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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Business case.
Give an algorithm to find the kth largest element in an unsorted vector of integers.
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Given a string of Rs and Gs, design an algorithm to produce a string with Rs in the front and Gs after that. The number of flips from Rs to Gs or otherwise should be minimum. The number of Rs and Gs in the end need not be same as that in the beginning, however the length of the entire string should be the same.
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