Let S1 and S2 be two sets of integers (they are not necessarily disjoint).
We know
that |S1| = |S2| = n (i.e. each set has n integers).
Each set is stored in an array of length n, where
its integers are sorted in ascending order.
Let k ≥ 1 be an integer.
Design an algorithm to find the
k smallest integers in S1 ∩ S2 in O(n) time.
This is what I have so far:
Create a new array called Intersection
For each e in S1 add e to hashset in O(n) time
For each e in S2 check if e exists in hashset in O(n) time
If e exists in hashset add e to Intersection
Once comparisons are done sort Intersection by count sort in O(n) time
return the first k integers
Thus O(n) + O(n) + O(n) = O(n)
Am I on the right track?
Yes, you're definitely on the right track but there's actually no need at all to generate a hash-table or extra set. As your two sets are already sorted, you can simply run an index/pointer through both of them, looking for the common elements.
For example, to find the first common element from the two sets, use the following pseudo-code:
start at first index of both sets
while more elements in both sets, and current values are different:
if set1 value is less than set2 value:
advance set1 index
else
advance set2 index
At the end of that, set1 index will refer to an intersect point provided that neither index has moved beyond the last element in their respective list. You can then just use that method in a loop to find the first x intersection values.
Here's a proof of concept in Python 3 that gives you the first three numbers that are in the two lists (multiples-of-two and multiples-of-three). The full intersection would be {0, 6, 12, 18, 24} but you will see that it will only extract the first three of those:
# Create the two lists to be used for intersection.
set1 = [i * 2 for i in range(15)] ; print(set1) # doubles
set2 = [i * 3 for i in range(15)] ; print(set2) # trebles
idx1 = 0 ; count1 = len(set1)
idx2 = 0 ; count2 = len(set2)
# Only want first three.
need = 3
while need > 0:
# Continue until we find next intersect or end of a list.
while idx1 < count1 and idx2 < count2 and set1[idx1] != set2[idx2]:
# Advance pointer of list with lowest value.
if set1[idx1] < set2[idx2]:
idx1 += 1
else:
idx2 += 1
# Break if reached end of a list with no intersect.
if idx1 >= count1 or idx2 >= count2:
break
# Otherwise print intersect and advance to next list candidate.
print(set1[idx1]) ; need -= 1
idx1 += 1 ; idx2 += 1
The output is, as expected:
[0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28]
[0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42]
0
6
12
If you needed a list at the end rather than just printing out the intersect points, you would simply initialise an empty container before the loop and the append the value to it rather than printing it. This then becomes a little more like your proposed solution but with the advantage of not needing hash tables or sorting.
Create two arrays, call them arr1 and arr2, of size array_size and populate them with integer values in ascending order. Create two indexes, call them i and j, that will be used to iterate over arr1 and arr2 respectively and initialize them to 0. Compare the first two values of each array: if arr1[0] is less than arr2[0] then increment i, else if arr1[0] is greater than arr2[0] increment j, else the values intersect and we can return this value. Once we have returned k intersecting values we can stop iterating. In the worst case scenario this will be i + j, O(n) if no intersections occur between both sets of values and we will have to iterate to the end of each array.
Here is the solution in bash:
#!/bin/bash
#-------------------------------------------------------------------------------
# Design an algorithm to find the k smallest integers in S1 ∩ S2 in O(n) time.
#-------------------------------------------------------------------------------
typeset -a arr1 arr2 arr_answer
typeset -i array_size=20 k=5
function populate_arrs {
typeset -i counter=0
while [[ ${counter} -lt ${array_size} ]]; do
arr1[${counter}]=$((${counter} * 2))
arr2[${counter}]=$((${counter} * 3))
counter=${counter}+1
done
printf "%8s" "Set1: "; printf "%4d" ${arr1[*]}; printf "\n"
printf "%8s" "Set2: "; printf "%4d" ${arr2[*]}; printf "\n\n"
}
function k_smallest_integers_main {
populate_arrs
typeset -i counter=0 i=0 j=0
while [[ ${counter} -lt ${k} ]]; do
if [[ ${arr1[${i}]} -eq ${arr2[${j}]} ]]; then
arr_answer[${counter}]=${arr1[${i}]}
counter=${counter}+1; i=${i}+1; j=${j}+1
elif [[ ${arr1[${i}]} -lt ${arr2[${j}]} ]]; then
i=${i}+1
else
j=${j}+1
fi
done
printf "%8s" "Answer: "; printf "%4d" ${arr_answer[*]}; printf "\n"
}
k_smallest_integers_main
Output:
Set1: 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38
Set2: 0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57
Answer: 0 6 12 18 24
In Python:
i1= 0; i2= 0
while k > 0 and i1 < n and i2 < n:
if S1[i1] < S2[i2]:
i1+= 1
elif S1[i1] > S2[i2]:
i2+= 1
else:
Process(S1[i1], S2[i2])
i1+= 1; i2+= 1
k-= 1
Execution will perform less than k calls to Process if there aren't sufficiently many elements in the intersection.
