Encuentre el número binario máximo posible de una string dada

Dada la string str que consta de los caracteres del conjunto {‘o’, ‘n’, ‘e’, ​​’z’, ‘r’} , la tarea es encontrar el mayor número binario posible que se puede formar reorganizando los caracteres de la string dada. Tenga en cuenta que la string formará al menos un número válido.

Ejemplos:  

Entrada: str = “roenenzooe” 
Salida: 110 
“oneonezero” es la string requerida.

Entrada: str = “cerocerocerouno” 
Salida: 1000  

Enfoque: Cree un mapa y almacene la frecuencia de ‘z’ y ‘n’ en él porque estos son los únicos caracteres que solo aparecerán en 0 o 1 y no en ambos. El número de unos en la string será igual a la frecuencia de ‘n’ y el número de ceros en la string será igual a la frecuencia de ‘z’ en el mapa. Ahora, para encontrar el número más grande, imprime todos los unos seguidos de todos los ceros.

A continuación se muestra la implementación del enfoque anterior: 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return maximum number
// that can be formed from the string
string maxNumber(string str, int n)
{
    // To store the frequency of 'z' and 'n'
    // in the given string
    int freq[2] = { 0 };
 
    for (int i = 0; i < n; i++) {
        if (str[i] == 'z') {
 
            // Number of zeroes
            freq[0]++;
        }
        else if (str[i] == 'n') {
 
            // Number of ones
            freq[1]++;
        }
    }
 
    // To store the required number
    string num = "";
 
    // Add all the ones
    for (int i = 0; i < freq[1]; i++)
        num += '1';
 
    // Add all the zeroes
    for (int i = 0; i < freq[0]; i++)
        num += '0';
 
    return num;
}
 
// Driver code
int main()
{
    string str = "roenenzooe";
    int n = str.length();
 
    cout << maxNumber(str, n);
 
    return 0;
}

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
    // Function to return maximum number
    // that can be formed from the string
    static String maxNumber(String str, int n)
    {
 
        // To store the frequency of 'z' and 'n'
        // in the given string
        int[] freq = new int[2];
 
        for (int i = 0; i < n; i++)
        {
            if (str.charAt(i) == 'z')
 
                // Number of zeroes
                freq[0]++;
                 
            else if (str.charAt(i) == 'n')
 
                // Number of ones
                freq[1]++;
        }
 
        // To store the required number
        String num = "";
 
        // Add all the ones
        for (int i = 0; i < freq[1]; i++)
            num += '1';
 
        // Add all the zeroes
        for (int i = 0; i < freq[0]; i++)
            num += '0';
 
        return num;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String str = "roenenzooe";
        int n = str.length();
 
        System.out.println(maxNumber(str, n));
    }
}
 
// This code is contributed by
// sanjeev2552

# Python3 implementation of the approach
 
# Function to return maximum number
# that can be formed from the string
def maxNumber(string , n) :
     
    # To store the frequency of 'z' and 'n'
    # in the given string
    freq = [0, 0]
     
    for i in range(n) :
        if (string[i] == 'z') :
 
            # Number of zeroes
            freq[0] += 1;
 
        elif (string[i] == 'n') :
 
            # Number of ones
            freq[1] += 1;
 
    # To store the required number
    num = "";
 
    # Add all the ones
    for i in range(freq[1]) :
        num += '1';
 
    # Add all the zeroes
    for i in range(freq[0]) :
        num += '0';
 
    return num;
 
# Driver code
if __name__ == "__main__" :
 
    string = "roenenzooe";
    n = len(string);
 
    print(maxNumber(string, n));
 
# This code is contributed by AnkitRai01

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to return maximum number
    // that can be formed from the string
    static string maxNumber(string str, int n)
    {
        // To store the frequency of 'z' and 'n'
        // in the given string
        int [] freq = new int[2];
 
        for (int i = 0; i < n; i++)
        {
            if (str[i] == 'z')
            {
     
                // Number of zeroes
                freq[0]++;
            }
            else if (str[i] == 'n')
            {
     
                // Number of ones
                freq[1]++;
            }
        }
     
        // To store the required number
        string num = "";
     
        // Add all the ones
        for (int i = 0; i < freq[1]; i++)
            num += '1';
     
        // Add all the zeroes
        for (int i = 0; i < freq[0]; i++)
            num += '0';
     
        return num;
    }
 
    // Driver code
    public static void Main()
    {
        string str = "roenenzooe";
        int n = str.Length;
        Console.Write(maxNumber(str, n));
    }
}
 
// This code is contributed by Sanjit Prasad

<script>
 
// Javascript implementation of the approach
 
// Function to return maximum number
// that can be formed from the string
function maxNumber(str, n)
{
    // To store the frequency of 'z' and 'n'
    // in the given string
    var freq = Array(2).fill(0);
 
    for (var i = 0; i < n; i++) {
        if (str[i] == 'z') {
 
            // Number of zeroes
            freq[0]++;
        }
        else if (str[i] == 'n') {
 
            // Number of ones
            freq[1]++;
        }
    }
 
    // To store the required number
    var num = "";
 
    // Add all the ones
    for (var i = 0; i < freq[1]; i++)
        num += '1';
 
    // Add all the zeroes
    for (var i = 0; i < freq[0]; i++)
        num += '0';
 
    return num;
}
 
// Driver code
var str = "roenenzooe";
var n = str.length;
document.write( maxNumber(str, n));
 
</script>

110

Complejidad de tiempo: O(N)