April 28, 2015 9:00 AM
Our first solution sorts the words and compares their “signatures:”
(define (anagram1 str1 str2)
(and (not (string=? str1 str2))
(equal? (sort charlist str1))
(sort charlist str2)))))
> (anagram1 "DEPOSIT" "DOPIEST")
#t
> (anagram1 "STAR" "MOON")
#f
> (anagram1 "ZEBRA" "ZEBRA")
#f
Wherever there is a sorting solution to a problem, there is also generally a searching solution also. Here, we increment an array of character counts when reading the first string, decrement it when reading the second string, and check that all the counts are zero:
(define (anagram2 str1 str2)
(let ((counts (make-vector 256 0)))
(define (add idx n)
(vector-set! counts idx
(+ (vector-ref counts idx) n)))
(do ((cs (string->list str1) (cdr cs))) ((null? cs))
(add (char->integer (car cs)) 1))
(do ((cs (string->list str2) (cdr cs))) ((null? cs))
(add (char->integer (car cs)) -1))
(let loop ((i 65))
(cond ((= i 91) (not (string=? str1 str2)))
((not (zero? (vector-ref counts i))) #f)
(else (loop (+ i 1)))))))
> (anagram2 "DEPOSIT" "DOPIEST")
#t
> (anagram2 "STAR" "MOON")
#f
> (anagram2 "ZEBRA" "ZEBRA")
#f
Our third solution maps each letter to a prime number, takes the product of the primes in each word, and compares them; beware of overflow in languages that don’t provide big integers natively:
(define (anagram3 str1 str2)
(let ((letters (vector 2 3 5 7 11 13 17 19 23 29 31
37 41 43 47 53 59 61 67 71 73 79 83 89 97 101)))
(define (lookup c)
(vector-ref letters (- (char->integer c) 65)))
(and (not (string=? str1 str2))
(= (apply * (map lookup (string->list str1)))
(apply * (map lookup (string->list str2)))))))
> (anagram3 "DEPOSIT" "DOPIEST")
#t
> (anagram3 "STAR" "MOON")
#f
> (anagram3 "ZEBRA" "ZEBRA")
#f
The first solution has time complexity O(n log n) for the sort; the other two are both O(n). But given its simplicity, I would probably prefer the first solution over the other two, changing to the second solution only if the first solution proved to be a bottleneck on my program. You can run the program at http://ideone.com/p1SAvl.
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Haven’t done the prime number version but the first two were the solutions I went for – both in perl tho’
sub m1 { my( $a,$b ) = @_; return (join q(), sort split q(),$a) eq (join q(), sort split q(),$b) ? 1 : 0 } sub m2 { my($a,$b)=@_; my %x; $x{$_}++ foreach split q(), $a; $x{$_}-- foreach split q(), $b; return 0 if grep { $_ } values %x; return 1; } print m1(@ARGV),"\n"; print m2(@ARGV),"\n";By James Curtis-Smith on April 28, 2015 at 9:12 AM
In Python.
from collections import Counter def is_anagram(a, b): return sorted(a) == sorted(b) def is_anagram(a, b): return Counter(a) == Counter(b)By Paul on April 28, 2015 at 10:47 AM
Same solutions as Paul in Python. Added additional method using mapping onto primes and taking the product.
By Rutger on April 28, 2015 at 1:52 PM
First version creates a bag out of each string and compares the bags for equality. The second solution removes all characters in the first string from the second string and checks that it results in the empty string, and vice versa.
By svenningsson on April 28, 2015 at 3:53 PM
Just one solution for the moment: a variant on the sorting method using two heaps. We save on work in the case that the two strings aren’t anagrams:
#include <algorithm> #include <iostream> #include <string.h> bool isanag(char *a, char *b) { size_t n = strlen(a); if (n != strlen(b)) return false; std::make_heap(a,a+n); std::make_heap(b,b+n); for ( ; n > 0; n--) { if (a[0] != b[0]) return false; std::pop_heap(a,a+n); std::pop_heap(b,b+n); } return true; } int main(int argc, char *argv[]) { std::cout << isanag(argv[1],argv[2]) << "\n"; }By matthew on April 28, 2015 at 9:27 PM
Here’s another one: generate all possible anagrams of a, and see if any are equal to b:
#include <algorithm> #include <iostream> #include <string.h> bool isanag(char *a, const char *b) { size_t n = strlen(a); std::sort(a,a+n); while(true) { if (strcmp(a,b) == 0) return true; if (!std::next_permutation(a,a+n)) return false; } } int main(int argc, char *argv[]) { std::cout << isanag(argv[1],argv[2]) << "\n"; }Not the most efficient way of solving the problem, but has a pleasant simplicity about it (and isanag only destructively modifies one of its arguments now).
By matthew on April 28, 2015 at 9:54 PM
Last one, pack the counts into a single 64 bit number. There are only 2 bits per character so eg. “AAAAC” is considered an anagram of “BBBBB”, but we always correctly detect a true anagram:
#include <iostream> #include <stdint.h> typedef uint64_t T; T sign(char *a) { T s = 0; for ( ; *a; a++) { s += T(1)<<(2*(*a-'A')); } return s; } int main(int argc, char *argv[]) { std::cout << (sign(argv[1]) == sign(argv[2])) << "\n"; }By matthew on April 28, 2015 at 10:48 PM
Your second solution doesn’t work too well on a Unicode system: you’d need an array of size 1,114,112. However, a hash table or similar map from characters to integers is the same in spirit.
