Find the maximum possible Binary Number from given string

Given string str consisting of the characters from the set {‘o’, ‘n’, ‘e’, ‘z’, ‘r’}, the task is to find the largest possible binary number that can be formed by rearranging the characters of the given string. Note that the string will form at least a valid number.
Examples:

Input: str = “roenenzooe” 
Output: 110 
“oneonezero” is the required string.

Input: str = “zerozerozeroone” 
Output: 1000  

Approach: Create a map and store the frequency of ‘z’ and ‘n’ in it because these are the only characters that will only appear either in 0 or 1 and not both. The number of ones in the string will be equal to the frequency of ‘n’ and the number of zeroes in the string will be equal to the frequency of ‘z’ in the map. Now to find the largest number, print all the ones followed by all the zeroes.

Below is the implementation of the above approach: 

C++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output:

110

Time Complexity: O(N)
Auxiliary Space: O(N)