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The following block of code assumes RxJava2 usage.Please, describe which scheduler will [map] execute on, [filter] execute on. fun theTest() { val io = Schedulers.io() Observable .just("long", "longer", "longest") .subscribeOn(Schedulers.io()) .map { println("map ${Thread.currentThread().name}") it.length } .observeOn(Schedulers.computation()) .filter { println("filter ${Thread.currentThread().name}") it > 6 } .blockingSubscribeBy { length -> println( "item length $length under " + Thread.currentThread().name ) } }
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Senior Android Developer

Interviewed at HungerStation

3.2★
Dec 21, 2023

The following block of code assumes RxJava2 usage.Please, describe which scheduler will [map] execute on, [filter] execute on. fun theTest() { val io = Schedulers.io() Observable .just("long", "longer", "longest") .subscribeOn(Schedulers.io()) .map { println("map ${Thread.currentThread().name}") it.length } .observeOn(Schedulers.computation()) .filter { println("filter ${Thread.currentThread().name}") it > 6 } .blockingSubscribeBy { length -> println( "item length $length under " + Thread.currentThread().name ) } }

It was a convoluted sorting question. Basically your given an array with a some leading words followed by some numbers and some more words at the end. You first sort them based on the leading words, then numbers, afterwards, if both are equal you sort by the remaining words. There was some edge cases, but it was basically a sorting problem
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Android Developer

Interviewed at LXE Hearing

3.6★
Dec 3, 2019

It was a convoluted sorting question. Basically your given an array with a some leading words followed by some numbers and some more words at the end. You first sort them based on the leading words, then numbers, afterwards, if both are equal you sort by the remaining words. There was some edge cases, but it was basically a sorting problem

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