We Took On a Tough SQL Interview Question! | Data Scientist Prep (Intermediate Level)

[@iqjayfeng]

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[@tahakhalid324]

Hi jay here is my solution, with next_positions as ( select id, start_date, end_date, user_id, position_name, lead(position_name,1,null) over(partition by user_id order by start_date) next_position from user_experiences ) , flags as ( select user_id, position_name, next_position, case when position_name = 'Data Analyst' and next_position = 'Data Scientist' then 1 else null end as flag_for_analyst from next_positions ) select (count(flag_for_analyst)/count(distinct user_id)) as percentage from flags

[@iqjayfeng]

Thanks for sharing this!