // Length of string S
    var N = S.length;
 
    var temp = [];
 
    // Traverse the string S
    for(var i = 0; i < N; i++)
    {
        if (S[i] != 'a' && S[i] != 'e' &&
            S[i] != 'i' && S[i] != 'o' &&
            S[i] != 'u')
            temp.push(S[i]);
    }
 
    // Sort the string temp
    if (temp.length != 0)
        temp.sort();
 
    // Pointer to traverse the
    // sorted string of consonants
    var ptr = 0;
 
    // Traverse the string S
    for(var i = 0; i < N; i++)
    {
        if (S[i] != 'a' && S[i] != 'e' &&
            S[i] != 'i' && S[i] != 'o' &&
            S[i] != 'u')
            S[i] = temp[ptr++];
    }
    var str = "";
    for(var i =0;i<S.length;i++)
    str+=S[i];
     
    document.write(str);
}
 
// Driver Code
 
 
    var S = "geeksforgeeks";
     
    sortStr(S);
 
// This code is contributed by Amit Katiyar
 
</script>

feeggkokreess

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