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 stringInput: 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++
// 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; } |
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 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 |
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 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 |
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 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 |
110
Time Complexity: O(N)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
