[LeetCode] 1693. Daily Leads and Partners

 

 

[문제]

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[MySQL]

WITH Unique_leads AS (
    SELECT date_id
        , make_name
        , COUNT(DISTINCT lead_id) AS unique_leads
    FROM DailySales
    GROUP BY date_id, make_name
) , Unique_partners AS (
    SELECT date_id
        , make_name
        , COUNT(DISTINCT partner_id) AS unique_partners
    FROM DailySales
    GROUP BY date_id, make_name
)

SELECT ul.*
     , up.unique_partners
FROM Unique_leads AS ul
    JOIN Unique_partners AS up ON ul.date_id = up.date_id
                                AND ul.make_name = up.make_name