I can sort alphabets or numbers using sort but how do I sort a mixture of alphabets and numbers.
(sort ["f" "g" "a" "b" "c"]) ; ==> ("a" "b" "c" "f" "g")
(sort [3 4 6 1 8 ]) ; ==> (1 3 4 6 8)
Question is, how do I sort this? ["g" "a" "c" 4 6 1] to get (1 4 6 "a" "c" "g")
The main issue is that you cannot compare a string with a number in a generic way: these are different types of values. When someone says "what's better: an apple or a house?", the first question that could probably pop into one's mind is "better in what way?" You can sort these two objects by many different properties, like size, cost or edibility. sort does not make the call about what property to use.
That's where sort-by function comes in. First it takes a keyfn, that when called on any element should produce its comparable property: in our case it's a string representation of a given element. Then it takes a collection, then (optionally) a comparator.
So you use str as your keyfn and you don't need a comparator, since comparison of strings is well-defined.
The resulting code is plain and simple:
(sort-by str ["g" "a" "c" 4 6 1]) ; => (1 4 6 "a" "c" "g")
You cannot compare a number and a string.
=> (sort ["g" "a" "c" 4 6 1])
ClassCastException java.lang.String cannot be cast to java.lang.Number
So to do what you want, you have to convert the numbers to strings then sort. For example:
=> (sort (map str ["g" "a" "c" 4 6 1]))
("1" "4" "6" "a" "c" "g")
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
I've already worked out this solution for myself with PHP, but I'm curious how it could be done differently - better even. The two languages I'm primarily interested in are PHP and Javascript, but I'd be interested in seeing how quickly this could be done in any other major language today as well (mostly C#, Java, etc).
Return only words with an occurrence greater than X
Return only words with a length greater than Y
Ignore common terms like "and, is, the, etc"
Feel free to strip punctuation prior to processing (ie. "John's" becomes "John")
Return results in a collection/array
Extra Credit
Keep Quoted Statements together, (ie. "They were 'too good to be true' apparently")Where 'too good to be true' would be the actual statement
Extra-Extra Credit
Can your script determine words that should be kept together based upon their frequency of being found together? This being done without knowing the words beforehand. Example:
*"The fruit fly is a great thing when it comes to medical research. Much study has been done on the fruit fly in the past, and has lead to many breakthroughs. In the future, the fruit fly will continue to be studied, but our methods may change."*
Clearly the word here is "fruit fly," which is easy for us to find. Can your search'n'scrape script determine this too?
Source text: http://sampsonresume.com/labs/c.txt
Answer Format
It would be great to see the results of your code, output, in addition to how long the operation lasted.
GNU scripting
sed -e 's/ /\n/g' | grep -v '^ *$' | sort | uniq -c | sort -nr
Results:
7 be
6 to
[...]
1 2.
1 -
With occurence greater than X:
sed -e 's/ /\n/g' | grep -v '^ *$' | sort | uniq -c | awk '$1>X'
Return only words with a length greater than Y (put Y+1 dots in second grep):
sed -e 's/ /\n/g' | grep -v '^ *$' | grep .... | sort | uniq -c
Ignore common terms like "and, is, the, etc" (assuming that the common terms are in file 'ignored')
sed -e 's/ /\n/g' | grep -v '^ *$' | grep -vf ignored | sort | uniq -c
Feel free to strip punctuation prior to processing (ie. "John's" becomes "John"):
sed -e 's/[,.:"\']//g;s/ /\n/g' | grep -v '^ *$' | sort | uniq -c
Return results in a collection/array: it is already like an array for shell: first column is count, second is word.
