Merging Missions II - codepath/compsci_guides GitHub Wiki

TIP102 Unit 7 Session 1 Standard (Click for link to problem statements)

Problem 9: Merging Missions II

Below is an iterative solution to the merge_missions() function from the previous problem. Compare your recursive solution to the iterative solution below.

Iterative Solution

def merge_missions_iterative(mission1, mission2):
    temp = Node()  # Temporary node to simplify the merging process
    tail = temp

    while mission1 and mission2:
        if mission1.value < mission2.value:
            tail.next = mission1
            mission1 = mission1.next
        else:
            tail.next = mission2
            mission2 = mission2.next
        tail = tail.next

    # Attach the remaining nodes, if any
    if mission1:
        tail.next = mission1
    elif mission2:
        tail.next = mission2

    return temp.next  # Return the head of the merged linked list

Recursive Solution

def merge_missions(mission1, mission2):
    if not mission1:
        return mission2
    if not mission2:
        return mission1

    if mission1.val < mission2.val:
        mission1.next = merge_missions(mission1.next, mission2)
        return mission1
    else:
        mission2.next = merge_missions(mission1, mission2.next)
        return mission2

Comparison

Similarities:

1. Purpose: Both the iterative and recursive solutions aim to merge two sorted linked lists into a single sorted linked list.
2. Functionality: Both approaches involve comparing the values of the nodes at the heads of the two lists and adding the smaller node to the merged list.
3. Final Output: Both methods yield a new linked list with nodes ordered according to their values, maintaining the sorted order.

Differences:

1. Approach:

   • The iterative solution uses a loop to traverse both linked lists, adjusting pointers iteratively until one of the lists is exhausted.
   • The recursive solution calls itself to handle the next node in the list, building the merged list through successive recursive calls.

2. 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, where N and M are the lengths of the two linked lists.
   • Time Complexity: Both solutions have a time complexity of O(N + M) because each solution must process all nodes from both lists.

3. Performance:

   • The iterative approach may be more efficient in practice due to its constant space complexity. This makes it preferable for very large lists where the depth of recursion could become a concern.
   • The recursive approach, while elegant and easier to understand conceptually, could lead to stack overflow errors if the lists are very long, especially in environments with limited stack size.

Preference Discussion:

• Personal Preference: The choice between iterative and recursive solutions depends on the context. For smaller lists or in environments where recursion depth isn't an issue, the recursive approach is elegant and straightforward. However, for larger lists or when concerned about memory usage, the iterative solution is generally preferred due to its O(1) space complexity and avoidance of potential recursion depth issues.

• Pod Discussion: In a group discussion, consider the trade-offs between readability and efficiency. While the recursive approach is often easier to understand and implement quickly, the iterative approach offers better performance and robustness for large inputs. Discussing these aspects will help determine which approach is more suitable for your specific use case.