Real Lyft interview questions
Reported questions from this company's loops, tagged by domain, round, and level.
SQLonsite sql· L42024
Given a drivers table (driver_id, name) and a rides table (ride_id, rider_id, driver_id, start_time, end_time, pick_up, drop_off, cost), write a query to calculate for each driver: total number of rides, average cost per ride, and number of rides costing over $20
Schema: drivers(driver_id INT, name VARCHAR), rides(ride_id INT, rider_id INT, driver_id INT, start_time TIMESTAMP, end_time TIMESTAMP, pick_up VARCHAR, drop_off VARCHAR, cost DECIMAL). Expected approach: JOIN drivers to rides ON driver_id, GROUP BY driver_id and name, compute COUNT(*) AS total_rides, AVG(cost) AS average_cost, and SUM(CASE WHEN cost > 20 THEN 1 ELSE 0 END) AS rides_over_20. DataLemur explicitly labels this as a Data Engineer question at Lyft.
Pythonphone screen python· L32025
Find the nth missing number in a sorted list without duplicates
Given a sorted list of unique integers and an integer n, return the nth number not present in the list. Iterate through the sorted list tracking gaps between consecutive elements, counting missing numbers until the nth is found. Tests loop control, arithmetic, and edge case handling.
SQLonsite sql· L52024
Given an orders table, find all customer_ids who placed their first-ever order on each calendar day (new customers per day: customer appears that day but not in any prior day)
SQLonsite sql· L42024
Given a table of customer orders, find each day's new customer_ids — customers who placed their first-ever order on that day.
Schema: orders(order_id, customer_id, order_date, ...). Approach: find each customer's MIN(order_date) as their first order date, then group by that date to count new customers per day. Alternative approach: LEFT JOIN customers against all prior orders and filter for NULL matches. Requires understanding of MIN aggregate with GROUP BY and self-join or window function alternatives. Lyft DE phone screen or onsite SQL question.
SQLonsite sql· L42024
Find customers who made purchases on two consecutive calendar days.
Schema: purchases(customer_id, purchase_date, ...). Approach 1: Self-join on customer_id where DATEDIFF(p2.purchase_date, p1.purchase_date) = 1. Approach 2: use LAG() window function to get previous purchase date per customer, then filter for date difference of 1 day. Need to handle duplicates (multiple purchases per day) by first deduplicating on (customer_id, DATE(purchase_date)). Edge cases: same-day multiple purchases, timezone differences. Lyft DE interview.
