Given an array of numbers of size n. It is also given that the array elements are in range from 0 to n^{2} – 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 question arises how to do this in 0(n)? Firstly, is it possible? Can we use data given in question? n numbers in range from 0 to n^{2} – 1?

The idea is to use Radix Sort. Following is 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 decimal system, b is 10. Since n^{2}-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[] = {00(0), 20(10), 12(7), 22(12), 23(13)} After second iteration, we get arr[] = {00(0),12(7),20(10),22(12),23(13)}

Following is C++ implementation to sort an array of size n where elements are in range from 0 to n^{2} – 1.

## C++

#include<iostream> using namespace std; // A function to do counting sort of arr[] according to // the digit represented by exp. int countSort(int arr[], int n, int exp) { int output[n]; // output array int i, count[n] ; for (int i=0; i < n; i++) count[i] = 0; // Store count of occurrences in count[] for (i = 0; i < n; i++) count[ (arr[i]/exp)%n ]++; // Change count[i] so that count[i] now contains actual // position of this digit in output[] for (i = 1; i < n; i++) count[i] += count[i - 1]; // Build the output array for (i = n - 1; i >= 0; i--) { output[count[ (arr[i]/exp)%n] - 1] = arr[i]; count[(arr[i]/exp)%n]--; } // Copy the output array to arr[], so that arr[] now // contains sorted numbers according to curent digit for (i = 0; i < n; i++) arr[i] = output[i]; } // The main function to that sorts arr[] of size n using Radix Sort void sort(int arr[], int n) { // Do counting sort for first digit in base n. Note that // instead of passing digit number, exp (n^0 = 0) is passed. countSort(arr, n, 1); // Do counting sort for second digit in base n. Note that // instead of passing digit number, exp (n^1 = n) is passed. countSort(arr, n, n); } // A utility function to print an array void printArr(int arr[], int n) { for (int i = 0; i < n; i++) cout << arr[i] << " "; } // Driver program to test above functions int main() { // Since array size is 7, elements should be from 0 to 48 int arr[] = {40, 12, 45, 32, 33, 1, 22}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Given array is n"; printArr(arr, n); sort(arr, n); cout << "nSorted array is n"; printArr(arr, n); return 0; }

## Java

// Java program to sort an array of size n where elements are // in range from 0 to n^2 – 1. class Sort1ToN2 { // A function to do counting sort of arr[] according to // the digit represented by exp. void countSort(int arr[], int n, int exp) { int output[] = new int[n]; // output array int i, count[] = new int[n] ; for (i=0; i < n; i++) count[i] = 0; // Store count of occurrences in count[] for (i = 0; i < n; i++) count[ (arr[i]/exp)%n ]++; // Change count[i] so that count[i] now contains actual // position of this digit in output[] for (i = 1; i < n; i++) count[i] += count[i - 1]; // Build the output array for (i = n - 1; i >= 0; i--) { output[count[ (arr[i]/exp)%n] - 1] = arr[i]; count[(arr[i]/exp)%n]--; } // Copy the output array to arr[], so that arr[] now // contains sorted numbers according to curent digit for (i = 0; i < n; i++) arr[i] = output[i]; } // The main function to that sorts arr[] of size n using Radix Sort void sort(int arr[], int n) { // Do counting sort for first digit in base n. Note that // instead of passing digit number, exp (n^0 = 0) is passed. countSort(arr, n, 1); // Do counting sort for second digit in base n. Note that // instead of passing digit number, exp (n^1 = n) is passed. countSort(arr, n, n); } // A utility function to print an array void printArr(int arr[], int n) { for (int i = 0; i < n; i++) System.out.print(arr[i]+" "); } // Driver program to test above functions public static void main(String args[]) { Sort1ToN2 ob = new Sort1ToN2(); // Since array size is 7, elements should be from 0 to 48 int arr[] = {40, 12, 45, 32, 33, 1, 22}; int n = arr.length; System.out.println("Given array"); ob.printArr(arr, n); ob.sort(arr, n); System.out.println("Sorted array"); ob.printArr(arr, n); } } /*This code is contributed by Rajat Mishra */

Output:

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

**How to sort if range is from 1 to n ^{2}?**

If range is from 1 to n n

^{2}, 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 n

^{2}, we can use following process.

1) Subtract all numbers by 1.

2) Since the range is now 0 to n

^{2}, 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 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.

This article is contributed by **Bateesh**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above