Algorithm-Sort-Radix

About Radix Sort

The lower bound for Comparison based sorting algorithm (Merge Sort, Heap Sort, Quick-Sort .. etc) is Ω(nLogn), i.e., they cannot do better than nLogn.

 * Radix sort algorithm !
 * Class	Sorting algorithm
 * Data structure	Array
 * Worst-case performance	О(n*k) comparisons, swaps
 * Best-case performance	O(n*k) comparisons, O(1) swaps
 * Average performance	О(n*k) comparisons, swaps
 * Worst-case space complexity
 *   where k is the length of the longest number and n is the
 *   size of the input array.
 *   Note: if k is greater than log(n) then an n*log(n) algorithm would be a
 *         better fit. In reality we can always change the radix to make k
 *         less than log(n).
 *
 * Author: Pooya Hatami

Installation

If you are using a browser, you can download node-sort-radix.js from GitHub or just bellow hotlink to it:

<script src="https://raw.githubusercontent.com/pooyahatami/Algorithm-Sort-Radix/master/node-sort-radix.js"></script>

If you are using node, you can install node-sort-radix with npm.

npm install node-sort-radix

Usage :

var nodesort = require('./node-sort-radix');
var displaymode = "No"; //"Yes";  // "Yes" for more details of algorithm progress
var base = 10;  //  (Decimal Base 10 , Hex Base 16 , Octal Base 8 , Binary Base 2 ).
...
nodesort(inputArray, displaymode, function(err,sortRef) {
        if (err) {
            // TODO error handeling 
            }
	      else {
           var result = sortRef.radixSort(inputArray,base);
           // TODO output 
	            }
    });

Ruls :

• Sort Array of integers (Decimal Base 10 , Hex Base 16 , Octal Base 8 , Binary Base 2 ).
• Array's element shoud be positive integers and not beager than 999,999 .
• Returns error mesage if not found valid input.
• Turn On details of Algorithms progress useing : displaymode = "Yes"

var displaymode = "No"; //"Yes";

Example

var nodesort = require('./node-sort-radix');
var displaymode = "No"; //"Yes";  // "Yes" for more details of algorithm progress 
var base = 10;  //  (Decimal Base 10 , Hex Base 16 , Octal Base 8 , Binary Base 2 ).

var arrin00 = [20, 8 , -11, 12, 22 , 9 , 10 ];
var arrin01 = [20, 8 , 48, 120, 220 , 390 , 1000 ];
var arrin02 = [20, 8 , 480 , 120, 220 , 390 , 1000 ];
var arrin03 = [1120, 800 , 480 , 120, 20 , 390 , 1000 ];
var arrin04 = ['g', 'e', 'e', 'k', 's', 'f', 'o',
                      'r', 'g', 'e', 'e', 'k', 's'];
var arrin05 = [1, 3, 7, 25, 12, 9, 8,
                      121, 221, 10, 18, 29, 49];
var arrin06 = [1, 3, -7, 25, 12, 9, 8,
                      121, 221, -10, 18, 29, 49];
var arrin07 = [1, 3, 7000000000000000000, 25, 12, 9, 8,
                      121, 221, 100000000000000000000000000 , 18, 290000000000000000000, 49];
var arrin08 = [1, 3, 75432, 25, 12, 9, 8,
                      121, 221, 976562 , 18, 299999, 49];
var arrin09 = [0.897, 0.565, 0.656, 0.1234, 0.665, 0.3434 , 0.611 , 0.621 ];
var arrin10 = [1,342, 14,293 , 0.897, 0.565, 0.656, 0.1234, 0.665, 0.3434 , 0.611 , 0.621 ];
var arrin11 = [5, 8 , 11, 12, 2 , 9 , 10 , 4 , 11, 10, 12, 7, 9 ];
var arrin12 = "";
//var arrin13 = [A7,02,22,77,37,15,00,40,B00,75,04,05,07,75,52,12,50,77,71,D07];    //base16
var arrin14 = [1001,101010,11,10,01,111,100,1000,11100,10110,101,100010,0111,101,11111,1000001,1,0,111,11010];   //base 2
var arrin15 = [7,2,22,77,37,15,10770,740,70,75,04,5,107,75,52,12,50,177,71,207];   //base 8


function solveSorting(inputArray,base) {
    var arr_original = inputArray.toString() ;
    var sortedArray = inputArray;

    nodesort(inputArray, displaymode,  function(err,sortRef) {
        if (err) {
	         console.log(err);
	                }
	      else {
           var result = sortRef.radixSort(inputArray,base);
	         console.log("Success attempt to sort array \r\n \t ["+arr_original+" ] \r\n and result is : \r\n \t [ "
                + result + " ]" );
  
	      sortedArray = result;
	            }
	      console.log("----------------------------------------------------------"); 
    });
    
    return sortedArray;
};

solveSorting(arrin01,10);
solveSorting(arrin02,10);
solveSorting(arrin06,10);
solveSorting(arrin11,10);
solveSorting(arrin12,10);
solveSorting(arrin14,2);
solveSorting(arrin15,8);

