Software Development Engineer Interview Questions

Any DFS would do the trick. Don't forget to label visited nodes!

One of the two simplest, most straightforward ways: - Depth-First search (http://en.wikipedia.org/wiki/Depth-first_search) - Breadth-First search (http://en.wikipedia.org/wiki/Breadth-first_search) If your maze is very, very long and deep, and has loops and crazies, and is complicated, I suggest the A* algorithm, which is the industry standard pathfinding algorithm which combines a Breadth-First search with heuristics...sort of like an "intelligent breadth-first search". See http://stackoverflow.com/questions/3097556/programming-theory-solve- a-maze?answertab=active#tab-top

Replace the maze with nodes at intersections and then perform A*

I honestly had no idea how to solve this in a reasonable amount of time.

HashMap map = new HashMap(); for(String str : strings) { Integer count = map.get(str) map.put(str, count != null ? count + 1: new Integer(1) ); } // then iterate through map to find the largest integer, the key will be the string with the most occurances

Its a similar answer to above, but I think using just a hash table is not correct if there is collision. Perfect hashing can be used for no collision, or an array in the hashmap.

Use recursion and build the tree from the root.

A sorted array is a perfect min-heap, and so a binary tree!

void reverse(String *str) { char *end = str; char temp; if (str) { end++; } end--; while (str

Same as the other solution posted, but Java. public String reverse(String str) { char[] arr = str.toCharArray(); int j = arr.length-1; for(int i=0;i

hi, did you actually complete the project?

I would solve this using a BST sorted by the start index of the range (end would work as well). Modify the insert to merge a node instead of inserting it at any comparison it detects an overlap. You would then need to check if you now overlap either of the child nodes and iterate down until you fix the tree.

It's similar to scheduling problems - Sort range based on range start time - Store the current start and end times in two variables - For each range, check - if range start time before the current end time print current start-end and update start-end - if range end time is bigger than current end time, update current end time void printMergedRanges(Range[] range){ Arrays.sort(range); if (range.lengthend){ System.out.println("["+start+","+end+"]"); start=range[i].start; end=range[i].end; }else if(range[i].end>end){ end=range[i].end; } } System.out.println("["+start+","+end+"]"); }

Given the number of references during the interview to compulsory and voluntary overtime I am guessing they would want a candidate to say they would work whatever hours are needed to get the job done.