# Find the maximum possible Binary Number from given string

• Difficulty Level : Easy
• Last Updated : 20 Jul, 2021

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 ``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

 ``
Output:
`110`

Time Complexity: O(N)

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