Perl in only 43 characters.
perl -MYAML -anE'$_{$_}++for#F;say Dump\%_'
Here is an example of it's use:
echo a a a b b c d e aa | perl -MYAML -anE'$_{$_}++for#F;say Dump \%_'
---
a: 3
aa: 1
b: 2
c: 1
d: 1
e: 1
If you need to list only the lowercase versions, it requires two more characters.
perl -MYAML -anE'$_{lc$_}++for#F;say Dump\%_'
For it to work on the specified text requires 58 characters.
curl http://sampsonresume.com/labs/c.txt |
perl -MYAML -F'\W+' -anE'$_{lc$_}++for#F;END{say Dump\%_}'
real 0m0.679s
user 0m0.304s
sys 0m0.084s
Here is the last example expanded a bit.
#! perl
use 5.010;
use YAML;
while( my $line = <> ){
for my $elem ( split '\W+', $line ){
$_{ lc $elem }++
}
END{
say Dump \%_;
}
}
F#: 304 chars
let f =
let bad = Set.of_seq ["and";"is";"the";"of";"are";"by";"it"]
fun length occurrence msg ->
System.Text.RegularExpressions.Regex.Split(msg, #"[^\w-']+")
|> Seq.countBy (fun a -> a)
|> Seq.choose (fun (a, b) -> if a.Length > length && b > occurrence && (not <| bad.Contains a) then Some a else None)
Ruby
When "minified", this implementation becomes 165 characters long. It uses array#inject to give a starting value (a Hash object with a default of 0) and then loop through the elements, which are then rolled into the hash; the result is then selected from the minimum frequency.
Note that I didn't count the size of the words to skip, that being an external constant. When the constant is counted too, the solution is 244 characters long.
Apostrophes and dashes aren't stripped, but included; their use modifies the word and therefore cannot be stripped simply without removal of all information beyond the symbol.
Implementation
CommonWords = %w(the a an but and is not or as of to in for by be may has can its it's)
def get_keywords(text, minFreq=0, minLen=2)
text.scan(/(?:\b)[a-z'-]{#{minLen},}(?=\b)/i).
inject(Hash.new(0)) do |result,w|
w.downcase!
result[w] += 1 unless CommonWords.include?(w)
result
end.select { |k,n| n >= minFreq }
end
Test Rig
require 'net/http'
keywords = get_keywords(Net::HTTP.get('www.sampsonresume.com','/labs/c.txt'), 3)
keywords.sort.each { |name,count| puts "#{name} x #{count} times" }
Test Results
code x 4 times
declarations x 4 times
each x 3 times
execution x 3 times
expression x 4 times
function x 5 times
keywords x 3 times
language x 3 times
languages x 3 times
new x 3 times
operators x 4 times
programming x 3 times
statement x 7 times
statements x 4 times
such x 3 times
types x 3 times
variables x 3 times
which x 4 times
C# 3.0 (with LINQ)
Here's my solution. It makes use of some pretty nice features of LINQ/extension methods to keep the code short.
public static Dictionary<string, int> GetKeywords(string text, int minCount, int minLength)
{
var commonWords = new string[] { "and", "is", "the", "as", "of", "to", "or", "in",
"for", "by", "an", "be", "may", "has", "can", "its"};
var words = Regex.Replace(text.ToLower(), #"[,.?\/;:\(\)]", string.Empty).Split(' ');
var occurrences = words.Distinct().Except(commonWords).Select(w =>
new { Word = w, Count = words.Count(s => s == w) });
return occurrences.Where(wo => wo.Count >= minCount && wo.Word.Length >= minLength)
.ToDictionary(wo => wo.Word, wo => wo.Count);
}
This is however far from the most efficient method, being O(n^2) with the number of words, rather than O(n), which is optimal in this case I believe. I'll see if I can creater a slightly longer method that is more efficient.
Here are the results of the function run on the sample text (min occurences: 3, min length: 2).
