Sort n numbers in range from 0 to n^2 - 1 in linear time

Given an array of numbers of size n. It is also given that the array elements are in the range from 0 to n² - 1. Sort the given array in linear time.

Examples:

Since there are 5 elements, the elements can be from 0 to 24.
Input: arr[] = {0, 23, 14, 12, 9}
Output: arr[] = {0, 9, 12, 14, 23}

Since there are 3 elements, the elements can be from 0 to 8.
Input: arr[] = {7, 0, 2}
Output: arr[] = {0, 2, 7}

Solution: If we use Counting Sort, it would take O(n^2) time as the given range is of size n^2. Using any comparison-based sorting like Merge Sort, Heap Sort, .. etc would take O(nLogn) time. 
Now the question arises how to do this in 0(n)? Firstly, is it possible? Can we use the data given in the question? n numbers in the range from 0 to n² - 1? 

The idea is to use Radix Sort. Following is the standard Radix Sort algorithm. 

1) Do following for each digit i where i varies from least
significant digit to the most significant digit.
…………..a) Sort input array using counting sort (or any stable sort) according to the i’th digit

Let there be d digits in input integers. Radix Sort takes O(d*(n+b)) time where b is the base for representing numbers, for example, for a decimal system, b is 10. Since n²-1 is the maximum possible value, the value of d would be O(log_b(n)). So overall time complexity is O((n+b)*O(log_b(n)). Which looks more than the time complexity of comparison-based sorting algorithms for a large k. The idea is to change base b. If we set b as n, the value of O(log_b(n)) becomes O(1) and overall time complexity becomes O(n). 

arr[] = {0, 10, 13, 12, 7}

Let us consider the elements in base 5. For example 13 in
base 5 is 23, and 7 in base 5 is 12.
arr[] = {00(0), 20(10), 23(13), 22(12), 12(7)}

After first iteration (Sorting according to the last digit in
base 5), we get.
arr[] = {0(0), 2(10), 1(7), 2(12), 2(13)}

After second iteration, we get
arr[] = {0(0), 2(7), 0(10), 2(12), 3(13)}

Following is the implementation to sort an array of size n where elements are in range from 0 to n² - 1. 

Given array is
40 12 45 32 33 1 22 
Sorted array is
1 12 22 32 33 40 45 

Time Complexity: O(n).
Auxiliary Space: O(n)

How to sort if the range is from 1 to n²?
If the range is from 1 to  n², the above process can not be directly applied, it must be changed. Consider n = 100 and range from 1 to 10000. Since the base is 100, a digit must be from 0 to 99 and there should be 2 digits in the numbers. But the number 10000 has more than 2 digits. So to sort numbers in a range from 1 to n², we can use the following process. 
1) Subtract all numbers by 1. 
2) Since the range is now 0 to n², do counting sort twice as done in the above implementation. 
3) After the elements are sorted, add 1 to all numbers to obtain the original numbers.

How to sort if the range is from 0 to n^3 -1?
Since there can be 3 digits in base n, we need to call counting sort 3 times.