Mid standard

• Defend your approach and an alternative: Start with active accounts and keep session filters in the join condition.
• Connect the mechanism to an edge case: Use an inclusive October start and an exclusive November start.
• State evidence for your conclusion: Check Orchard 2, Ridge 0 and Vale 0, with exactly one row per active account.

Lessons to prepare

Mid standard

• Defend your approach and an alternative: Order by updated_at descending and event_id descending within each order.
• Connect the mechanism to an edge case: Keep all orders and substitute pending when no eligible event exists.
• State evidence for your conclusion: Check both pending orders and the tied event, then reconcile one output row per order.

Lessons to prepare

Mid standard

• Connect the assumptions, method and result: One row represents one session because repeated event rows are delivery copies. I need a start and a completion in the same session, with completion time at or after the start. Sessions without that ordered pair remain false.
• Test the consequential edge case: I check that sessions 1 and 4 are true, while 2, 3 and 5 are false. The same-time pair qualifies because the contract allows at or after; an earlier completion cannot qualify merely because a later start exists.
• Explain an owned validation or operating check: I reconcile the output count and uniqueness to the sessions table because missing and duplicate session rows are both defects. I check event identities and timestamp meaning before choosing an index, and I monitor unknown event types.

Lessons to prepare

Mid standard

• I use two grouped CTEs and LEFT JOIN both to orders because an order without either child still needs an output row with zero totals.
• Order 1 has billed 100 and paid 100, order 2 has billed 30 and paid zero, and order 3 has both zero because missing child rows remain visible.
• I compare each output total with its independent source sum because matching billed and paid balances can both be inflated by the same fan-out.

Lessons to prepare

Mid standard

• I use the calendar as a LEFT JOIN spine and aggregate each day before a window sum because a window over raw ledger entries would return several balances for one day.
• The balances are 80, 80, 50 and 50 because October 1 nets to 80 and the October 3 withdrawal reduces the carried balance by 30.
• I compare the last balance with the total signed ledger change because the opening balance is zero, and I check output dates against the supplied calendar.

Lessons to prepare

Mid standard

• I create one retained flag per customer using EXISTS because repeated activity must not multiply the numerator or denominator when I aggregate each cohort.
• August has two customers and one retained customer for a rate of 0.5, while September has two and zero for 0.0 because activity in the wrong month doesn't qualify.
• I compare cohort size with the customer source and require retained customers to be no larger than that size because duplicate activity can otherwise inflate retention.

Lessons to prepare

Mid standard

• I subtract five hours before casting the timestamp to a date and LEFT JOIN to report days because UTC dates split this local day at the wrong boundary.
• October 1 has two visits and October 2 has zero because the exact 05:00 UTC boundary starts a new local date while the earlier instant stays on the prior day.
• I would use the named regional timezone rather than a fixed offset because daylight saving changes can make local days shorter or longer than twenty-four hours.

Lessons to prepare

Mid standard

• I rank every delivery before filtering deletion flags because dropping tombstones first would revive an obsolete live transaction.
• Transaction 1 returns 35 and transaction 3 returns zero, while transaction 2 is absent because its latest version is a deletion rather than a missing amount.
• I check uniqueness of the ingestion identity and the source version contract because equal versions with an undefined tie-breaker cannot produce reproducible current state.

Lessons to prepare

Mid standard

• I aggregate completed jobs at worker grain before ranking and LEFT JOIN from workers because an employee with no October jobs still has a zero total.
• Workers 2 and 3 both have seven jobs and share second position because worker 1 has ten; West worker 5 remains first at zero.
• I compare worker totals with October source jobs and check team membership because a transfer can change the partition even when the arithmetic still balances.

Lessons to prepare

Mid standard

• I rank by timestamp and order_id, then join positions one and two because the contract concerns the first two completed orders rather than a later qualifying pair.
• Customer 1 qualifies at seven days and customer 4 at zero days, while customer 3 misses at eight because the upper boundary includes exactly seven.
• I check customers with one completed order, tied timestamps and a canceled first order because each can expose a different ordering or filter error.

Lessons to prepare