SELECT

    customer_id

FROM orders

GROUP BY customer_id

HAVING COUNT(DISTINCT product_id) = (

    SELECT COUNT(*)

    FROM products

);

📌 Question 100: Rank Customers by Lifetime Revenue 

Table: orders (customer_id, amount)

WITH customer_revenue AS (

    SELECT

        customer_id,

        SUM(amount) AS lifetime_revenue

    FROM orders

    GROUP BY customer_id

)

SELECT

    customer_id,

    lifetime_revenue,

    DENSE_RANK() OVER (

        ORDER BY lifetime_revenue DESC

    ) AS revenue_rank

FROM customer_revenue;

💡 Pro Tip: Practice these regularly, understand the business logic behind each solution, and you'll be well-prepared for SQL interviews at product companies, startups, fintech firms, and MNCs.

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