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This is the technical phone interview question from Google. Google has offices in London but the call was from Google Switzerland (+41). The interview lasts for 45 minutes.

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The engineer also asks me the test cases if I want to writeunit testsfor it. I can think of three cases:

Did you cleared the round? As after the phone interview they get back within 2 days.

It would be good that someone can make testing data for this question. To adapt it for Online Judge, we can re-write the function signature to:

time key.first = 10 // then remove from queue and cache.

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We can use queue to store the messages in last 10 seconds. We can usehashmap/set to store the messages printed, in order to speed up lookup. This method is the combination of both queue and hashmap/set. The following C++ code demonstrates the idea. We use theinttype to represent the time type for simplification.

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Given a list of messages and the date/time, print each message if it is not printed in the last 10 seconds. It is possible that several messages arrive roughly at the same time (within 1 second)

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The engineer didnt provide any interface (so you can come up with the function signature). We dont need to store all the messages and the frequency for the function call could be several per second.

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I did make some mistakes in writing the code on the Google Docs without an actualProgramming IDE!

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I am getting a bit confused here . Probably I am wrong, but shouldnt the condition be ?

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Using a hashmap to store the printed messages and their corresponding time is trivial but this will have a memory issue (Out-Of-Memory) if this function is running for weeks. Hashmap will grow, especially that all the messages are unique!

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// compute the string hash as a 32-bit integer

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// member variables unordered_setint cache; queuepairtime, int last10; void PrintMessage(int time, string msg) // compute the string hash as a 32-bit integer int hash = compute_hash(msg); // remove invalid entries while (!last10.empty()) auto key = last10.front(); if (time key.first = 10) last10.pop(); cache.erase(key.second); // remove from hash set else break; if unt(hash)) return; // printed in last 10 seconds // we can print the message now cout msg endl; // inserting the entry sert(hash); last10.push(make_pair(time, hash));

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