3 x such
4 x code
4 x which
4 x declarations
5 x function
4 x statements
3 x new
3 x types
3 x keywords
7 x statement
3 x language
3 x expression
3 x execution
3 x programming
4 x operators
3 x variables
And my test program:
static void Main(string[] args)
{
string sampleText;
using (var client = new WebClient())
sampleText = client.DownloadString("http://sampsonresume.com/labs/c.txt");
var keywords = GetKeywords(sampleText, 3, 2);
foreach (var entry in keywords)
Console.WriteLine("{0} x {1}", entry.Value.ToString().PadLeft(3), entry.Key);
Console.ReadKey(true);
}
#! perl
use strict;
use warnings;
while (<>) {
for my $word (split) {
$words{$word}++;
}
}
for my $word (keys %words) {
print "$word occurred $words{$word} times.";
}
That's the simple form. If you want sorting, filtering, etc.:
while (<>) {
for my $word (split) {
$words{$word}++;
}
}
for my $word (keys %words) {
if ((length($word) >= $MINLEN) && ($words{$word) >= $MIN_OCCURRENCE) {
print "$word occurred $words{$word} times.";
}
}
You can also sort the output pretty easily:
...
for my $word (keys %words) {
if ((length($word) >= $MINLEN) && ($words{$word) >= $MIN_OCCURRENCE) {
push #output, "$word occurred $words{$word} times.";
}
}
$re = qr/occurred (\d+) /;
print sort {
$a = $a =~ $re;
$b = $b =~ $re;
$a <=> $b
} #output;
A true Perl hacker will easily get these on one or two lines each, but I went for readability.
Edit: this is how I would rewrite this last example
...
for my $word (
sort { $words{$a} <=> $words{$b} } keys %words
){
next unless length($word) >= $MINLEN;
last unless $words{$word) >= $MIN_OCCURRENCE;
print "$word occurred $words{$word} times.";
}
Or if I needed it to run faster I might even write it like this:
for my $word_data (
sort {
$a->[1] <=> $b->[1] # numerical sort on count
} grep {
# remove values that are out of bounds
length($_->[0]) >= $MINLEN && # word length
$_->[1] >= $MIN_OCCURRENCE # count
} map {
# [ word, count ]
[ $_, $words{$_} ]
} keys %words
){
my( $word, $count ) = #$word_data;
print "$word occurred $count times.";
}
It uses map for efficiency,
grep to remove extra elements,
and sort to do the sorting, of course. ( it does so it in that order )
This is a slight variant of the Schwartzian transform.
Another Python solution, at 247 chars. The actual code is a single line of highly dense Python line of 134 chars that computes the whole thing in a single expression.
x=3;y=2;W="and is the as of to or in for by an be may has can its".split()
from itertools import groupby as gb
d=dict((w,l)for w,l in((w,len(list(g)))for w,g in
gb(sorted(open("c.txt").read().lower().split())))
if l>x and len(w)>y and w not in W)
A much longer version with plenty of comments for you reading pleasure:
# High and low count boundaries.
x = 3
y = 2
# Common words string split into a list by spaces.
Words = "and is the as of to or in for by an be may has can its".split()
# A special function that groups similar strings in a list into a
# (string, grouper) pairs. Grouper is a generator of occurences (see below).
from itertools import groupby
# Reads the entire file, converts it to lower case and splits on whitespace
# to create a list of words
sortedWords = sorted(open("c.txt").read().lower().split())
# Using the groupby function, groups similar words together.
# Since grouper is a generator of occurences we need to use len(list(grouper))
# to get the word count by first converting the generator to a list and then
# getting the length of the list.
wordCounts = ((word, len(list(grouper))) for word, grouper in groupby(sortedWords))
# Filters the words by number of occurences and common words using yet another
# list comprehension.
filteredWordCounts = ((word, count) for word, count in wordCounts if word not in Words and count > x and len(word) > y)
# Creates a dictionary from the list of tuples.
result = dict(filteredWordCounts)
print result
The main trick here is using the itertools.groupby function to count the occurrences on a sorted list. Don't know if it really saves characters, but it does allow all the processing to happen in a single expression.
