Find the maximum possible Binary Number from given string
Last Updated :
12 Oct, 2022
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 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>
Time Complexity: O(N)
Auxiliary Space: O(N)
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