SQL и Анализ данных: “Microsoft SQL Interview Question — разбор №2: Early Responders Задача:…”

Microsoft SQL Interview Question — разбор №2: Early Responders

Задача: найти топ-3 пользователей, которые чаще всего отвечали на сообщения в течение 5 минут в августе 2022.
Нужно вывести их sender_id и число «быстрых ответов».

Данные: таблица messages(message_id, sender_id, receiver_id, content, sent_date).

Подход:
— Фильтруем август полуинтервалом [2022-08-01, 2022-09-01).
— Группируем переписку по паре пользователей (conv_id = min(sender,receiver)–max(sender,receiver)).
— Используем LAG по conv_id, чтобы взять предыдущее сообщение в диалоге.
— Быстрый ответ — когда направление сменилось и прошло ≤ 5 минут.

Решение (T-SQL):


WITH aug AS (
  SELECT message_id, sender_id, receiver_id, sent_date
  FROM   messages
  WHERE  sent_date >= '2022-08-01'
     AND sent_date <  '2022-09-01'
),
conv AS (
  SELECT *,
         CONCAT(
           IIF(sender_id < receiver_id, sender_id, receiver_id), '-',
           IIF(sender_id < receiver_id, receiver_id, sender_id)
         ) AS conv_id
  FROM aug
),
seq AS (
  SELECT *,
         LAG(sender_id) OVER (PARTITION BY conv_id ORDER BY sent_date, message_id) AS prev_sender,
         LAG(sent_date) OVER (PARTITION BY conv_id ORDER BY sent_date, message_id) AS prev_time
  FROM conv
),
fast AS (
  SELECT sender_id
  FROM   seq
  WHERE  prev_sender IS NOT NULL
     AND sender_id <> prev_sender
     AND DATEDIFF(minute, prev_time, sent_date) <= 5
)
SELECT TOP (3)
       sender_id,
       COUNT(*) AS fast_reply_count
FROM   fast
GROUP BY sender_id
ORDER BY COUNT(*) DESC, sender_id;

Вариант с учётом ничьих (возьмёт всех на 1-3 местах):


WITH aug AS (
  SELECT message_id, sender_id, receiver_id, sent_date
  FROM   messages
  WHERE  sent_date >= '2022-08-01'
     AND sent_date <  '2022-09-01'
),
conv AS (
  SELECT *,
         CONCAT(
           IIF(sender_id < receiver_id, sender_id, receiver_id), '-',
           IIF(sender_id < receiver_id, receiver_id, sender_id)
         ) AS conv_id
  FROM aug
),
seq AS (
  SELECT *,
         LAG(sender_id) OVER (PARTITION BY conv_id ORDER BY sent_date, message_id) AS prev_sender,
         LAG(sent_date) OVER (PARTITION BY conv_id ORDER BY sent_date, message_id) AS prev_time
  FROM conv
),
fast AS (
  SELECT sender_id
  FROM   seq
  WHERE  prev_sender IS NOT NULL
     AND sender_id <> prev_sender
     AND DATEDIFF(minute, prev_time, sent_date) <= 5
),
agg AS (
  SELECT sender_id, COUNT(*) AS fast_reply_count
  FROM   fast
  GROUP BY sender_id
),
ranked AS (
  SELECT sender_id, fast_reply_count,
         DENSE_RANK() OVER (ORDER BY fast_reply_count DESC) AS rnk
  FROM   agg
)
SELECT sender_id, fast_reply_count
FROM   ranked
WHERE  rnk <= 3
ORDER BY fast_reply_count DESC, sender_id;

Почему так:
— Диапазон дат без функций сохраняет использование индекса по sent_date.
— LAG по conv_id гарантирует, что сравниваем соседние сообщения в одном диалоге.
— Проверяем смену направления (sender_id ≠ prev_sender) и порог по времени (≤ 5 минут).