Sort an Array alphabetically when each number is converted into words

Given an array arr[] containing N non-negative integers, the task is to sort these integers alphabetically when each number is converted into words.

Examples:

Input: arr[] = {12, 10, 102, 31, 15}
Output: 15 102 10 31 12
Explanation:
The above set of numbers are sorted alphabetically. That is:
15 -> Fifteen
102 -> One hundred and two
10 -> Ten
31 -> Thirty-one
12 -> Twelve

Input: arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }
Output: 8 5 4 9 1 7 6 10 3 2
Explanation:
The above set of numbers are sorted alphabetically. That is:
8 -> eight
5 -> five
4 -> four
9 -> nine
1 -> one
7 -> seven
6 -> six
10 -> ten
3 -> three
2 -> two

Approach: In order to sort the numbers alphabetically, we first need to convert the number to its word form. Therefore, the idea is to store each element along with its word form in a vector pair and then sort all the elements of the vector according to the corresponding words of the number. Therefore:



  • Precompute and store the word forms of all the units digits in an array.
  • Precompute and store the word forms of all the tens digits in another array.
  • For all the remaining numbers greater than 2 digits, the number is divided and the word form is added.
  • Iterate through every digit of the number in the array arr[] and store the corresponding word form of the numbers in a vector as a pair.
  • Iterate through the vector and sort the vector according to the words.
  • Finally, print the sorted order.

Below is the implementation of the above approach:

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// C++ program to sort an array of
// integers alphabetically
  
#include <bits/stdc++.h>
using namespace std;
  
// Variable to store the word form of
// units digit and up to twenty
string one[]
    = { "", "one ", "two ", "three ",
        "four ", "five ", "six ",
        "seven ", "eight ", "nine ", "ten ",
        "eleven ", "twelve ", "thirteen ",
        "fourteen ", "fifteen ", "sixteen ",
        "seventeen ", "eighteen ", "nineteen " };
  
// Variable to store the word form of
// tens digit
string ten[]
    = { "", "", "twenty ",
        "thirty ", "forty ",
        "fifty ", "sixty ",
        "seventy ", "eighty ",
        "ninety " };
  
// Function to convert a two digit number
// to the word by using the above defined arrays
string numToWords(int n, string s)
{
    string str = "";
  
    // If n is more than 19, divide it
    if (n > 19)
        str += ten[n / 10] + one[n % 10];
    else
        str += one[n];
  
    // If n is non-zero
    if (n)
        str += s;
  
    return str;
}
  
// Function to print a given number in words
string convertToWords(int n)
{
    // Stores the word representation
    // of the given number n
    string out;
  
    // Handles digits at ten millions
    // and hundred millions places
    out += numToWords((n / 10000000),
                      "crore ");
  
    // Handles digits at hundred thousands
    // and one millions places
    out += numToWords(((n / 100000) % 100),
                      "lakh ");
  
    // Handles digits at thousands and
    // tens thousands places
    out += numToWords(((n / 1000) % 100),
                      "thousand ");
  
    // Handles digit at hundreds places
    out += numToWords(((n / 100) % 10),
                      "hundred ");
  
    if (n > 100 && n % 100)
        out += "and ";
  
    // Call the above function to convert
    // the number into words
    out += numToWords((n % 100), "");
  
    return out;
}
  
// Function to sort the array according to
// the albhabetical order
void sortArr(int arr[], int n)
{
    // Vector to store the number in words
    // with respective elements
    vector<pair<string, int> > vp;
  
    // Inserting number in words
    // with respective elements in vector pair
    for (int i = 0; i < n; i++) {
        vp.push_back(make_pair(
            convertToWords(arr[i]), arr[i]));
    }
  
    // Sort the vector, this will sort the pair
    // according to the alphabetical order.
    sort(vp.begin(), vp.end());
  
    // Print the sorted vector content
    for (int i = 0; i < vp.size(); i++)
        cout << vp[i].second << " ";
}
  
// Driver code
int main()
{
    int arr[] = { 12, 10, 102, 31, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    sortArr(arr, n);
  
    return 0;
}

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Output:

15 102 10 31 12

Time Complexity: O(N * log(N)), where N is the size of the array

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