Software Engineer Interview Questions

Anonymous Unicorn - "input = [ {"topic": 1, "chapter": 1, "section": 1}, {"topic": 2, "chapter": 2, "section": 1}, {"topic": 3, "chapter": 2, "section": 2}, {"topic": 4, "chapter": 1, "section": 1}, {"topic": 5, "chapter": 1, "section": 1}, {"topic": 6, "chapter": 2, "section": 2}, {"topic": 7, "chapter": 2, "section": 2}, {"topic": 8, "chapter": 2, "section": 3}, ] expected_output = [ {'chapter': 1, 'sections': [ {'section': 1, 'topics': [ {'top"See full answer

Hyder A. - "Control plane has few component, it's like the brain of the cluster, the data plane is the work horse, CP has etcd, api-server, DP has kublet and kube-proxy"See full answer

B. T. - "Example: bucket A: 3 liters capacity bucket B: 5 liters capacity goal: 4 liters You are asked to print the logical sequence to get to the 4 liters of water in one bucket. Follow up: How would you solve the problem if you have more than 2 buckets of water?"See full answer

Farhan -. - "Sorting is a technique to arrange data in either increasing order or decreasing order, and, the function that implements this functionality is called sort function"See full answer

Sue G. - "In Java, GC is the process of automatically identifying and reclaming memory occupied by objects that are no longer reachable.. the process involves the 3 steps Mark phase -- identify the objects that are still in use. i.e reachable sweep Phase -- removes the unreachable objects compact phase -- rearragnes objects to prevent fragmentation 4 types of Garbage collection Serial GC -- single threaded simple and compacting -- best for small applications Parrallel GC -- throughput GC"See full answer

Idan R. - " read_dir(path: str) -> list[str] returns the full path of all files and sub- directories of a given directory. is_file(path: str) -> bool: returns true if the path points to a regular file. is_dir(path: str) -> bool: returns true if the path points to a directory. read_file(path: str) -> str: reads and returns the content of the file. The algorithm: notice that storing all the file contents' is too space intensive, so we can't read all the files' contents to store and compare with each"See full answer

Anonymous Wolf - "Construct a min-heap either inplace, or by making a copy of the array and then applying heapify on that copy. This is done in O(n) time. Maintain two zero-initialised variables - sum and count. Keep popping off the heap while sum < k, and update count. In the worst case you will have to do n pops, and each pop is O(log n), so the algorithm would take O(n log n) in total. Space complexity depends on whether you're allowed to modify inplace or not, so either O(1) or O(n) respectively."See full answer

Abdurhman M. - "Basic computer skills. Languages you are not fluent in. Irrelevant skills. Skills you don't possess. Generalized skills."See full answer