1. Exam: Using classses of name, address, which build a person class, buiuld those classses and build a graph, connecting the person nodes - using name and address(if they are the same they are related). Write a function to find the minimum level between two people. 2.Interview - background and then a question. Given a class, print out its attribute names and values. Aftwerwards questions about it, about infinitive loops, complexity, other data structures and how to use it as an api.
# n=3 * * * @ * @ * @ * * * #n=5 * * * @ * * * @ * * * @ * * * @ * * * @ * * * somewhat similar to these !
What is your current job?
What good ideas would you bring to the Training Department if extended the job?
1000bulbs to be shipped. Should all be tested or not
What Passion can you deliver over the phone?
During panel they asked what were the details of the software outage from 2013.
how to find the second smallest number out of iven n integers
Most of the technical questions focused on power supply design. For this position this is a one of the more difficult tasks so they want to know how switching regulators work and different trade offs involved.
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)
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