This notebook was prepared by Donne Martin. Source and license info is on GitHub.

# Solution Notebook¶

## Constraints¶

• Can we assume this is a non-circular, singly linked list?
• Yes
• Do we expect the return to be in reverse order too?
• Yes
• What if one of the inputs is None?
• Return None for an invalid operation
• How large are these numbers--can they fit in memory?
• Yes
• Can we assume we already have a linked list class that can be used for this problem?
• Yes
• Can we assume this fits in memory?
• Yes

## Test Cases¶

• Empty list(s) -> None
• Add values of different lengths
• Input 1: 6->5->None
• Input 2: 9->8->7
• Result: 5->4->8
• Add values of same lengths
• Exercised from values of different lengths
• Done here for completeness

## Algorithm¶

We could solve this with an iterative or a recursive algorithm, both are well suited for this exercise. We'll use a recursive algorithm for practice with recursion. Note this takes an extra space of O(m) where m is the recursion depth.

• Base case:
• if first and second lists are None AND carry is zero
• Return None
• Recursive case:
• Set value to carry
• Add both nodes' data to value
• Set the carry to 1 if value >= 10, else 0
• Set the remainder to value % 10
• Create a node with the remainder
• Set node.next to a recursive call on the next nodes, passing in the carry
• Return node

Complexity:

• Time: O(n)
• Space: O(m), extra space for result and recursion depth

Notes:

• Careful with adding if the lists differ
• Only add if a node is not None
• Alternatively, we could add trailing zeroes to the smaller list

## Code¶

In [1]:
%run ../linked_list/linked_list.py

In [2]:
class MyLinkedList(LinkedList):

# Base case
if first_node is None and second_node is None and not carry:
return None

# Recursive case
value = carry
value += first_node.data if first_node is not None else 0
value += second_node.data if second_node is not None else 0
carry = 1 if value >= 10 else 0
value %= 10
node = Node(value)
first_node.next if first_node is not None else None,
second_node.next if first_node is not None else None,
carry)
return node

# See constraints
if first_list is None or second_list is None:
return None


## Unit Test¶

In [3]:
%%writefile test_add_reverse.py
import unittest

print('Test: Empty list(s)')

print('Test: Add values of different lengths')
# Input 1: 6->5->None
# Input 2: 9->8->7
# Result: 5->4->8
first_list.append(5)
second_list.append(8)
second_list.append(7)
self.assertEqual(result.get_all_data(), [5, 4, 8])

print('Test: Add values of same lengths')
# Input 1: 6->5->4
# Input 2: 9->8->7
# Result: 5->4->2->1
first_list.append(5)
first_list.append(4)
second_list.append(8)
second_list.append(7)
self.assertEqual(result.get_all_data(), [5, 4, 2, 1])

def main():

if __name__ == '__main__':
main()

Overwriting test_add_reverse.py

In [4]:
%run -i test_add_reverse.py

Test: Empty list(s)
Test: Add values of different lengths
Test: Add values of same lengths