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kyra-ptn edited this page Aug 25, 2024 · 3 revisions

TIP102 Unit 5 Session 1 Advanced (Click for link to problem statements)

Problem Highlights

  • 💡 Difficulty: Easy
  • Time to complete: 10 mins
  • 🛠️ Topics: Linked List, Node Swapping

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

  • Established a set (2-3) of test cases to verify their own solution later.
  • Established a set (1-2) of edge cases to verify their solution handles complexities.
  • Have fully understood the problem and have no clarifying questions.
  • Have you verified any Time/Space Constraints for this problem?
  • What should the function do if the target index is 1?
    • Return the original list since there's no previous node to swap with.
  • How should the function handle an empty list?
    • Return None.
HAPPY CASE
Input: Linked List: mario -> peach -> luigi -> daisy, target = 3
Output: mario -> luigi -> peach -> daisy
Explanation: The nodes at indices 2 and 3 are swapped.

EDGE CASE
Input: Linked List: None, target = 1
Output: None
Explanation: The list is empty.

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Linked List problems, we want to consider the following approaches:

  • Swapping nodes at given positions.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Traverse the linked list to find the nodes at indices target - 1 and target, and swap them.

1) If the target is 1, return the head as there is no previous node to swap with.
2) Initialize a pointer to traverse the list and keep track of the current index.
3) Traverse the linked list to find the nodes at indices `target - 1` and `target`.
4) Swap the player names of the nodes at `target - 1` and `target`.
5) Return the head of the modified list.

⚠️ Common Mistakes

  • Forgetting to handle the case where the target is the first node.
  • Incorrectly swapping the nodes' data instead of swapping the nodes themselves.

4: I-mplement

Implement the code to solve the algorithm.

class Node:
    def __init__(self, player, next=None):
        self.player = player
        self.next = next

# For testing
def print_linked_list(head):
    current = head
    while current:
        print(current.player, end=" -> " if current.next else "\n")
        current = current.next

def increment_rank(head, target):
    if target <= 1 or head is None or head.next is None:
        return head

    index = 1
    prev = None
    current = head
    
    while index < target:
        prev = current
        current = current.next
        index += 1

    temp = prev.player
    prev.player = current.player
    current.player = temp

    return head

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

  • Trace through your code with an input to check for the expected output.
  • Catch possible edge cases and off-by-one errors.

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work. Assume N represents the number of nodes in the linked list.

  • Time Complexity: O(N) because we need to traverse up to the target node.
  • Space Complexity: O(1) because we only need a fixed amount of extra space for pointers.
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