Unit 8 Session 2 Standard (Click for link to problem statements)
Unit 8 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: pearls = Pearl(4, Pearl(2, Pearl(1), Pearl(3)), Pearl(6, None, Pearl(8)))
Output: 1
Explanation: The smallest difference is between the sizes 3 and 4 or between 2 and 3, both of which have a difference of 1.
EDGE CASE
Input: pearls = Pearl(4)
Output: Infinity or some large value depending on implementation
Explanation: With only one pearl, there is no difference to calculate.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 Binary Search Tree (BST) problems, we want to consider the following approaches:
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Perform an inorder traversal of the BST to obtain the nodes in sorted order. As we traverse, keep track of the previous node and compute the difference with the current node, updating the minimum difference found.
1) Initialize `prev` to `None` and `min_diff` to infinity.
2) Define a helper function `inorder_traversal(node, prev, min_diff)` that:
- If the node is `None`, return the current `prev` and `min_diff`.
- Recursively traverse the left subtree.
- Update `min_diff` with the difference between the current node value and `prev` if `prev` is not `None`.
- Set `prev` to the current node value.
- Recursively traverse the right subtree.
3) Call `inorder_traversal(pearls, prev, min_diff)` starting from the root of the tree.
4) Return `min_diff`.⚠️ Common Mistakes
prev value as we traverse the tree.Implement the code to solve the algorithm.
class Pearl:
def __init__(self, size=0, left=None, right=None):
self.val = size
self.left = left
self.right = right
def min_diff_in_pearl_sizes(pearls):
def inorder_traversal(node, prev, min_diff):
if not node:
return prev, min_diff
# Traverse the left subtree
prev, min_diff = inorder_traversal(node.left, prev, min_diff)
# Update the minimum difference
if prev is not None:
min_diff = min(min_diff, node.val - prev)
# Update the previous node value
prev = node.val
# Traverse the right subtree
return inorder_traversal(node.right, prev, min_diff)
# Initialize prev as None and min_diff as infinity
prev = None
min_diff = float('inf')
# Perform the inorder traversal and get the result
_, min_diff = inorder_traversal(pearls, prev, min_diff)
return min_diffReview the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
- Example 1:
- Input: `pearls = Pearl(4, Pearl(2, Pearl(1), Pearl(3)), Pearl(6, None, Pearl(8)))`
- Execution:
- Perform inorder traversal: 1 -> 2 -> 3 -> 4 -> 6 -> 8.
- Differences: (2-1), (3-2), (4-3), (6-4), (8-6).
- Minimum difference: 1.
- Output: `1`- Example 2:
- Input: `pearls = Pearl(4)`
- Execution:
- With only one pearl, no difference can be calculated.
- Output: Could return infinity or another large value depending on the implementation.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.
H of the tree. In a balanced tree, H is O(log N), but in the worst case (skewed tree), it could be O(N).