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A movie theatre sells the three concession stand items listed below: · Popcorn = $3 · Snickers = $4 (or five for the price of three) · Soda = $2 Implement a web application for the movie theatre that allows a user to add the above items to a shopping cart and calculates the total for the given collection of items. For example, the following basket should total up to $23. · 3x Popcorn · 5x Snickers · 1x Soda Deliver your solution as a web application using a modern client-side framework such as Angular, Ember, Backbone, etc. Your solution should be able to run in Chrome.
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Senior Software Engineer

Interviewed at Comscore

3.6★
Jul 19, 2018

A movie theatre sells the three concession stand items listed below: · Popcorn = $3 · Snickers = $4 (or five for the price of three) · Soda = $2 Implement a web application for the movie theatre that allows a user to add the above items to a shopping cart and calculates the total for the given collection of items. For example, the following basket should total up to $23. · 3x Popcorn · 5x Snickers · 1x Soda Deliver your solution as a web application using a modern client-side framework such as Angular, Ember, Backbone, etc. Your solution should be able to run in Chrome.

/* Write a query to find all duplicate jobs in the People table */ select count(*) from People p1 join People p2 on p1.Job = p2.job and p1.FirstName <> p2.FirstName; /* Write a query to find all duplicate FirstName, and Job values in the People table and number of duplicates */ with cte as( select row_number() over(order by FirstName, job) as num, FirstName, job from People ) select count(*) from cte p1 join cte p2 on p1.Job = p2.job and p1.FirstName = p2.FirstName and p1.num <> p2.num; /* Write a query to find all duplicate First Names with different jobs in the People table */ select p1.FirstName from People p1 join People p2 on p1.Job < p2.job and p1.FirstName = p2.FirstName;
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Senior Software Engineer

Interviewed at WideOrbit

3.6★
Nov 18, 2020

/* Write a query to find all duplicate jobs in the People table */ select count(*) from People p1 join People p2 on p1.Job = p2.job and p1.FirstName <> p2.FirstName; /* Write a query to find all duplicate FirstName, and Job values in the People table and number of duplicates */ with cte as( select row_number() over(order by FirstName, job) as num, FirstName, job from People ) select count(*) from cte p1 join cte p2 on p1.Job = p2.job and p1.FirstName = p2.FirstName and p1.num <> p2.num; /* Write a query to find all duplicate First Names with different jobs in the People table */ select p1.FirstName from People p1 join People p2 on p1.Job < p2.job and p1.FirstName = p2.FirstName;

Summing up the individual digits for each number from 0 to k (0&lt;=k&lt;=10000000), return how many times the most common sum occurs. Examples: k=10 gives 2 (since 1 and 10 both sum up to 1) k=50 gives 6 (since 5, 14, 23, 32, 41, 50 all sum up to 5) k=7777 gives 555
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Senior Software Engineer

Interviewed at Medallia

3.1★
Jun 9, 2011

Summing up the individual digits for each number from 0 to k (0&lt;=k&lt;=10000000), return how many times the most common sum occurs. Examples: k=10 gives 2 (since 1 and 10 both sum up to 1) k=50 gives 6 (since 5, 14, 23, 32, 41, 50 all sum up to 5) k=7777 gives 555

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