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Python 1 #1.returns the number of times a given character occurs in the given string s1='missisipi' #print(s1.find('s')) res=[] for i in range(len(s1)): #print(s1[i]) if s1[i]=='s': res.append('s') print(len(res)) #2.[1,None,1,2,None} --> [1,1,1,2,2] arr=[None,1,2,None] new_l=[] for i in range(0,len(arr)): if arr[i] != None: new_l.append(arr[i]) else: new_l.append(arr[i-1]) print(new_l) #2. (python) Given two sentences, construct an array that has the words that appear in one sentence and not the other. A = "Geeks for Geeks" B = "Learning from Geeks for Geeks" d={} for w in A.split(): if w in d: d[w]=d.get(w,0)+1 else: d[w]=1 for w in B.split(): if w in d: d[w]=d.get(w,0)+1 else: d[w]=1 unmatchedW=[w for w in d if d[w]==1] print (unmatchedW) 3. d = {"a": 4, "c": 3, "b": 12} [(k, v) for k, v in sorted(d.items(), key=lambda x: x[1], reverse=True)] #[('b', 12), ('a', 4), ('c', 3)] SQL # # sales # products # +------------------+---------+ +---------------------+---------+ # | product_id | INTEGER |>--------| product_id | INTEGER | # | store_id | INTEGER | +---<| product_class_id | INTEGER | # | customer_id | INTEGER | | | brand_name | VARCHAR | # +---<| promotion_id | INTEGER | | | product_name | VARCHAR | # | | store_sales | DECIMAL | | | is_low_fat_flg | TINYINT | # | | store_cost | DECIMAL | | | is_recyclable_flg |… Show More 1. find top 5 sales products having promotions Select Sum(s.store_sales), brand_name, count(p.product_id) from products p inner join sales s p.product_id = s.product_id where promotion_id is not null group by brand_name having count(p.product_id) =1 /* single-channel media type */ order by 1 desc limit 5 2. # -- % Of sales that had a valid promotion, the VP of marketing # -- wants to know what % of transactions occur on either # -- the very first day or the very last day of a promotion campaign. select sum(case when valid_promotion = 1 then 1 else 0 end)/count(*) * 100 as percentage from sales where day = First_day(date) or day = last_day(date) or select sum(case when transaction_date = (select min(transaction_date) from sales) then 1 else 0)/count(*) as first_day_sales, sum(case when transaction_date = (select max(transaction_date) from sales) then 1 else 0)/count(*) as last_day_sales from sales or select avg(transaction_date in (p.start_date,p.end_date))*100 as first_last_pct from sales s join promotions p using(promotion_id)
avatar

Data Engineer

Interviewed at Meta

3.5
Aug 25, 2020

Python 1 #1.returns the number of times a given character occurs in the given string s1='missisipi' #print(s1.find('s')) res=[] for i in range(len(s1)): #print(s1[i]) if s1[i]=='s': res.append('s') print(len(res)) #2.[1,None,1,2,None} --> [1,1,1,2,2] arr=[None,1,2,None] new_l=[] for i in range(0,len(arr)): if arr[i] != None: new_l.append(arr[i]) else: new_l.append(arr[i-1]) print(new_l) #2. (python) Given two sentences, construct an array that has the words that appear in one sentence and not the other. A = "Geeks for Geeks" B = "Learning from Geeks for Geeks" d={} for w in A.split(): if w in d: d[w]=d.get(w,0)+1 else: d[w]=1 for w in B.split(): if w in d: d[w]=d.get(w,0)+1 else: d[w]=1 unmatchedW=[w for w in d if d[w]==1] print (unmatchedW) 3. d = {"a": 4, "c": 3, "b": 12} [(k, v) for k, v in sorted(d.items(), key=lambda x: x[1], reverse=True)] #[('b', 12), ('a', 4), ('c', 3)] SQL # # sales # products # +------------------+---------+ +---------------------+---------+ # | product_id | INTEGER |>--------| product_id | INTEGER | # | store_id | INTEGER | +---<| product_class_id | INTEGER | # | customer_id | INTEGER | | | brand_name | VARCHAR | # +---<| promotion_id | INTEGER | | | product_name | VARCHAR | # | | store_sales | DECIMAL | | | is_low_fat_flg | TINYINT | # | | store_cost | DECIMAL | | | is_recyclable_flg |… Show More 1. find top 5 sales products having promotions Select Sum(s.store_sales), brand_name, count(p.product_id) from products p inner join sales s p.product_id = s.product_id where promotion_id is not null group by brand_name having count(p.product_id) =1 /* single-channel media type */ order by 1 desc limit 5 2. # -- % Of sales that had a valid promotion, the VP of marketing # -- wants to know what % of transactions occur on either # -- the very first day or the very last day of a promotion campaign. select sum(case when valid_promotion = 1 then 1 else 0 end)/count(*) * 100 as percentage from sales where day = First_day(date) or day = last_day(date) or select sum(case when transaction_date = (select min(transaction_date) from sales) then 1 else 0)/count(*) as first_day_sales, sum(case when transaction_date = (select max(transaction_date) from sales) then 1 else 0)/count(*) as last_day_sales from sales or select avg(transaction_date in (p.start_date,p.end_date))*100 as first_last_pct from sales s join promotions p using(promotion_id)

They gave me a problem in which I had to find the next largest number of a given integer, where the next largest number can consist only of a combination of 3s and 5s. ex: number: 23 output: 33, number: 129 output: 333, number: 432 output: 533
avatar

QA Engineer

Interviewed at Hike

3.7
Apr 20, 2015

They gave me a problem in which I had to find the next largest number of a given integer, where the next largest number can consist only of a combination of 3s and 5s. ex: number: 23 output: 33, number: 129 output: 333, number: 432 output: 533

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