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Given two unsigned integer values, write a function that returns the first divided by the second. You cannot (of course) use div or mod operators - only addition, subtraction and multiplication. Discuss the strengths/weaknesses/algorithmic complexity of your solution. Is there a better way to do it? If so, what, and what is its complexity?
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Software Developer

Interviewed at Meta

3.6★
Jul 30, 2015

Given two unsigned integer values, write a function that returns the first divided by the second. You cannot (of course) use div or mod operators - only addition, subtraction and multiplication. Discuss the strengths/weaknesses/algorithmic complexity of your solution. Is there a better way to do it? If so, what, and what is its complexity?

Given the following list of objects {user, loginTime, logoutTime}. What is the maximum number of concurrent users logged in at the same time? Input: [ {user: A, login: 1, logout: 3}, {user: B, login: 3, logout: 4}, {user: C, login: 1, logout: 2}, {user: D, login: 123123123, logout: 987987987}, {user: E, login: 1, logout: 3} ] Output: 3
avatar

Software Developer

Interviewed at Booking.com

4.1★
Aug 24, 2016

Given the following list of objects {user, loginTime, logoutTime}. What is the maximum number of concurrent users logged in at the same time? Input: [ {user: A, login: 1, logout: 3}, {user: B, login: 3, logout: 4}, {user: C, login: 1, logout: 2}, {user: D, login: 123123123, logout: 987987987}, {user: E, login: 1, logout: 3} ] Output: 3

Considering a 2-dimension matrix that can only be traversed by 1 adjacent position at a time and never diagonally. Create an algorithm to traverse that matrix from its upper-left corner to its lower-right corner using the shorter possible path in the most efficient way.
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Senior Software Engineer

Interviewed at Oracle

3.5★
Mar 19, 2009

Considering a 2-dimension matrix that can only be traversed by 1 adjacent position at a time and never diagonally. Create an algorithm to traverse that matrix from its upper-left corner to its lower-right corner using the shorter possible path in the most efficient way.

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