Steve Yegge’s Phone-Screen Coding Exercises

June 30, 2009

All of these are easy, and very much shorter than the Java versions that Yegge provides:

  1. Most Scheme systems provide a string-reverse function, although it is not required by the Standard. Here is a purely-Standard version:

    (define (string-reverse s)
      (list->string (reverse (string->list s))))

  2. Yegge gives this function to compute fibonacci numbers:

    (define (fib n)
      (if (<= n 1) n
        (+ (fib (- n 1)) (fib (- n 2)))))

    That function takes exponential time to compute, due to the constant re-computation of previously-computed results. If Yegge gave me that function during a phone screen, he wouldn’t pass to the next stage of the hiring process. Here is a linear-time version of the fibonacci function:

    (define (fib n)
      (let fib ((n n) (f1 1) (f2 0))
        (if (zero? n) f2
          (fib (- n 1) (+ f1 f2) f1))))

  3. Standard Scheme doesn’t provide formatted output, but most Scheme systems provide a version of Lisp’s format function, which provides a wealth of options:

    (define (times-table n)
      (do ((i 1 (+ i 1))) ((> i n))
        (do ((j 1 (+ j 1))) ((> j n) (newline))
          (format #t "~4d" (* i j)))))

  4. Read gets the next object from the input; if the input consists of integers separated by newlines, each call to read will return the next number on the input:

    (define (sum-file file-name)
      (with-input-from-file file-name
        (lambda ()
          (let loop ((n (read)) (sum 0))
            (if (eof-object? n) sum
              (loop (read) (+ n sum)))))))

    If you prefer, you can use the fold-input function from the Standard Prelude:

    (define (sum-file file-name)
      (fold-input read + 0 file-name))

  5. Things don’t get much simpler than this:

    (define (print-odds)
      (do ((i 1 (+ i 2))) ((> i 100))
        (display i) (newline)))

  6. Well, maybe they do:

    (define (largest xs) (apply max xs))

  7. Standard Scheme provides number->string, which takes an optional radix argument but doesn’t write leading zeros, so we make our own to-hex function:

    (define (format-rgb r g b)
      (define (to-hex n)
        (string
          (string-ref "0123456789ABCDEF" (quotient n 16))
          (string-ref "0123456789ABCDEF" (modulo n 16))))
      (string-append (to-hex r) (to-hex g) (to-hex b)))

    If you like format, it has an option to print hexadecimal numbers.

It’s hard to imagine that coding exercises like this would be a barrier for any practicing programmer, but apparently they are; Yegge got the fibonacci question wrong, and if you read the comments on his original article, he also got the print-odds program wrong initially (he since changed his article, claiming the mistake was due to a failed optimization), printing even numbers instead of odd numbers. Of course, that’s the reason this blog exists: to give savvy programmers a chance to practice their skills, so the don’t make dumb coding mistakes, either during a phone screening or in their working programs.

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10 Responses to “Steve Yegge’s Phone-Screen Coding Exercises”

  1. [...] Praxis – Steve Yegge’s Phone-Screen Coding Exercises By Remco Niemeijer Today’s Programming Praxis problem is an easy one. In 2005, Steve Yegge posted an article about [...]

  2. Remco Niemeijer said

    My Haskell solution (see http://bonsaicode.wordpress.com/2009/06/30/programming-praxis-steve-yegge%E2%80%99s-phone-screen-coding-exercises/ for a version with comments):

    import Text.Printf
    import Data.Word
    
    reverse' :: String -> String
    reverse' = foldr (:) []
    
    fib :: (Num a) => Int -> a
    fib n = fibs !! n where fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
    
    timesTable :: IO ()
    timesTable = mapM_ putStrLn [concat
        [printf "%4d" (a * b) | a <- [1..12]] | b <- [1..12 :: Int]]
    
    sumFile :: FilePath -> IO ()
    sumFile path = print . sum . map read . lines =<< readFile path
    
    printOdds :: IO ()
    printOdds = print $ filter odd [1..99]
    
    largest :: [Int] -> Int
    largest = foldl max 0
    
    toHex :: Word8 -> Word8 -> Word8 -> String
    toHex = printf "%02x%02x%02x"
    
  3. Scott Pedersen said

    C# Solution

    I couldn’t resist writing smartass responses for questions three and five to just return fixed strings. I also wrote more general answers that I thought were more in keeping with the spirit of the exercise.

    The basic recursive implementation for Fibonacci numbers is a classic, but is terribly slow. It’s the answer I’d expect out of most interview candidates as a first pass since it is taught so regularly as an example of recursion. However, they’d be in trouble if they couldn’t quickly identify the problems with such an implementation. I wrote out a couple of alternatives that are a bit more efficient.

  4. Remco Niemeijer said

    Thanks to nonowarn for catching a bug:

    The correct code for exercise 1 should be

    reverse' :: String -> String
    reverse' = foldl (flip (:)) []
    
  5. liesen said

    I think you’ve got them all, Remco. My Haskell solutions are similar. Here is one simpler solution to #5:

    printOdds = print [1,3..99]

    There’s one more bug in Remco’s code, though: largest assumes positive integers, but it should read:

    largest = foldl1 max

  6. Remco Niemeijer said

    Yeah, I noticed that one too, but didn’t get round to fixing it. However, if you treat Steve Yegge’s solution as the spec then the correct implementation is actually

    largest = foldl max minBound

  7. Connochaetes said
    def reverse(str)
      # 1. Write a function to reverse a string.
      len = str.length - 1
      len.downto(0).collect {|n| str[n].chr}.to_s
    end
    
    def nth_fib(n)
      # 2. Write a function to compute the Nth fibonacci number.
      # NB. assumes fibonacci numbers are zero-indexed.
      x,y = 0,1
      n.times {x,y = y,x+y}
      y
    end
    
    def multi_table
      # 3. Print out the grade-school multiplication table up to 12 x 12.
      size = 12
      1.upto(size).each do |x|
        1.upto(size).each do |y|
          printf "%4d", x*y
        end
        printf "\n"
      end
    end
    
    def sum_ints_from_file(filename)
      # 4. Write a function that sums up integers from a text file, one per line.
      sum = 0
      File.open(filename).each {|line| sum += line.to_i}
      sum
    end
    
    def print_odd_numbers
      # 5. Write a function to print the odd numbers from 1 to 99.
      (1..99).step(2).each {|n| puts n} ; nil
    end
    
    def find_largest_value(arr)
      # 6. Find the largest int value in an int array.
      arr.inject {|x,y| x > y ? x : y}
    end
    
    def format_RGB(r,g,b)
      # 7. Format an RGB value (three 1-byte numbers) as a 6-digit hexadecimal string.
      format "%02X%02X%02X", r,g,b
    end
    
    
  8. [...] for the exercise, I picked Steve Yegge’s Phone-Screen Coding Exercises, figuring seven different small problems might cover slightly more ground than one bigger [...]

  9. Graham said

    I’m never sure how much of a language’s built-in functionality we’re supposed to use for a question.
    Python solution
    Common Lisp solution

  10. brooknovak said

    For the RGB => Hex conversion you can always take advantage API knowledge (such as string formatting features), but its not as fun!

    public static string ConvertToHex(byte r, byte g, byte b) {
    	char[] buffer = new char[6];
    	int rgb = (r << 16) | (g << 8) | b;
    	for (int i = 0; i < 6; i++) {
    		int val = rgb & 15;
    		if (val <= 9)
    			buffer [5 - i] = (char)('0' + val);
    		else 
    			buffer [5 - i] = (char)('A' + (val - 10));
    		rgb >>= 4;
    	}
    	return new string(buffer);
    }
    

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