Counting sort is a linear time sorting algorithm that sort in O(n+k) time when elements are in range from 1 to k.

What if the elements are in range from 1 to n2?

We can’t use counting sort because counting sort will take O(n2) which is worse than comparison based sorting algorithms. Can we sort such an array in linear time? Radix Sort is the answer. The idea of Radix Sort is to do digit by digit sort starting from least significant digit to most significant digit. Radix sort uses counting sort as a subroutine to sort.

The 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 us understand it with the help of an example.

Original, unsorted list:

170, 45, 75, 90, 802, 24, 2, 66
Sorting by least significant digit (1s place) gives: [*Notice that we keep 802 before 2, because 802 occurred before 2 in the original list, and similarly for pairs 170 & 90 and 45 & 75.]

170, 90, 802, 2, 24, 45, 75, 66
Sorting by next digit (10s place) gives: [*Notice that 802 again comes before 2 as 802 comes before 2 in the previous list.]

802, 2, 24, 45, 66, 170, 75, 90
Sorting by most significant digit (100s place) gives:

2, 24, 45, 66, 75, 90, 170, 802

What is the running time of Radix Sort?

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. What is the value of d? If k is the maximum possible value, then d would be O(logb(k)). So overall time complexity is O((n+b) * logb(k)). Which looks more than the time complexity of comparison based sorting algorithms for a large k. Let us first limit k. Let k <= nc where c is a constant. In that case, the complexity becomes O(nLogb(n)). But it still doesn’t beat comparison based sorting algorithms. What if we make value of b larger? What should be the value of b to make the time complexity linear? If we set b as n, we get the time complexity as O(n). In other words, we can sort an array of integers with range from 1 to nc if the numbers are represented in base n (or every digit takes log2(n) bits).

Is Radix Sort preferable to Comparison based sorting algorithms like Quick-Sort?

If we have log2n bits for every digit, the running time of Radix appears to be better than Quick Sort for a wide range of input numbers. The constant factors hidden in asymptotic notation are higher for Radix Sort and Quick-Sort uses hardware caches more effectively. Also, Radix sort uses counting sort as a subroutine and counting sort takes extra space to sort numbers.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Implementation of Radix Sort Following is a simple C++ implementation of Radix Sort. For simplicity, the value of d is assumed to be 10. We recommend you to see Counting Sort for details of countSort() function in below code.

// C++ implementation of Radix Sort
#include<iostream>
using namespace std;
 
// A utility function to get maximum value in arr[]
int getMax(int arr[], int n)
{
    int mx = arr[0];
    for (int i = 1; i < n; i++)
        if (arr[i] > mx)
            mx = arr[i];
    return mx;
}
 
// 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[n]; // output array
    int i, count[10] = {0};
 
    // Store count of occurrences in count[]
    for (i = 0; i < n; i++)
        count[ (arr[i]/exp)%10 ]++;
 
    // Change count[i] so that count[i] now contains actual
    //  position of this digit in output[]
    for (i = 1; i < 10; i++)
        count[i] += count[i - 1];
 
    // Build the output array
    for (i = n - 1; i >= 0; i--)
    {
        output[count[ (arr[i]/exp)%10 ] - 1] = arr[i];
        count[ (arr[i]/exp)%10 ]--;
    }
 
    // Copy the output array to arr[], so that arr[] now
    // contains sorted numbers according to current 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 radixsort(int arr[], int n)
{
    // Find the maximum number to know number of digits
    int m = getMax(arr, n);
 
    // Do counting sort for every digit. Note that instead
    // of passing digit number, exp is passed. exp is 10^i
    // where i is current digit number
    for (int exp = 1; m/exp > 0; exp *= 10)
        countSort(arr, n, exp);
}
 
// A utility function to print an array
void print(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Driver program to test above functions
int main()
{
    int arr[] = {170, 45, 75, 90, 802, 24, 2, 66};
    int n = sizeof(arr)/sizeof(arr[0]);
    radixsort(arr, n);
    print(arr, n);
    return 0;
}

Following is a simple Java implementation of Radix Sort. For simplicity, the value of d is assumed to be 10. We recommend you to see Counting Sort for details of countSort() function in below code.

