Google L4 | Interview Experience Aug 2026

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September 16, 2026 · 1 reads

Summary

I completed a phone screen, Googlyness round, and two onsite rounds for a Google L4 position; the process is currently in team matching.

Full Experience

Phone Screening

Given an n x m matrix, you start at (n-1,0) and must reach (n-1, m-1). Allowed moves: diagonally up, right, diagonally down.

  • Follow-up 1: There are some checkpoints - count the number of ways that include visiting all checkpoints.
  • Follow-up 2: You must visit the checkpoints in order (doesn't fully make sense since only one checkpoint can be visited per column (Since we are changing column in all moves)). Also optimize for space complexity.

Feedback: SH (Based on HR's response, as they didn't share the exact rating)

Googlyness Round

  • How would you handle a conflict where someone is taking credit for your work?
  • Standard questions on juggling multiple tasks/priorities.

Feedback: SH (Based on HR's response, as they didn't share the exact rating)

Onsite Round 1

Given an undirected, unweighted graph, Alice is at one node and needs to reach a destination node. Find the shortest path.

  • Follow-up 1: Find all nodes that are part of some shortest path.
  • Follow-up 2: Bob is at another node. Alice picks one shortest path and starts moving; Bob starts moving at the same time too. Determine whether Alice can reach the destination without being caught by Bob.

Feedback: SH (Based on HR's response, as they didn't share the exact rating)

Onsite Round 2

Given a list of log entries:

timestamp, userA connects to userB
timestamp, userB connects to userC
...

along with the total number of users, find the timestamp at which all users become connected to each other.

  • Follow-up 1: Find the timestamp after which the maximum number of connections between any two nodes is <= k (fixed), and all users are connected. (Similar to the original problem.)
  • Follow-up 2: Logs can now also include:
timestamp, userB unfriends userC

Find the timestamp when all users are connected. (I wasn't able to give a fully optimal solution - this seems related to dynamic connectivity / DSU with rollback.)

Feedback: H (Based on HR's response, as they didn't share the exact rating)

Current Status

In team matching.

Interview Questions (5)

1.

Matrix Path with Diagonal Moves

Data Structures & Algorithms

Given an n x m matrix, start at cell (n-1,0) and reach cell (n-1,m-1). Allowed moves are:

  • Diagonally up (to (i-1, j+1))
  • Right (to (i, j+1))
  • Diagonally down (to (i+1, j+1))

Follow-up 1: There are some checkpoints; count the number of ways that include visiting all checkpoints.

Follow-up 2: You must visit the checkpoints in order (only one checkpoint can be visited per column). Optimize for space complexity.

2.

Handling Credit Conflict

Behavioral

How would you handle a conflict where someone is taking credit for your work?

3.

Prioritization and Multitasking

Behavioral

Standard questions on juggling multiple tasks and priorities.

4.

Shortest Path and Nodes on All Shortest Paths

Data Structures & Algorithms

Given an undirected, unweighted graph, Alice starts at a source node and needs to reach a destination node. Find the shortest path.

Follow-up 1: Find all nodes that are part of some shortest path.

Follow-up 2: Bob starts at another node at the same time. Alice picks one shortest path and moves; Bob also moves. Determine whether Alice can reach the destination without being caught by Bob.

5.

Earliest Time All Users Connected

Data Structures & Algorithms

Given a list of log entries of the form timestamp, userA connects to userB and the total number of users, find the timestamp at which all users become connected to each other.

Follow-up 1: Find the timestamp after which the maximum number of connections between any two nodes is <= k (fixed), and all users are connected.

Follow-up 2: Logs may also include timestamp, userB unfriends userC. Find the timestamp when all users are connected again, considering dynamic connectivity (similar to DSU with rollback).

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