Design a Scalable, Fault-Tolerant Distributed Order System
Company: Amazon
Role: Data Scientist
Category: Coding & Algorithms
Difficulty: Medium
Interview Round: Onsite
##### Scenario
Round covering distributed-system design and algorithm implementation.
##### Question
Design a highly available, low-latency distributed order-processing system. Explain architecture, scalability levers, and fault-tolerance mechanisms. Implement LeetCode 987 ‘Vertical Order Traversal of a Binary Tree’ and analyze its time–space complexity.
##### Hints
Explain component interactions clearly; for LC987 discuss traversal strategy and sorting keys.
Quick Answer: This question evaluates system-design and algorithmic competencies, combining distributed-systems concepts — high availability, low-latency architecture, scalability levers, and fault-tolerance mechanisms — with practical binary-tree traversal and time–space complexity analysis.
Given a binary tree represented by a level-order list where null denotes a missing child, return its vertical order traversal. Assign coordinates so the root is at row 0, column 0; the left child of a node at (r, c) is at (r+1, c-1) and the right child is at (r+1, c+1). The output should list columns from left to right. Within each column, nodes are ordered by row ascending; if multiple nodes share the same row and column, order them by value ascending.
Constraints
- 0 <= n <= 10^4 where n is the number of list entries
- Node values are integers in the range [-10^9, 10^9]
- Input is a level-order (breadth-first) representation using null for missing nodes
- Duplicates are allowed
- Return an empty list if the input is empty or the root is null
Solution
from typing import List, Optional
from collections import deque
class TreeNode:
__slots__ = ("val", "left", "right")
def __init__(self, val: int, left: 'TreeNode' = None, right: 'TreeNode' = None):
self.val = val
self.left = left
self.right = right
def _build_tree(values: List[Optional[int]]) -> Optional[TreeNode]:
n = len(values)
if n == 0 or values[0] is None:
return None
root = TreeNode(values[0])
q = deque([root])
i = 1
while q and i < n:
node = q.popleft()
if i < n:
v = values[i]
i += 1
if v is not None:
node.left = TreeNode(v)
q.append(node.left)
if i < n:
v = values[i]
i += 1
if v is not None:
node.right = TreeNode(v)
q.append(node.right)
return root
def vertical_traversal(values: List[Optional[int]]) -> List[List[int]]:
root = _build_tree(values)
if root is None:
return []
triples = [] # (col, row, val)
dq = deque([(root, 0, 0)]) # node, row, col
while dq:
node, row, col = dq.popleft()
triples.append((col, row, node.val))
if node.left is not None:
dq.append((node.left, row + 1, col - 1))
if node.right is not None:
dq.append((node.right, row + 1, col + 1))
triples.sort() # sorts by col, then row, then value (all ascending)
result: List[List[int]] = []
prev_col = None
for col, row, val in triples:
if prev_col != col:
result.append([])
prev_col = col
result[-1].append(val)
return result
Explanation
Build the tree from the level-order list, then assign each node (row, col) coordinates using BFS starting at (0,0). Collect (col, row, value) for all nodes, sort by (col asc, row asc, value asc), and group by column to form the output. This ordering enforces column ordering left-to-right, row ordering top-to-bottom, and value-based tie-breaking when row and column are the same.
Time complexity: O(n log n). Space complexity: O(n).
Hints
- Assign each node (row, col) coordinates via BFS or DFS starting with (0,0) at the root.
- Collect triples (col, row, value), sort by (col asc, row asc, value asc), then group by column.
- Building the tree from the level-order list can be done with a queue.
- Be careful with tie-breaking: same column and row must be ordered by value.