// Radix sort Java implementation
import java.io.*;
import java.util.*;
 
class Radix {
 
    // A utility function to get maximum value in arr[]
    static int getMax(int arr[], int n)
    {
        int mx = arr[0];
        for (int i = 1; i < n; i++)
            if (arr[i] > mx)
                mx = arr[i];
        return mx;
    }
 
    // A function to do counting sort of arr[] according to
    // the digit represented by exp.
    static void countSort(int arr[], int n, int exp)
    {
        int output[] = new int[n]; // output array
        int i;
        int count[] = new int[10];
        Arrays.fill(count,0);
 
        // Store count of occurrences in count[]
        for (i = 0; i < n; i++)
            count[ (arr[i]/exp)%10 ]++;
 
        // Change count[i] so that count[i] now contains
        // actual position of this digit in output[]
        for (i = 1; i < 10; i++)
            count[i] += count[i - 1];
 
        // Build the output array
        for (i = n - 1; i >= 0; i--)
        {
            output[count[ (arr[i]/exp)%10 ] - 1] = arr[i];
            count[ (arr[i]/exp)%10 ]--;
        }
 
        // 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
    static void radixsort(int arr[], int n)
    {
        // Find the maximum number to know number of digits
        int m = getMax(arr, n);
 
        // Do counting sort for every digit. Note that instead
        // of passing digit number, exp is passed. exp is 10^i
        // where i is current digit number
        for (int exp = 1; m/exp > 0; exp *= 10)
            countSort(arr, n, exp);
    }
 
    // A utility function to print an array
    static void print(int arr[], int n)
    {
        for (int i=0; i<n; i++)
            System.out.print(arr[i]+" ");
    }
 
 
    /*Driver function to check for above function*/
    public static void main (String[] args)
    {
        int arr[] = {170, 45, 75, 90, 802, 24, 2, 66};
        int n = arr.length;
        radixsort(arr, n);
        print(arr, n);
    }
}
/* This code is contributed by Devesh Agrawal */

Following is a simple Python implementation of Radix Sort. For simplicity, the value of d is assumed to be 10. We recommend you to see Counting Sort for details of countSort() function in below code.

# Python program for implementation of Radix Sort
 
# A function to do counting sort of arr[] according to
# the digit represented by exp.
def countingSort(arr, exp1):
 
    n = len(arr)
 
    # The output array elements that will have sorted arr
    output = [0] * (n)
 
    # initialize count array as 0
    count = [0] * (10)
 
    # Store count of occurrences in count[]
    for i in range(0, n):
        index = (arr[i]/exp1)
        count[ (index)%10 ] += 1
 
    # Change count[i] so that count[i] now contains actual
    #  position of this digit in output array
    for i in range(1,10):
        count[i] += count[i-1]
 
    # Build the output array
    i = n-1
    while i>=0:
        index = (arr[i]/exp1)
        output[ count[ (index)%10 ] - 1] = arr[i]
        count[ (index)%10 ] -= 1
        i -= 1
 
    # Copying the output array to arr[],
    # so that arr now contains sorted numbers
    i = 0
    for i in range(0,len(arr)):
        arr[i] = output[i]
 
# Method to do Radix Sort
def radixSort(arr):
 
    # Find the maximum number to know number of digits
    max1 = max(arr)
 
    # Do counting sort for every digit. Note that instead
    # of passing digit number, exp is passed. exp is 10^i
    # where i is current digit number
    exp = 1
    while max1/exp > 0:
        countingSort(arr,exp)
        exp *= 10
 
# Driver code to test above
arr = [ 170, 45, 75, 90, 802, 24, 2, 66]
radixSort(arr)
 
for i in range(len(arr)):
    print(arr[i]),
 
# This code is contributed by Mohit Kumra

Output: 2 24 45 66 75 90 170 802

Other Sorting Algorithms :

• Selection Sort
• Bubble Sort
• Insertion Sort
• Merge Sort
• Heap Sort
• QuickSort
• Counting Sort
• Radix Sort
• Bucket Sort
• ShellSort

References:

• https://en.wikipedia.org/wiki/Radix_sort
• http://www.geeksforgeeks.org/Radix-sort
• http://www.geekviewpoint.com/java/sorting/radixsort