By John Cowan on April 29, 2015 at 2:30 AM
#include <map>
#include <string>
#include <algorithm>
#include <iostream>
void normalize(std::string& s)
{
s.erase(s.begin(), std::find_if_not(s.begin(), s.end(), ::isspace));
s.erase(std::find_if_not(s.rbegin(), s.rend(), ::isspace).base(), s.end());
std::transform(s.begin(), s.end(), s.begin(), ::toupper);
}
bool common(std::string& a, std::string& b)
{
normalize(a);
normalize(b);
if (a.size() != b.size()) return false;
if (a == b) return false;
return true;
}
std::map<char, int> analyze(std::string word)
{
std::map<char, int> data;
for (auto c: word) {
auto iter = data.find(c);
if (data.end() == iter) data[c] = 1;
else ++(iter->second);
}
return data;
}
bool are_anagrams1(std::string a, std::string b)
{
if (!common(a, b)) return false;
return analyze(a) == analyze(b);
}
bool are_anagrams2(std::string a, std::string b)
{
if (!common(a, b)) return false;
std::sort(a.begin(), a.end());
std::sort(b.begin(), b.end());
return a == b;
}
void test(const std::string& a, const std::string& b, std::ostream& out)
{
out << a << " and " << b << ": " <<
are_anagrams1(a, b) << ", " <<
are_anagrams2(a, b) << ‘\n’;
}
int main(int argc, char** argv)
{
std::cout.setf(std::ios_base::boolalpha);
test("deposit ", " dopiest", std::cout);
test("STOP", "pots", std::cout);
test("rite", "write", std::cout);
test("right", "write", std::cout);
test("same", "same", std::cout);
}
By fisherro on April 29, 2015 at 2:34 PM
Sorry for the bad formatting in my previous post. Is there a way to delete it?
#include <map> #include <string> #include <algorithm> #include <iostream> void normalize(std::string& s) { s.erase(s.begin(), std::find_if_not(s.begin(), s.end(), ::isspace)); s.erase(std::find_if_not(s.rbegin(), s.rend(), ::isspace).base(), s.end()); std::transform(s.begin(), s.end(), s.begin(), ::toupper); } bool common(std::string& a, std::string& b) { normalize(a); normalize(b); if (a.size() != b.size()) return false; if (a == b) return false; return true; } std::map<char, int> analyze(std::string word) { std::map<char, int> data; for (auto c: word) { auto iter = data.find(c); if (data.end() == iter) data[c] = 1; else ++(iter->second); } return data; } bool are_anagrams1(std::string a, std::string b) { if (!common(a, b)) return false; return analyze(a) == analyze(b); } bool are_anagrams2(std::string a, std::string b) { if (!common(a, b)) return false; std::sort(a.begin(), a.end()); std::sort(b.begin(), b.end()); return a == b; } void test(const std::string& a, const std::string& b, std::ostream& out) { out << a << " and " << b << ": " << are_anagrams1(a, b) << ", " << are_anagrams2(a, b) << '\n'; } int main(int argc, char** argv) { std::cout.setf(std::ios_base::boolalpha); test("deposit ", " dopiest", std::cout); test("STOP", "pots", std::cout); test("rite", "write", std::cout); test("right", "write", std::cout); test("same", "same", std::cout); }By fisherro on April 29, 2015 at 2:41 PM
def anagram?(a, b) dict = ('a'..'z').each_with_object({}) { |v,h| h[v] = 0 } a.split('').each{|let| dict[let] += 1} b.split('').each{|let| dict[let] -= 1} dict.values.select{|v| v != 0}.empty? end a, b = 'stop', 'pots' puts anagram?(a, b) # Approach 1 puts a.chars.sort.join == b.chars.sort.join # Approach 2By Scott on May 1, 2015 at 6:45 PM
My discussion and solution in Java here http://www.capacode.com/?p=7
By truongkhanhit on May 9, 2015 at 3:11 PM
fun merge lt (xs, ys) = let fun loop(out, [], []) = List.rev out | loop(out, x::xs, []) = loop (x::out, xs, []) | loop(out, [], y::ys) = loop (y::out, [], ys) | loop(out, left as x::xs, right as y::ys) = if lt (x, y) then loop (x::out, xs, right) else loop (y::out, left, ys) in loop([], xs, ys) end fun mergesort lt xs = let val merge' = merge lt (* splits a list into two semi-equal halves in Linear time *) fun split ns = let fun loop([], xs, ys) = (xs, ys) | loop(x::y::zs, xs, ys) = loop(zs, x::xs, y::ys) | loop(x::[], xs, ys) = loop([], x::xs, ys) in loop(List.rev(ns), [], []) end fun ms [] = [] | ms [x] = [x] | ms xs = let val (l, r) = split xs in merge'(ms l, ms r) end in ms xs end fun anagram1 a b = if a = b then true else let val a_list = explode a val b_list = explode b val ms = mergesort (op <) in (ms a_list) = (ms b_list) end fun anagram2 a b = let val array = Array.tabulate(128, fn _ => 0) in CharVector.app (fn char => Array.update (array, ord char, Array.sub (array, ord char) + 1)) a; CharVector.app (fn char => Array.update (array, ord char, Array.sub (array, ord char) - 1)) b; Array.all (fn e => e = 0) array endBy mcmillhj on May 12, 2015 at 6:01 PM
By wert310 on October 25, 2015 at 6:59 PM