Address Calculation Sort using Hashing

Difficulty Level : Easy
Last Updated : 10 Oct, 2018

In this sorting algorithm, Hash Function f is used with the property of Order Preserving Function which states that if $x <= y, f(x) <= f(y)$ .

Hash Function:

f(x) = floor( (x/maximum) * SIZE )
where maximum => maximum value in the array,
      SIZE => size of the address table (10 in our case),
      floor => floor function

This algorithm uses an address table to store the values which is simply a list (or array) of Linked lists. The Hash function is applied to each value in the array to find its corresponding address in the address table. Then the values are inserted at their corresponding addresses in a sorted manner by comparing them to the values which are already present at that address.

Examples:

Input : arr = [29, 23, 14, 5, 15, 10, 3, 18, 1] 
Output:
After inserting all the values in the address table, the address table looks like this:

ADDRESS 0: 1 --> 3 
ADDRESS 1: 5 
ADDRESS 2: 
ADDRESS 3: 10 
ADDRESS 4: 14 --> 15 
ADDRESS 5: 18 
ADDRESS 6: 
ADDRESS 7: 23 
ADDRESS 8: 
ADDRESS 9: 29

The below figure shows the representation of the address table for the example discussed above:

After insertion, the values at each address in the address table are sorted. Hence we iterate through each address one by one and insert the values at that address in the input array.

Below is the implementation of the above approach

# Python3 code for implementation of 
# Address Calculation Sorting using Hashing
  
# Size of the address table (In this case 0-9)
SIZE = 10
  
class Node(object):
  
    def __init__(self, data = None):
        self.data = data
        self.nextNode = None
  
class LinkedList(object):
  
    def __init__(self):
        self.head = None
  
    # Insert values in such a way that the list remains sorted
    def insert(self, data):
        newNode = Node(data)
  
        # If there is no node or new Node's value
        # is smaller than the first value in the list,
  
        # Insert new Node in the first place
        if self.head == None or data < self.head.data:
            newNode.nextNode = self.head
            self.head = newNode
  
        else:
            current = self.head
              
            # If the next node is null or its value
            # is greater than the new Node's value,
   
            # Insert new Node in that place
            while current.nextNode != None \
                    and \
                    current.nextNode.data < data:
                current = current.nextNode
  
            newNode.nextNode = current.nextNode
            current.nextNode = newNode
              
# This function sorts the given list 
# using Address Calculation Sorting using Hashing
def addressCalculationSort(arr):
  
    # Declare a list of Linked Lists of given SIZE
    listOfLinkedLists = []
    for i in range(SIZE):
        listOfLinkedLists.append(LinkedList())
  
    # Calculate maximum value in the array
    maximum = max(arr)
  
    # Find the address of each value
    # in the address table 
    # and insert it in that list
    for val in arr:
        address = hashFunction(val, maximum)
        listOfLinkedLists[address].insert(val)
      
    # Print the address table 
    # after all the values have been inserted
    for i in range(SIZE):
        current = listOfLinkedLists[i].head
        print("ADDRESS " + str(i), end = ": ")
  
        while current != None:
            print(current.data, end = " ")
            current = current.nextNode
  
        print()
      
    # Assign the sorted values to the input array
    index = 0
    for i in range(SIZE):
        current = listOfLinkedLists[i].head
  
        while current != None:
            arr[index] = current.data
            index += 1
            current = current.nextNode
              
  
# This function returns the corresponding address
# of given value in the address table
def hashFunction(num, maximum):
  
    # Scale the value such that address is between 0 to 9
    address = int((num * 1.0 / maximum) * (SIZE-1))
    return address
  
# -------------------------------------------------------
# Driver code
  
# giving the input address as follows
arr = [29, 23, 14, 5, 15, 10, 3, 18, 1]
  
# Printing the Input array
print("\nInput array: " + " ".join([str(x) for x in arr]))
  
# Performing address calculation sort
addressCalculationSort(arr)
  
# printing the result sorted array
print("\nSorted array: " + " ".join([str(x) for x in arr]))

Output:

Input array: 29 23 14 5 15 10 3 18 1

ADDRESS 0: 1 3 
ADDRESS 1: 5 
ADDRESS 2: 
ADDRESS 3: 10 
ADDRESS 4: 14 15 
ADDRESS 5: 18 
ADDRESS 6: 
ADDRESS 7: 23 
ADDRESS 8: 
ADDRESS 9: 29 

Sorted array: 1 3 5 10 14 15 18 23 29

Time Complexity:
The time complexity of this algorithm is $O(n)$ in the best case. This happens when the values in the array are uniformly distributed within a specific range.

Whereas the worst case time complexity is $O(n^2)$ . This happens when most of the values occupy 1 or 2 addresses because then significant work is required to insert each value at its proper place.

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Address Calculation Sort using Hashing