Reorder characters of a string to valid English representations of digits

Given a string S of length N, consisting of lowercase characters containing reordered English representations of digits [0 - 9], the task is to print those digits in ascending order.

Examples:

Input: S = "fviefuro"
Output: 45
Explanation: The given string can be reshuffled to "fourfive", Therefore, the digits represented by the strings are 4 and 5.

Input: S = "owoztneoer"
Output: 012
Explanation: The given string can be reshuffled to get "zeroonetwo", Therefore, the digits represented by the strings are 0, 1 and 2.

Naive Approach: The simplest approach is to generate all permutations of the given string and for each permutation, check if it is possible to find valid digits represented by the string. If found to be true, then print the set of digits in ascending order.

Time Complexity: O(N * N!)
Auxiliary Space: O(1)

Efficient Approach: The idea is based on the observation that some characters only appear in one number.

In 'zero', character 'z' is unique.
In 'two', character 'w' is unique.
In 'four', character 'u' is unique.
In 'six', character 'x' is unique.  
In 'eight', character 'g' is unique.
In 'three', character 'h' is unique since word "eight" having character 'h' has already been considered.
In 'one', character 'o' is unique since words having character 'o' have already been considered.
In 'five', character f' is unique since word "four" having character 'f' has already been considered.
In 'seven', character 'v' is unique.
In 'nine', character 'i' is unique since words having character 'i' have already been considered.

Follow the steps below to solve the problem:

• Initialize an empty string, ans to store the required result.
• Store the frequency of each character of the string in M.
• Create a mapping of a unique character to its corresponding string.
• Traverse the map, and perform the following steps:
  • Store the unique character corresponding to the digit in a variable x.
  • Get the occurrence of x in M, and store it in a variable freq.
  • Append the corresponding digit, freq number of times to ans.
  • Traverse the word corresponding to x and decrement the frequency of its characters by freq in M.
• Print the string, ans as the result. 

Below is the implementation of the above approach:

// C++ program to implement the above approach

// Function to construct the original set of digits
// from the string in ascending order
#include <bits/stdc++.h>
using namespace std;

string construct_digits(string s)
{

    // Store the unique characters
    // corresponding to word and number
    vector<char>k = { 'z', 'w', 'u', 'x', 'g',
                'h', 'o', 'f', 'v', 'i' };

    vector<string>l = { "zero", "two", "four", "six", "eight",
                "three", "one", "five", "seven", "nine" };

    vector<int>c = { 0, 2, 4, 6, 8, 3, 1, 5, 7, 9 };

    // Store the required result
    vector<int> ans = {};

    // Store the frequency of
    // each character of S
    unordered_map<char,int>d;
    for(int i = 0; i < s.length(); i++)
    {
        d[s[i]]++;
    }

    // Traverse the unique characters
    for(int i = 0; i < k.size(); i++)
    {
        
        // Store the count of k[i] in S
        int x = 0;
        if (d.find(k[i]) != d.end())
            x = d[k[i]];

        // Traverse the corresponding word
        for(int j = 0; j < l[i].length(); j++)
        {
            
            // Decrement the frequency
            // of characters by x
            if (d.find(l[i][j]) != d.end())
                d[l[i][j]]-= x;
        }

        // Append the digit x times to ans
        for(int j = 0; j < x; j++)
            ans.push_back(c[i]);
    }

    // Sort the digits in ascending order
    sort(ans.begin(),ans.end());
    
    string res;
    for(auto x: ans)res+=(x+'0');
    return res;
}

// Driver Code
int main()
{
  
   // Given string, s
   string s = "fviefuro";
  
   // Function Call
   cout<<(construct_digits(s));
}

// This code is contributed by shinjanpatra

// Java program to implement the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG{

// Function to construct the original set of digits
// from the string in ascending order
static String construct_digits(String s)
{
    
    // Store the unique characters
    // corresponding to word and number
    char[] k = { 'z', 'w', 'u', 'x', 'g',
                 'h', 'o', 'f', 'v', 'i' };

    String[] l = { "zero", "two", "four", "six", "eight",
                   "three", "one", "five", "seven", "nine" };

    int[] c = { 0, 2, 4, 6, 8, 3, 1, 5, 7, 9 };

    // Store the required result
    List<Integer> ans = new ArrayList<>();

    // Store the frequency of
    // each character of S
    HashMap<Character, Integer> d = new HashMap<>();
    for(int i = 0; i < s.length(); i++)
    {
        d.put(s.charAt(i),
              d.getOrDefault(s.charAt(i), 0) + 1);
    }

    // Traverse the unique characters
    for(int i = 0; i < k.length; i++)
    {
        
        // Store the count of k[i] in S
        int x = 0;
        if (d.containsKey(k[i]))
            x = d.get(k[i]);

        // Traverse the corresponding word
        for(int j = 0; j < l[i].length(); j++)
        {
            
            // Decrement the frequency
            // of characters by x
            if (d.containsKey(l[i].charAt(j)))
                d.put(l[i].charAt(j),
                d.get(l[i].charAt(j)) - x);
        }

