Find The Difference
January 15, 2019
Today’s question comes from a programming student:
You are given two strings s and t that consist only of lower-case letters. String t was created by adding one letter chosen at random to string s, then shuffling the resulting string. Find the letter that was added to t. For instance, the difference between the two strings “abcdef” and “bfxdeac” is the character “x”.
Your task is to write a program to find the random letter that was added to the string. 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.
Programming Praxis 
Scala: (t.toSet–(s.toSet)).iterator.next //> res0: Char = x
Nice one for perl… in the case we are only looking for one known difference it is just finding the letter in the two strings which appears an odd number of times….
use feature 'say'; my( $x, $y, %f ) = ( 'abcdef', 'bfxdeac' ); $f{$_}=!$f{$_} foreach split //, $x.$y; say grep {$f{$_}} keys %f;'If you were looking for all added/removed letters you could change this to:
use feature 'say'; my( $x, $y, %f ) = ( 'abcdef', 'bfxdeac' ); $f{$_}++ foreach split //, $x; $f{$_}-- foreach split //, $y; say grep {$f{$_}} keys %f;'The %f contains +ve values for letters in $x but not $y and then -ve values for letters in $y but not in $x if that makes sense (and the absolute value is the count of each).
Cool one for basic text processing. Here is my take with Julia 1.0.2:
function StringDifference(s::AbstractString, t::AbstractString)
S = sort(split(s, “”))
T = sort(split(t, “”))
end
#include
#include
#include
using namespace std;
int main()
{
string s1;
string s2;
string s3;
}
Fun exercise! I wouldn’t have thought of the xor on my own.
I also didn’t trust myself to always put the shorter string first, so my solutions allow for supplying the strings in either order.
module Main where
import Data.Bits (xor)
import Data.Char (chr, ord)
import Data.Foldable (foldl')
import qualified Data.Set as S
data Solution a b = S { _in :: String -> a
, _compute :: a -> a -> b
, _out :: b -> Char
}
run :: Solution a b -> String -> String -> Char
run (S i c o) x y = o $ c (i x) (i y)
s1 :: Solution (S.Set Char) (S.Set Char)
s1 =
S S.fromList (\x y -> (x
S.uniony) S.\\ (xS.intersectiony)) S.findMins2 :: Solution Int Int
s2 = S (sum . fmap ord) (-) (chr . abs)
s3 :: Solution Int Int
s3 = S (foldl' xor 0 . fmap ord) xor chr
main :: IO ()
main = do
let x = "abcdef"
y = "bfxdeac"
print $ all (== 'x') [run s1 x y, run s2 x y, run s3 x y]
Sorry, I tried to HTML-escape & post in code tags, but that didn’t work right. Retrying with the WordPress sourcecode attribute:
module Main where import Data.Bits (xor) import Data.Char (chr, ord) import Data.Foldable (foldl') import qualified Data.Set as S data Solution a b = S { _in :: String -> a , _compute :: a -> a -> b , _out :: b -> Char } run :: Solution a b -> String -> String -> Char run (S i c o) x y = o $ c (i x) (i y) s1 :: Solution (S.Set Char) (S.Set Char) s1 = S S.fromList (\x y -> (x `S.union` y) S.\\ (x `S.intersection` y)) S.findMin s2 :: Solution Int Int s2 = S (sum . fmap ord) (-) (chr . abs) s3 :: Solution Int Int s3 = S (foldl' xor 0 . fmap ord) xor chr main :: IO () main = do let x = "abcdef" y = "bfxdeac" print $ all (== 'x') [run s1 x y, run s2 x y, run s3 x y]Clearly, it’s been too long since I visited!
More syntax woes, apologies.
My answer in Python
s = ‘abcdef’
t = ‘cbdefxa’
for c in t:
if c not in s:
print(c)
My C++ is in danger of getting rusty, so let’s use it for a solution. See https://godbolt.org/z/qgGfAe for how the main function compiles.