Results:
{'function': 4, 'operators': 4, 'declarations': 4, 'which': 4, 'statement': 5}
C# code:
IEnumerable<KeyValuePair<String, Int32>> ProcessText(String text, int X, int Y)
{
// common words, that will be ignored
var exclude = new string[] { "and", "is", "the", "as", "of", "to", "or", "in", "for", "by", "an", "be", "may", "has", "can", "its" }.ToDictionary(word => word);
// regular expression to find quoted text
var regex = new Regex("\"[^\"]\"", RegexOptions.Compiled);
return
// remove quoted text (it will be processed later)
regex.Replace(text, "")
// remove case dependency
.ToLower()
// split text by all these chars
.Split(".,'\\/[]{}()`~##$%^&*-=+?!;:<>| \n\r".ToCharArray())
// add quoted text
.Concat(regex.Matches(text).Cast<Match>().Select(match => match.Value))
// group words by the word and count them
.GroupBy(word => word, (word, words) => new KeyValuePair<String, Int32>(word, words.Count()))
// apply filter(min word count and word length) and remove common words
.Where(pair => pair.Value >= X && pair.Key.Length >= Y && !exclude.ContainsKey(pair.Key));
}
Output for ProcessText(text, 3, 2) call:
3 x languages
3 x such
4 x code
4 x which
3 x based
3 x each
4 x declarations
5 x function
4 x statements
3 x new
3 x types
3 x keywords
3 x variables
7 x statement
4 x expression
3 x execution
3 x programming
3 x operators
In C#:
Use LINQ, specifically groupby, then filter by group count, and return a flattened (selectmany) list.
Use LINQ, filter by length.
Use LINQ, filter with 'badwords'.Contains.
REBOL
Verbose, perhaps, so definitely not a winner, but gets the job done.
min-length: 0
min-count: 0
common-words: [ "a" "an" "as" "and" "are" "by" "for" "from" "in" "is" "it" "its" "the" "of" "or" "to" "until" ]
add-word: func [
word [string!]
/local
count
letter
non-letter
temp
rules
match
][
; Strip out punctuation
temp: copy {}
letter: charset [ #"a" - #"z" #"A" - #"Z" #" " ]
non-letter: complement letter
rules: [
some [
copy match letter (append temp match)
|
non-letter
]
]
parse/all word rules
word: temp
; If we end up with nothing, bail
if 0 == length? word [
exit
]
; Check length
if min-length > length? word [
exit
]
; Ignore common words
ignore:
if find common-words word [
exit
]
; OK, its good. Add it.
either found? count: select words word [
words/(word): count + 1
][
repend words [word 1]
]
]
rules: [
some [
{"}
copy word to {"} (add-word word)
{"}
|
copy word to { } (add-word word)
{ }
]
end
]
words: copy []
parse/all read %c.txt rules
result: copy []
foreach word words [
if string? word [
count: words/:word
if count >= min-count [
append result word
]
]
]
sort result
foreach word result [ print word ]
The output is:
act
actions
all
allows
also
any
appear
arbitrary
arguments
assign
assigned
based
be
because
been
before
below
between
braces
branches
break
builtin
but
C
C like any other language has its blemishes Some of the operators have the wrong precedence some parts of the syntax could be better
call
called
calls
can
care
case
char
code
columnbased
comma
Comments
common
compiler
conditional
consisting
contain
contains
continue
control
controlflow
criticized
Cs
curly brackets
declarations
define
definitions
degree
delimiters
designated
directly
dowhile
each
effect
effects
either
enclosed
enclosing
end
entry
enum
evaluated
evaluation
evaluations
even
example
executed
execution
exert
expression
expressionExpressions
expressions
familiarity
file
followed
following
format
FORTRAN
freeform
function
functions
goto
has
high
However
identified
ifelse
imperative
include
including
initialization
innermost
int
integer
interleaved
Introduction
iterative
Kernighan
keywords
label
language
languages
languagesAlthough
leave
limit
lineEach
loop
looping
many
may
mimicked
modify
more
most
name
needed
new
next
nonstructured
normal
object
obtain
occur
often
omitted
on
operands
operator
operators
optimization
order
other
perhaps
permits
points
programmers
programming
provides
rather
reinitialization
reliable
requires
reserve
reserved
restrictions
results
return
Ritchie
say
scope
Sections
see
selects
semicolon
separate
sequence
sequence point
sequential
