Sunday, August 14, 2011

Basic coding question

When interviewing candidates I always like to include a basic coding question. The goal is to have something simple that can be done in just about any language to see if the candidate can actually write code. When I used to interview C programmers the question I used was to implement the strtok function. This function useful because you can then follow up with questions about memory management, modifying the input parameters, thread safety etc.

These days, I mostly interview Java programmers, but I still use string tokenization as the basic question and follow up with questions about such as regular expressions and unicode. One of the things that has always surprised me is the amount of variety in the answers. It seems everyone can find different ways of tokenizing strings. So when a colleague said that just about all candidates fail trying the same approach to his basic coding question, I couldn't help but wonder why he was seeing such consistency. The question is to write a function that will print a diamond to the console. The function should take a single integer parameter that will be the width of the diamond. The width must be odd or else the function should print an error.

I thought about it for a bit and had a general sketch of a program within about five minutes. Within another five minutes I had a working implementation in python:
#!/usr/bin/env python

import sys

def printDiamond(width):
    Print a diamond to the console. The diamond must have an odd width, if the
    width is even and error will be printed to stderr and the program will exit.
    if width % 2 == 0:
        sys.stderr.write("ERROR: width must be odd\n")

    # Print top half of diamond
    numSpaces = width / 2
    numAsterisks = 1
    while numAsterisks <= width:
        sys.stdout.write(" " * numSpaces)
        sys.stdout.write("*" * numAsterisks)
        numSpaces -= 1
        numAsterisks += 2

    # Print bottom half of dimaond
    numSpaces = 1
    numAsterisks = width - 2
    while numAsterisks >= 1:
        sys.stdout.write(" " * numSpaces)
        sys.stdout.write("*" * numAsterisks)
        numSpaces += 1
        numAsterisks -= 2

if len(sys.argv) < 2:
    print "Usage: %s <width>" % sys.argv[0]
So what was the common mistake he was seeing? Apparently most people start out with nested for loops and try to figure out some equations to indicate whether or not there should be an asterisk at position (i, j). At this point a lot of candidates just get bogged down trying to figure out the math and never step back to think about whether there is an easier way. I suppose the consistency is just that the basic trap is so easy to fall into for this question.

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