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

Everyone who knows what an of statement is has written fizz-buzz. Check x%3, then check x%5, and write out your four if/else cases. You can use only a single branch with a switch on from x%15 from -14 to 14, or no branches at all by hardcoding 30 strings in an array and accessing that index.

Pretty common question everyone should have learned in high school, except this one is only for base 10, so it is even easier. I made sure I handled 0 and -2^31 correctly. -2^31 I special-cases out, but in hindsight a cleaner solution would be to just store x as a long long before doing anything. You need to watch out for -2^31 because if you multiply it by -1, you get the same thing.

Literally 2 lines: If (!node) return null; return 1+max(height(node->left), height(node->right));