several
side
single
skip
sometimes
source
specify
statement
statements
storage
struct
Structured
structuresAs
such
supported
switch
syntax
testing
textlinebased
than
There
This
turn
type
types
union
Unlike
unspecified
use
used
uses
using
usually
value
values
variable
variables
variety
which
while
whitespace
widespread
will
within
writing
Python (258 chars as is, including 66 chars for first line and 30 chars for punctuation removal) :
W="and is the as of to or in for by an be may has can its".split()
x=3;y=2;d={}
for l in open('c.txt') :
for w in l.lower().translate(None,',.;\'"!()[]{}').split() :
if w not in W: d[w]=d.get(w,0)+1
for w,n in d.items() :
if n>y and len(w)>x : print n,w
output :
4 code
3 keywords
3 languages
3 execution
3 each
3 language
4 expression
4 statements
3 variables
7 statement
5 function
4 operators
4 declarations
3 programming
4 which
3 such
3 types
Here is my variant, in PHP:
$str = implode(file('c.txt'));
$tok = strtok($str, " .,;()\r\n\t");
$splitters = '\s.,\(\);?:'; // string splitters
$array = preg_split( "/[" . $splitters . "]*\\\"([^\\\"]+)\\\"[" . $splitters . "]*|[" . $splitters . "]+/", $str, 0, PREG_SPLIT_DELIM_CAPTURE );
foreach($array as $key) {
$res[$key] = $res[$key]+1;
}
$splitters = '\s.,\(\)\{\};?:'; // string splitters
$array = preg_split( "/[" . $splitters . "]*\\\"([^\\\"]+)\\\"[" . $splitters . "]*|[" . $splitters . "]+/", $str, 0, PREG_SPLIT_DELIM_CAPTURE );
foreach($array as $key) {
$res[$key] = $res[$key]+1;
}
unset($res['the']);
unset($res['and']);
unset($res['to']);
unset($res['of']);
unset($res['by']);
unset($res['a']);
unset($res['as']);
unset($res['is']);
unset($res['in']);
unset($res['']);
arsort($res);
//var_dump($res); // concordance
foreach ($res AS $word => $rarity)
echo $word . ' <b>x</b> ' . $rarity . '<br/>';
foreach ($array as $word) { // words longer than n (=5)
// if(strlen($word) > 5)echo $word.'<br/>';
}
And output:
statement x 7
be x 7
C x 5
may x 5
for x 5
or x 5
The x 5
as x 5
expression x 4
statements x 4
code x 4
function x 4
which x 4
an x 4
declarations x 3
new x 3
execution x 3
types x 3
such x 3
variables x 3
can x 3
languages x 3
operators x 3
end x 2
programming x 2
evaluated x 2
functions x 2
definitions x 2
keywords x 2
followed x 2
contain x 2
several x 2
side x 2
most x 2
has x 2
its x 2
called x 2
specify x 2
reinitialization x 2
use x 2
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var_dump statement simply displays concordance. This variant preserves double-quoted expressions.
For supplied file this code finishes in 0.047 seconds. Though larger file will consume lots of memory (because of file function).
This is not going to win any golfing awards but it does keep quoted phrases together and takes into account stop words (and leverages CPAN modules Lingua::StopWords and Text::ParseWords).
In addition, I use to_S from Lingua::EN::Inflect::Number to count only the singular forms of words.
You might also want to look at Lingua::CollinsParser.
#!/usr/bin/perl
use strict; use warnings;
use Lingua::EN::Inflect::Number qw( to_S );
use Lingua::StopWords qw( getStopWords );
use Text::ParseWords;
my $stop = getStopWords('en');
my %words;
while ( my $line = <> ) {
chomp $line;
next unless $line =~ /\S/;
next unless my #words = parse_line(' ', 1, $line);
++ $words{to_S $_} for
grep { length and not $stop->{$_} }
map { s!^[[:punct:]]+!!; s![[:punct:]]+\z!!; lc }
#words;
}
print "=== only words appearing 4 or more times ===\n";
print "$_ : $words{$_}\n" for sort {
$words{$b} <=> $words{$a}
} grep { $words{$_} > 3 } keys %words;
print "=== only words that are 12 characters or longer ===\n";
print "$_ : $words{$_}\n" for sort {
$words{$b} <=> $words{$a}
} grep { 11 < length } keys %words;
Output:
=== only words appearing 4 or more times ===
statement : 11
function : 7
expression : 6
may : 5
code : 4
variable : 4
operator : 4
declaration : 4
c : 4
type : 4
=== only words that are 12 characters or longer ===
reinitialization : 2
control-flow : 1
sequence point : 1
optimization : 1
curly brackets : 1
text-line-based : 1
non-structured : 1
column-based : 1
initialization : 1