        // Append the digit x times to ans
        for(int j = 0; j < x; j++)
            ans.add(c[i]);
    }

    // Sort the digits in ascending order
    Collections.sort(ans);
    String str = "";
    for(int val : ans)
        str += val;

    return str;
}

// Driver Code
public static void main(String[] args)
{
    
    // Given string, s
    String s = "fviefuro";

    // Function Call
    System.out.println(construct_digits(s));
}
}

// This code is contributed by Kingash

# Python program to implement the above approach
from collections import Counter

# Function to construct the original set of digits
# from the string in ascending order
def construct_digits(s):
  
    # Store the unique characters
    # corresponding to word and number
    k = ["z", "w", "u", "x", "g", 
         "h", "o", "f", "v", "i"]

    l = ["zero", "two", "four", "six", "eight",
         "three", "one", "five", "seven", "nine"]

    c = [0, 2, 4, 6, 8, 3, 1, 5, 7, 9]
    
    # Store the required result
    ans = []
    
    # Store the frequency of
    # each character of S
    d = Counter(s)

    # Traverse the unique characters 
    for i in range(len(k)):

        # Store the count of k[i] in S
        x = d.get(k[i], 0)
        
        # Traverse the corresponding word
        for j in range(len(l[i])):
              
            # Decrement the frequency
            # of characters by x
            d[l[i][j]]-= x
            
        # Append the digit x times to ans
        ans.append(str(c[i])*x)

    # Sort the digits in ascending order
    ans.sort()

    return "".join(ans)

# Driver Code

# Given string, s
s = "fviefuro"

# Function Call
print(construct_digits(s))

// C# program to implement the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to construct the original set of digits
// from the string in ascending order
static string construct_digits(string s)
{
    
    // Store the unique characters
    // corresponding to word and number
    char[] k = { 'z', 'w', 'u', 'x', 'g',
                 'h', 'o', 'f', 'v', 'i' };

    string[] l = { "zero", "two", "four", "six", "eight",
                   "three", "one", "five", "seven", "nine" };

    int[] c = { 0, 2, 4, 6, 8, 3, 1, 5, 7, 9 };

    // Store the required result
    List<string> ans = new List<string>();

    // Store the frequency of
    // each character of S
    Dictionary<char, 
               int> d = new Dictionary<char,
                                       int>();
    for(int i = 0; i < s.Length; i++) 
    {
        if (!d.ContainsKey(s[i]))
            d[s[i]] = 0;
            
        d[s[i]] += 1;
    }

    // Traverse the unique characters
    for(int i = 0; i < k.Length; i++)
    {
        
        // Store the count of k[i] in S
        int x = 0;
        if (d.ContainsKey(k[i]))
            x = d[k[i]];

        // Traverse the corresponding word
        for(int j = 0; j < l[i].Length; j++)
        {
            
            // Decrement the frequency
            // of characters by x
            if (d.ContainsKey(l[i][j]))
                d[l[i][j]] -= x;
        }
        
        // Append the digit x times to ans
        ans.Add(((c[i]) * x).ToString());
    }
    
    // Sort the digits in ascending order
    ans.Sort();

    string str = (String.Join("", ans.ToArray()));
    return str.Replace("0", "");
}

// Driver Code
public static void Main(string[] args)
{
    
    // Given string, s
    string s = "fviefuro";

    // Function Call
    Console.WriteLine(construct_digits(s));
}
}

// This code is contributed by ukasp

<script>
// Javascript program to implement the above approach

// Function to construct the original set of digits
// from the string in ascending order
function construct_digits(s)
{

    // Store the unique characters
    // corresponding to word and number
    let k = [ 'z', 'w', 'u', 'x', 'g',
                 'h', 'o', 'f', 'v', 'i' ];
 
    let l = [ "zero", "two", "four", "six", "eight",
                   "three", "one", "five", "seven", "nine" ];
 
    let c = [ 0, 2, 4, 6, 8, 3, 1, 5, 7, 9 ];
 
    // Store the required result
    let ans = [];
 
    // Store the frequency of
    // each character of S
    let d = new Map();
    for(let i = 0; i < s.length; i++)
    {
        if(!d.has(s[i]))
            d.set(s[i],0);
        d.set(s[i],
              d.get(s[i]) + 1);
    }
 
    // Traverse the unique characters
    for(let i = 0; i < k.length; i++)
    {
         
        // Store the count of k[i] in S
        let x = 0;
        if (d.has(k[i]))
            x = d.get(k[i]);
 
        // Traverse the corresponding word
        for(let j = 0; j < l[i].length; j++)
        {
             
            // Decrement the frequency
            // of characters by x
            if (d.has(l[i][j]))
                d.set(l[i][j],
                d.get(l[i][j]) - x);
        }
 
        // Append the digit x times to ans
        for(let j = 0; j < x; j++)
            ans.push(c[i]);
    }
 
    // Sort the digits in ascending order
    ans.sort();
    
 
    return ans.join("");
}

// Driver Code
// Given string, s
let s = "fviefuro";
// Function Call
document.write(construct_digits(s));

// This code is contributed by avanitrachhadiya2155
</script>

45

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