K’th Largest Item

August 1, 2014

Today’s exercise is a common interview question, and also interesting in its own right:

Find the kth largest integer in a file of n integers, where n is large (defined as too large to fit into memory all at once) and k is small (defined as small enough to all fit in memory).

Your task is to write the program described above. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

About these ads

Pages: 1 2

8 Responses to “K’th Largest Item”

  1. #!/usr/local/bin/perl
    
    use strict;
    use warnings;
    
    my $k = shift @ARGV;
    
    my @nos = ();
    foreach my $i (1..$k) {
      my $l = <>;
      $l*=1;
      push @nos, $l;
    }
    @nos = sort { $b<=>$a } @nos;
    
    while(<>) {
      $_*=1;
      next if $_ < $nos[$k-1];
      foreach my $i (0..($k-1)) {
        next unless $_ > $nos[$i];
        splice @nos,$i,0,$_;
        pop @nos;
        last;
      }
    }
    
    print "$nos[$k-1]\n";
    
  2. Josef said

    Too lazy to do the file processing. The program below finds the kth largest in a list. In Haskell.

    code{white-space: pre;}

    table.sourceCode, tr.sourceCode, td.lineNumbers, td.sourceCode {
    margin: 0; padding: 0; vertical-align: baseline; border: none; }
    table.sourceCode { width: 100%; line-height: 100%; }
    td.lineNumbers { text-align: right; padding-right: 4px; padding-left: 4px; color: #aaaaaa; border-right: 1px solid #aaaaaa; }
    td.sourceCode { padding-left: 5px; }
    code > span.kw { color: #007020; font-weight: bold; }
    code > span.dt { color: #902000; }
    code > span.dv { color: #40a070; }
    code > span.bn { color: #40a070; }
    code > span.fl { color: #40a070; }
    code > span.ch { color: #4070a0; }
    code > span.st { color: #4070a0; }
    code > span.co { color: #60a0b0; font-style: italic; }
    code > span.ot { color: #007020; }
    code > span.al { color: #ff0000; font-weight: bold; }
    code > span.fu { color: #06287e; }
    code > span.er { color: #ff0000; font-weight: bold; }

    module Kth where
    
    import Data.List
    
    findKth :: Ord a => Int -> [a] -> Maybe a
    findKth k [] = Nothing
    findKth k as = Just (loop (sort ls) rs)
      where (ls,rs) = splitAt k as
    
    loop (k:_) [] = k
    loop (k:ls) (r:rs)
      | r > k     = loop (insert r ls) rs
      | otherwise = loop (k:ls) rs
  3. svenningsson said

    Sorry about the weird looking comment. I experimented with getting syntax highlighting for Haskell. It seems there is no way for me to edit the comment to fix the problem.

  4. James Curtis-Smith’s answer is the best approach I can think of, which means to always keep a list of the top k elements and adjust that list when necessary. It requires only one pass on the original list and worst case scenario happens when the original list is sorted so that everytime you read a new element you need to “splice”. In that case it’s O(k*n). There’s another approach that’s also O(k*n), which is removing from the original list the biggest element and doing this k times, but since the original list is so big you probably can’t “remove” things from it. At best you can store them in a hash to discard them later, but it seems more complicated and (worst of all) it’s o(k*n) always, not just in the worst case scenario.

    Anyway, just wanted to share my analysis. No need to share an actual solution, since it’s all there already.

  5. Paul said

    In Python.

    from heapq import heappush, heapreplace
    
    def k_largest(seq, k):
        """number of items in seq is much larger than k
           return the k-th largest
        """
        it = iter(seq)
        heap = [] # keeps the argest k items
        for i in range(k):
            heappush(heap, next(it))
        for item in it:
            if item > heap[0]:
                heapreplace(heap, item)
        return heap[0]
    
  6. Mike said

    The python library ‘heapq’ provides a function nlargest().

    import heapq
    
    def kthlargest(k, iterable):
        return heapq.nlargest(k, iterable)[0]
    

    It basically uses Paul’s algorithm. I think using a heap reduces the complexity from O(n*k) to O(n*log k).

  7. MS said

    Haskell, so compile with ghc, then call with k and filename as command-line arguments. Integers should be one per line in the file.

    import System.Environment
    import Data.List

    main = do
    (k:file:_) <- getArgs
    ints [Integer] -> [Integer]
    klargest k = foldl (\a -> take k . sortBy (flip compare) . (:a)) []

  8. MS said

    Some of that got eaten. Try again.

    import System.Environment
    import Data.List

    main = do
    (k:file:_) <- getArgs
    ints [Integer] -> [Integer]
    klrg k = foldl (\a -> take k . sortBy (flip compare) . (:a)) []

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 634 other followers

%d bloggers like this: