← Snowflake Interview Insights
The binary search setup is clean once you see it but I fumbled the decision rule explanation at first.
Recognize the log as a sorted boolean array where entries before the first [Error] are non-[Error] and entries from the first [Error] onward are [Error]. Use binary search to find the first true (i.e., first [Error]) by comparing the midpoint entry: if it's [Error], search left; otherwise, search right. Explain the decision rule, prove correctness via loop invariant, and state O(log n) time and O(1) space.
Pro tip: Explicitly connect the problem to binary search on a monotonic predicate, and mention that the guarantee about the first [Error] being preceded by a [Warn] ensures the array is non-empty and the predicate is well-defined. This shows you can abstract the problem and handle edge cases.
Define a boolean function f(i) = true if log[i] is [Error], else false. The problem guarantees f is monotonic: all false then all true. The goal is to find the smallest index i where f(i) is true.
Initialize low = 0, high = n-1. While low < high, compute mid = low + (high - low) // 2. If f(mid) is true, set high = mid; else set low = mid + 1. Return low as the first [Error] index.
At each midpoint, if the entry is [Error], the first [Error] must be at or before mid, so discard the right half. If it's not [Error], the first [Error] must be after mid, so discard the left half.
Maintain the invariant that the first [Error] is always within [low, high]. Initially true. When f(mid) is true, the first [Error] ≤ mid, so high = mid preserves the invariant. When f(mid) is false, the first [Error] > mid, so low = mid + 1 preserves it. Loop ends when low == high, which must be the first [Error].
Each iteration halves the search space, so at most ⌈log₂ n⌉ comparisons. Space is O(1) extra. Mention that the guarantee about the first [Error] being preceded by a [Warn] ensures the array is non-empty and the predicate is well-defined.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
I went with BFS and built a reverse adjacency list first, which the interviewer seemed to like.
Clarify the input format (e.g., adjacency list) and output format (e.g., list of service IDs). Then, perform a reverse BFS/DFS starting from the failing service to find all upstream callers, using a visited set to handle cycles and avoid redundant work. Explain the time and space complexity, and discuss how the solution scales for large graphs.
Pro tip: Mention that in a real system, you'd also consider the direction of failure propagation and whether the graph is static or dynamic; for Snowflake's scale, an iterative BFS avoids recursion depth limits and is more cache-friendly.
Confirm the graph representation (e.g., adjacency list) and the expected output (e.g., set of service IDs). Ask if the graph is static and if the failing service is guaranteed to exist.
Decide between BFS and DFS; BFS is often preferred for shortest path but here we need all reachable nodes. Since edges point from caller to callee, we must traverse reverse edges (from callee to caller) to find upstream services.
Initialize a queue with the failing service and a visited set. While the queue is not empty, pop a node, add it to the result, and for each incoming edge (caller), if not visited, mark and enqueue.
Use the visited set to ensure each service is processed once, preventing infinite loops in cycles. This also avoids redundant work when multiple paths lead to the same service.
State time complexity O(V+E) and space O(V). Discuss edge cases: no upstream callers, self-loops, disconnected components, and very large graphs (consider iterative BFS for memory).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one tripped me up more than I expected.
Start by clarifying the problem: we need a longest simple path from a given source in a directed graph, which is NP-hard in general. Propose a DFS-based backtracking solution that explores all simple paths, tracking the current path and the best (longest) path found so far. Then discuss complexity (exponential worst-case) and assumptions (e.g., small graph, DAG, or bounded path length) that make it tractable.
Pro tip: Mention that if the graph is a DAG, we can solve it in O(V+E) using DP, but for general graphs, we must use backtracking and may need to rely on pruning or heuristics. Also, clarify whether the graph is directed and whether cycles are possible.
Confirm that the graph is directed, that we need a simple path (no repeated nodes), and that we start from a specific failing service s. Ask about graph size, whether cycles exist, and if any additional constraints (e.g., DAG) apply.
For a general graph, use DFS with backtracking to explore all simple paths from s. Maintain the current path and the longest path found so far. If the graph is a DAG, use topological sort + DP for efficiency.
Recursively visit neighbors, adding nodes to the current path and marking them as visited. When a dead end is reached, compare the current path length to the best and update if longer. Backtrack by unmarking and removing the node.
Worst-case time is O(V!) for a complete graph, but can be much better with pruning. Space is O(V) for recursion and visited set. Assumptions for tractability: small V, sparse graph, DAG, or bounded path length.
Mention pruning techniques (e.g., if current path length + remaining nodes <= best, prune), memoization for DAGs, or using heuristics for large graphs. Trade-off: exact solution vs. approximation.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.