Google·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Google SWE interview with a tree propagation problem that looks deceptively like a BFS question but really isn't. Took me a while to even figure out what they were actually asking.

Questions Asked (1)

You have a large n-ary tree. The root can pass a message to one child per iteration, and in each subsequent iteration every already-informed node can pass the message to one of its children simultaneously. What is the minimum number of iterations needed to inform every node in the tree?

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

Spent the first few minutes thinking it was just a level-order traversal thing and almost said the answer was the height of the tree.

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AI HintsAI Generated

Suggested Approach

Model the problem as a bottom-up dynamic programming problem where for each node, you compute the minimum iterations required to inform its entire subtree. The key insight is that a node can only start informing its children after it receives the message, and children with larger subtree requirements should be informed first to minimize overall time.

Pro tip: After deriving the algorithm, mention that this is a classic problem solvable with a greedy strategy: sort children by their required iterations in descending order and assign them to consecutive time slots. This shows you understand both the DP and the greedy optimization.

1. Understand the process

Clarify that the root starts at iteration 0, and in each iteration, every informed node can inform one child. The goal is to minimize the total iterations until all nodes are informed.

2. Define subproblem

For each node, define f(node) as the minimum number of iterations needed to inform all nodes in its subtree, assuming the node is informed at time 0. The answer for the root is f(root).

3. Derive recurrence

For a node with children c1, c2, ..., ck, if we inform child ci at time ti (where ti are distinct positive integers starting from 1), then the total time is max_i (ti + f(ci)). To minimize this, sort f(ci) in descending order and assign ti = 1, 2, ..., k respectively.

4. Compute bottom-up

Perform a post-order traversal to compute f for each node. For each node, collect f values of children, sort them descending, and compute f(node) = max_{i=1..k} (i + f(ci)).

5. Analyze complexity

The time complexity is O(N log N) due to sorting at each node, where N is the number of nodes. Space complexity is O(N) for recursion stack and storage.

Key Points to Mention

The problem is equivalent to scheduling children with different processing times on a single machine to minimize makespan, where each child's processing time is its subtree's required iterations.
Greedy strategy: inform children with larger f(ci) first to avoid them becoming the bottleneck.
The recurrence f(node) = max_{i} (i + f(ci)) after sorting f(ci) descending.
Base case: leaf node has f(leaf) = 0.
Time complexity O(N log N) and space O(N).
This approach is optimal because any schedule that informs a child with smaller f later cannot improve the maximum.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.