Find the maximum possible Binary Number from given string

Given a string str consisting of the characters from the set {‘o’, ‘n’, ‘e’, ‘z’, ‘r’}, the task is to 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 which 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++

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// 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 requried 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;
}

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Java














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// 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 requried 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 



















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Python3














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# 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 requried 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 



















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C#














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// 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 requried 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 



















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Output:


110





Time Complexity: O(N)


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