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Python Program To Subtract Two Numbers Represented As Linked Lists

  • Last Updated : 03 Jan, 2022

Given two linked lists that represent two large positive numbers. Subtract the smaller number from the larger one and return the difference as a linked list. Note that the input lists may be in any order, but we always need to subtract smaller from the larger ones.
It may be assumed that there are no extra leading zeros in input lists.
Examples:

Input: l1 = 1 -> 0 -> 0 -> NULL,  l2 = 1 -> NULL
Output: 0->9->9->NULL
Explanation: Number represented as 
lists are 100 and 1, so 100 - 1 is 099

Input: l1 = 7-> 8 -> 6 -> NULL,  l2 = 7 -> 8 -> 9 NULL
Output: 3->NULL
Explanation: Number represented as 
lists are 786 and  789, so 789 - 786 is 3, 
as the smaller value is subtracted from 
the larger one.

Approach: Following are the steps.

  1. Calculate sizes of given two linked lists.
  2. If sizes are not the same, then append zeros in the smaller linked list.
  3. If the size is the same, then follow the below steps:
    1. Find the smaller valued linked list.
    2. One by one subtract nodes of the smaller-sized linked list from the larger size. Keep track of borrow while subtracting.

Following is the implementation of the above approach.

Python




# Python program to subtract smaller 
# valued list from larger valued list 
# and return result as a list.
  
# A linked List Node
class Node: 
    def __init__(self, new_data): 
        self.data = new_data 
        self.next = None
  
# A utility which creates Node.
def newNode(data):
    temp = Node(0)
    temp.data = data
    temp.next = None
    return temp
  
# A utility function to get 
# length of linked list 
def getLength(Node):
    size = 0
  
    while (Node != None):    
        Node = Node.next
        size = size + 1
      
    return size
  
# A Utility that padds zeros in 
# front of the Node, with the 
# given diff 
def paddZeros(sNode, diff):
    if (sNode == None):
        return None
  
    zHead = newNode(0)
    diff = diff - 1
    temp = zHead
    while (diff > 0):
        diff = diff - 1
        temp.next = newNode(0)
        temp = temp.next
      
    temp.next = sNode
    return zHead
  
borrow = True
  
# Subtract LinkedList Helper is a 
# recursive function, move till the 
# last Node, and subtract the digits 
# and create the Node and return the 
# Node. If d1 < d2, we borrow the number 
# from previous digit. 
def subtractLinkedListHelper(l1, l2):
    global borrow
      
    if (l1 == None and 
        l2 == None and not borrow ):
        return None
  
    l3 = None
    l4 = None
    if(l1 != None):
        l3 = l1.next
    if(l2 != None):
        l4 = l2.next
    previous = 
    subtractLinkedListHelper(l3, l4)
  
    d1 = l1.data
    d2 = l2.data
    sub = 0
  
    # If you have given the value value 
    # to next digit then reduce the d1 by 1 
    if (borrow):
        d1 = d1 - 1
        borrow = False
      
    # If d1 < d2, then borrow the number 
    # from previous digit. Add 10 to d1 
    # and set borrow = True 
    if (d1 < d2):
        borrow = True
        d1 = d1 + 10
  
    # Subtract the digits 
    sub = d1 - d2
  
    # Create a Node with sub value 
    current = newNode(sub)
  
    # Set the Next pointer as Previous 
    current.next = previous
  
    return current
  
# This API subtracts two linked lists 
# and returns the linked list which 
# shall have the subtracted result. 
def subtractLinkedList(l1, l2):
  
    # Base Case.
    if (l1 == None and l2 == None):
        return None
  
    # In either of the case, get the 
    # lengths of both
    # Linked list.
    len1 = getLength(l1)
    len2 = getLength(l2)
  
    lNode = None
    sNode = None
  
    temp1 = l1
    temp2 = l2
  
    # If lengths differ, calculate the 
    # smaller Node and padd zeros for 
    # smaller Node and ensure both larger 
    # Node and smaller Node has equal length.
    if (len1 != len2):
        if(len1 > len2):
            lNode = l1
        else:
            lNode = l2
          
        if(len1 > len2):
            sNode = l2
        else:
            sNode = l1
        sNode = paddZeros(sNode, 
                          abs(len1 - len2))
      
    else:
      
        # If both list lengths are equal, then 
        # calculate the larger and smaller list. 
        # If 5-6-7 & 5-6-8 are linked list, then 
        # walk through linked list at last Node 
        # as 7 < 8, larger Node is 5-6-8 and 
        # smaller Node is 5-6-7.
        while (l1 != None and l2 != None):        
            if (l1.data != l2.data):
                if(l1.data > l2.data ):
                    lNode = temp1 
                else:
                    lNode = temp2
                  
                if(l1.data > l2.data ):
                    sNode = temp2 
                else:
                    sNode = temp1
                break
              
            l1 = l1.next
            l2 = l2.next
          
    global borrow
      
    # After calculating larger and smaller 
    # Node, call subtractLinkedListHelper 
    # which returns the subtracted
    # linked list.
    borrow = False
    return subtractLinkedListHelper(lNode, 
                                    sNode)
  
# A utility function to print 
# linked list 
def printList(Node):
    while (Node != None):    
        print (Node.data, 
               end =" ")
        Node = Node.next    
    print(" ")
  
# Driver code
head1 = newNode(1)
head1.next = newNode(0)
head1.next.next = newNode(0)
head2 = newNode(1)
result = subtractLinkedList(head1, head2)
printList(result)
# This code is contributed by Arnab Kundu

Output:

0 9 9 

Complexity Analysis:

  • Time complexity: O(n). 
    As no nested traversal of linked list is needed.
  • Auxiliary Space: O(n). 
    If recursive stack space is taken into consideration O(n) space is needed.

Please refer complete article on Subtract Two Numbers represented as Linked Lists for more details!


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