← Google Interview Insights

Google·Software Engineer·Technical Phone Screen·Junior

JuniorRejected
Apr 2025Remote

Summary

Interviewed for a software engineer role at Google with about two weeks to prep after a recruiter reached out in late April. The phone screen had one algorithmic question and I fumbled it badly enough that they passed on me. Twelve month cooldown now, which stings.

Questions Asked (1)

Q1

Given a binary tree with weighted edges, find the minimum cost to disconnect the root from all leaf nodes by cutting edges along root-to-leaf paths.

Algorithms & Data Structures
Author's notes

My first instinct was memoization and I pitched an N-squared solution.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Use a post-order DFS to compute the minimum cost to disconnect each subtree from its leaves. At each node, compare the cost of cutting the edge to its parent versus cutting edges within its subtree, and return the minimum. The root's result is the sum of the minimum costs of its children.

Pro tip: Clarify edge cases upfront: what if the root is a leaf? What if edge weights are negative? Handling these shows thoroughness and prevents incorrect assumptions.

1. Clarify problem constraints and edge cases

Ask about tree size, edge weight ranges (negative?), and whether the root can be a leaf. Confirm that cutting an edge disconnects the subtree below it.

2. Define recursive subproblem

For a node, define f(node) as the minimum cost to disconnect the subtree rooted at node from all its leaves, assuming the edge to its parent is not cut.

3. Derive recurrence relation

For a leaf, f(leaf) = 0. For an internal node, f(node) = min(edge_weight_to_parent, sum of f(child) for all children). This captures the choice of cutting the parent edge or cutting within the subtree.

4. Implement post-order DFS

Traverse the tree recursively, computing f for each node bottom-up. At the root, the answer is the sum of f(child) for all children (since the root has no parent edge).

5. Analyze complexity and test

Time complexity is O(n) as each node is visited once. Space is O(h) for recursion stack. Test with simple trees, skewed trees, and negative weights if allowed.

Key Points to Mention

  • Post-order traversal (DFS) to process children before parent
  • Dynamic programming recurrence: f(node) = min(edge_to_parent, sum of f(children))
  • Handling leaf nodes: f(leaf) = 0
  • Root case: sum of f(children) since no parent edge
  • Time and space complexity analysis: O(n) time, O(h) space
  • Edge cases: root as leaf, negative edge weights, single-node tree

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