Senior Software Engineer Interview Questions

There are obviously multiple solutions: Solution 1: Set sum = 0 Find the remainder by dividing by 2. Divide by 2 for the next iteration. Solution 2--much better. Set sum = 0 Start loop Set mask = 1 sum += mask & number Loop return sum

There is another way, that is explained here: http://en.wikipedia.org/wiki/Hamming_weigh (with just 24 operation and without any cycle you can find the number of bit set to 1)

An example in Java with 10000 results in answer 5 int number = 10000; int numberOfBitsSet = 0; for (int i = 1; i <= number;) { int result = (i & number); if (result != 0) { numberOfBitsSet++; } i = i << 1; } System.out.println(numberOfBitsSet);

Use MergeSort.

You don't need merge sort here. This is a O(n) task, because the lists are already sorted. In fact, this merge routine is part of merge sort. No sorting takes place here, only merging, and it could be done in linear time. Node* merge(Node* list1, Node* list2) { Node* merged = null; Node** tail = &merged; while (list1 && list2) { if (list1->data data) { *tail = list1; list1 = list1->next; } else { *tail = list2; list2 = list2->next; } tail = &((*tail)->next); } *tail = list1 ? list1 : list2; return merged; }

Put me in a very unconformable situation.

simple algorithm

Interview was great, to the point, they know what they are doing

It is 1 and the reason the size is non-zero is "To ensure that the addresses of two different objects will be different." And the size can be 1 because alignment doesn't matter here, as there is nothing to actually look at.

Converting between the private IP/port and public IP/port so that you can access Intnert from internal network