Business Days Between — Solution & ExplanationProblems / Business Days Between / EditorialApproachInstead of iterating over each date range, count weekdays from a single fixed anchor and subtract.

Anchor on a Monday. DATE '2000-01-03' is a Monday earlier than every date in the table. For any date d, let days = datediff('day', anchor, d) + 1 — the number of calendar days in [anchor, d], a span that *starts on a Monday*.

Count weekdays in that span. Full weeks contribute exactly 5 weekdays each: 5 * (days // 7). The leftover days % 7 days begin on a Monday, so they are all weekdays until they run past Friday: least(days % 7, 5). So f(d) = 5 * (days // 7) + least(days % 7, 5) is the number of weekdays in [anchor, d].

Subtract. Weekdays in (order_date, delivery_date] = f(delivery_date) - f(order_date). The order date drops out of the count (exclusive) and the delivery date stays in (inclusive) — exactly the required semantics, with no CASE ladder over start/end weekdays.

Sanity checks: Friday order, Monday delivery -> 1 (only the Monday). Saturday order, Sunday delivery -> 0. Monday order, same-week Friday delivery -> 4 (Tue, Wed, Thu, Fri).Key InsightThis is the *prefix-sum trick* applied to calendars: any "count X between a and b" problem becomes trivial if you can compute "count X from a fixed origin up to d" in closed form. Anchoring the origin on a Monday is what makes the closed form simple — the days % 7 remainder always starts the week, so its weekday count is just least(remainder, 5). The same idea powers np.busday_count in NumPy, which is how the expected output is verified independently.Hints

Counting weekdays between two dates directly is fiddly. Counting weekdays from one FIXED reference point up to each date is easy — and the answer is just the difference of two such counts.

Pick a reference Monday earlier than all data (DATE '2000-01-03'). If a date is `days` days after it (counting the date itself, so days = datediff + 1), the full weeks contribute 5 * (days // 7) weekdays.

The leftover days % 7 days start on a Monday (because the reference is a Monday), so at most 5 of them are weekdays: least(days % 7, 5). So f(d) = 5 * (days // 7) + least(days % 7, 5).

business_days in (order_date, delivery_date] = f(delivery_date) - f(order_date). The subtraction naturally excludes the order date and includes the delivery date.

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