TIP102 Unit 9 Session 2 Advanced (Click for link to problem statements)
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?
HAPPY CASE
Input:
room_nums = [5, 1, 2, 3, None, 6, 4]
start_value = 3
dest_value = 6
Output:
"UURL"
Explanation:
The shortest path is: 3 -> 1 -> 5 -> 2 -> 6
EDGE CASE
Input:
room_nums = [2, 1]
start_value = 2
dest_value = 1
Output:
"L"
Explanation:
The shortest path is: 2 -> 1Match 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 Tree Pathfinding problems, we want to consider the following approaches:
Plan the solution with appropriate visualizations and pseudocode.
General Idea:
1) Define a helper function `find_path(root, target, path)`:
- Base Case: If `root` is `None`, return `False`.
- If `root.val` equals the target, return `True`.
- Recur for the left subtree by appending 'L' to the path. If successful, return `True`.
- Recur for the right subtree by appending 'R' to the path. If successful, return `True`.
- If the target is not found, backtrack by popping the last direction and return `False`.
2) In the main function `get_directions(root, start_value, dest_value)`:
- Use `find_path` to determine the path from the root to the start node and the path from the root to the destination node.
- Identify the LCA by comparing the paths until they diverge.
- Calculate the directions by adding 'U' for each step back to the LCA and then appending the directions to the destination node.
- Return the resulting direction string.
⚠️ Common Mistakes
Implement the code to solve the algorithm.
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def find_path(root, target, path):
if not root:
return False
if root.val == target:
return True
# Try to find the target in the left subtree
path.append('L')
if find_path(root.left, target, path):
return True
path.pop() # Backtrack if not found
# Try to find the target in the right subtree
path.append('R')
if find_path(root.right, target, path):
return True
path.pop() # Backtrack if not found
return False
def get_directions(root, start_value, dest_value):
start_path = []
dest_path = []
# Find paths from the root to the start and destination nodes
find_path(root, start_value, start_path)
find_path(root, dest_value, dest_path)
# Find the lowest common ancestor (LCA)
i = 0
while i < len(start_path) and i < len(dest_path) and start_path[i] == dest_path[i]:
i += 1
# 'U' moves to go up from start to LCA
directions = 'U' * (len(start_path) - i)
# Append the remaining path to the destination
directions += ''.join(dest_path[i:])
return directions
# Example Usage:
room_nums = [5, 1, 2, 3, None, 6, 4]
hotel1 = build_tree(room_nums)
print(get_directions(hotel1, 3, 6)) # Output: "UURL"Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
- Example 1:
- Input:
`room_nums = [5, 1, 2, 3, None, 6, 4]`
`start_value = 3`
`dest_value = 6`
- Execution:
- Find the paths to 3 and 6.
- Identify the LCA (node 5).
- Construct directions: "UURL".
- Output:
"UURL"- Example 2:
- Input:
`room_nums = [2, 1]`
`start_value = 2`
`dest_value = 1`
- Execution:
- Find the paths to 2 and 1.
- Identify the LCA (node 2).
- Construct directions: "L".
- Output:
"L"Evaluate the performance of your algorithm and state any strong/weak or future potential work.
O(N) where N is the number of nodes in the tree.
O(H) where H is the height of the tree.
