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Day 5: Solving Walmart SQL Interview Questions | 100 Days Challenge

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Welcome to Day 5 of my 100 Days Challenge! Today, we're diving into Walmart SQL interview questions to help you ace your next tech interview. Join me as I walk through each problem step-by-step, providing detailed explanations and practical tips. Don't forget to like, subscribe, and hit the bell icon to stay updated with daily challenges!
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11 июл 2024

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Комментарии : 6

@anirbanbiswas7624 Месяц назад

with cte as(select *,sum(sales_amount) over(partition by store_name rows between unbounded preceding and current row)as running_sales, lag(sales_amount,1,sales_amount) over(partition by store_name)as prev_month_sales from sales) select *,round((sales_amount-prev_month_sales)/prev_month_sales*100,2) as growth_rate from cte i think null value on the 1st row or in the other cases for the first month we may assume the initial value as the value instead of null & to get that i have used--- lag(sales_amount,1,sales_amount)

@Chathur732 Месяц назад

select *, case when lagging is not null then round(((sales_amount - lagging)/lagging) * 100,2) else 0 end as growth_ratio from( select * ,sum(sales_amount) over(partition by store_name order by sale_date) as sum, lag(sales_amount) over(partition by store_name order by store_name) as lagging from sales) as Subquery

@devarajululanka6427 27 дней назад

SOLUTION IN SQL SERVER : with cte as ( select store_name,DATEPART (month,sale_date) as sale_month,sum(sales_amount) as total_sale_amount from sales group by store_name,DATEPART (month,sale_date) ), cte2 as (select *, sum(total_sale_amount) over (partition by store_name order by total_sale_amount ) as running_sale_amount, lag(total_sale_amount,1) over (partition by store_name order by total_sale_amount ) as previous_sale_amount from cte ) select *, (total_sale_amount - previous_sale_amount)/previous_sale_amount*100 as diff from cte2;