Day 83 — FULL MOCK: Google
Coding problem
| Problem | FULL MOCK: Google |
| LeetCode ID(s) | — |
| Difficulty | Mixed |
| Pattern | Mock |
| Company tags | |
| Suggested time | 45m |
Solution outline (coding)
- Same structure as Meta mock: verbalize thought process continuously.
- Prioritize correctness on the first problem before speed on the second.
- Note Google-flavored emphasis: scale, testing, and clear abstractions.
Time complexity: Varies.
Space complexity: Varies.
Show Python solution
class ReviewDay:
"""Practice / review: FULL MOCK: Google."""
def practice_plan(self):
return [
"Pick 2–3 problems from this phase; re-solve timed without notes.",
"For each: pattern name, time/space complexity, one alternative approach.",
]
# Input: (your choice of problems from this week or phase)
# Output: a short list of gaps to drill before the next sessionSQL interview practice
1. Interview question
Companies / track: Google
Google: SQL screens usually assume BigQuery, Ads / Search / YouTube-style fact tables, and talking through bytes processed and partition pruning.
What you are asked to write (SQL prompt):
Frame this as metrics work for **Google**-scale surfaces (ads, product, or engagement — as the tables suggest).
Google-style mock: one BigQuery query for tree/graph analysis and one for product metrics, recording time/bytes and reasoning commentary.
Tables implied by the prompt:
- Infer schemas from the prompt and state them before coding.
Engine: BigQuery — use its date, array, and approximate functions as documented.
2. Solution outline
- Clarify out loud: result grain (one row per what?), join keys, time zone, and any
ORDER BY/LIMIT/ tie-breakers. - Map Mock to SQL: say the relational equivalent (e.g. hash map →
GROUP BY+ key; two pointers → ordered window + filter). - Cost: selective columns, partition pruning, avoid
SELECT *when tables are huge. - Structure: CTEs (
WITH) — one step per CTE; validate on a tiny slice (counts, nulls, duplicates).
Show SQL solution (BigQuery)
Main query
SELECT 'tree' AS section, COUNT(*) FROM tree_stats
UNION ALL SELECT 'metrics', COUNT(*) FROM product_metrics;