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Weaving Spells II
Jessica Sang edited this page Sep 14, 2024
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1 revision
TIP102 Unit 7 Session 1 Standard (Click for link to problem statements)
Below is an iterative solution to the weave_spells()
function from the previous problem. Compare your recursive solution to the iterative solution below.
class Node:
def __init__(self, value, next=None):
self.value = value
self.next = next
def weave_spells(spell_a, spell_b):
# If either list is empty, return the other
if not spell_a:
return spell_b
if not spell_b:
return spell_a
# Start with the first node of spell_a
head = spell_a
# Loop through both lists until one is exhausted
while spell_a and spell_b:
# Store the next pointers
next_a = spell_a.next
next_b = spell_b.next
# Weave spell_b after spell_a
spell_a.next = spell_b
# If there's more in spell_a, weave it after spell_b
if next_a:
spell_b.next = next_a
# Move to the next nodes
spell_a = next_a
spell_b = next_b
# Return the head of the new woven list
return head
class Node:
def __init__(self, value, next=None):
self.value = value
self.next = next
def weave_spells(spell_a, spell_b):
# Base cases
if not spell_a:
return spell_b
if not spell_b:
return spell_a
# Weave the current nodes
spell_a.next = weave_spells(spell_b, spell_a.next)
# Return the new head of the list
return spell_a
- Purpose: Both the iterative and recursive solutions aim to weave two linked lists by alternating nodes from each list.
-
Functionality: Both solutions ensure that the nodes from
spell_a
andspell_b
are merged into a single linked list in an alternating pattern.
-
Approach:
- The iterative solution uses a loop to traverse both linked lists, weaving nodes together in each iteration.
- The recursive solution achieves the same by using recursive function calls to process the nodes one pair at a time.
-
Complexity:
-
Space Complexity:
- The iterative approach has a space complexity of
O(1)
because it uses a constant amount of extra space beyond the input lists. - The recursive approach has a space complexity of
O(N + M)
due to the recursion stack, whereN
andM
are the lengths of the two linked lists.
- The iterative approach has a space complexity of
-
Time Complexity: Both solutions have a time complexity of
O(N + M)
because they process each node exactly once.
-
Space Complexity:
-
Performance:
- The iterative approach is generally more efficient in terms of memory usage, as it does not involve the overhead of recursive calls. It is also less prone to stack overflow errors, especially when dealing with large lists.
- The recursive approach is often considered more elegant and easier to understand, especially for those familiar with recursive patterns, but it may be less practical for very large lists.
- Personal Preference: The choice between iterative and recursive solutions depends on the context. For larger inputs where stack depth might be a concern, the iterative solution is more efficient and practical. However, for smaller or moderate-sized inputs, the recursive approach can be more intuitive and easier to write.
- Pod Discussion: When discussing with your podmates, consider the trade-offs between readability, maintainability, and performance. The recursive approach is more concise and aligns with the recursive nature of linked lists, while the iterative approach offers better performance and avoids potential issues with recursion limits.