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <numeric> #include <functional> template <typename Iter> auto xorseq(const Iter &from, const Iter &to) { return std::accumulate(from,to,0,std::bit_xor<>()); } template <typename Iter> auto xordiff(const Iter &from0, const Iter &to0, const Iter &from1, const Iter &to1) { return xorseq(from0,to0)^xorseq(from1,to1); } auto xordiff(const char *s0, const char *s1) { return xordiff(s0,s0+strlen(s0),s1,s1+strlen(s1)); } int main(int argc, char *argv[]) { if (argc != 3) abort(); printf("%c\n",xordiff(argv[1],argv[2])); }That needs C++14 features, so build with eg:
Python solution using sets (similar to Haskell s1 solution above).
Three Python solutions.
Python solution.
Works when s or t have duplicate letters (set solutions don’t).
Stops when the new letter is found, so it may not process the entire string.
def diff(s, t): return next(b for a,b in zip(s+' ',t) if a!=b)@Mike: doesn’t that assume that the characters haven’t been shuffled? Good point on use of sets though.
Here’s a more streamlined version of my C++ solution:
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <numeric> #include <functional> template <typename Iter> auto xorseq(const Iter &from, const Iter &to, typename std::iterator_traits<Iter>::value_type init = 0) { return std::accumulate(from,to,init,std::bit_xor<>()); } auto xordiff(const char *s0, const char *s1) { return xorseq(s1,s1+strlen(s0),xorseq(s0,s0+strlen(s0),0)); } int main(int argc, char *argv[]) { if (argc != 3) abort(); printf("%c\n",xordiff(argv[1],argv[2])); }Another Haskell version.
import Data.List ((\\), find, sort) import Data.Maybe (fromMaybe) -- Return the extra elements that are in one list, but not in the other. diff :: Ord a => [a] -> [a] -> [a] diff xs ys = let (xs', ys') = (sort xs, sort ys) in fromMaybe [] $ find (not . null) [xs' \\ ys', ys' \\ xs'] main :: IO () main = do putStrLn $ diff "abcdef" "bfxdeac" -- the random letter is a duplicate (here 'e') putStrLn $ diff "abcdef" "bfedeac"@Graham, I use the following shell script to create comments here.
#!/bin/sh -eu case $# in 0) echo "Usage: ${0##*/} path-to-Haskell-source [command]" 1>&2; exit 1;; 1) src=$1; shift; cmd=;; *) src=$1; shift; cmd=$@;; esac cat <<EOF [code] $(cat "$src")EOF
test -z "$cmd" || {
echo "
" echo \$ "$cmd" sh -c "$cmd" echo ""
}
[/code]
I run it like
then just paste the output in the comment widget.
LOL! I feel your pain. :-)
Here are a few approaches in C. The table-driven approach was my initial implementation. It counts each letter and then compares the counts. The other approaches were inspired by the solution: an XOR approach, a sum approach, and a sorting approach.
#include <assert.h> #include <stdio.h> #include <stdlib.h> #include <string.h> char find_letter_table(char* s, char* t) { int* s_counts = calloc(26, sizeof(int)); int* t_counts = calloc(26, sizeof(int)); for (char c = *s; c; c = *(++s)) ++(s_counts[c - 'a']); for (char c = *t; c; c = *(++t)) ++(t_counts[c - 'a']); for (int i = 0; i < 26; ++i) if (t_counts[i] > s_counts[i]) return i + 'a'; return '\0'; } char find_letter_xor(char* s, char* t) { char result = '\0'; for (char c = *s; c; c = *(++s)) result ^= c; for (char c = *t; c; c = *(++t)) result ^= c; return result; } char find_letter_sum_diff(char* s, char* t) { int result = 0; for (char c = *t; c; c = *(++t)) result += c; for (char c = *s; c; c = *(++s)) result -= c; return (char)result; } int compare(const void* a, const void* b) { char a_char = *(char*)a; char b_char = *(char*)b; return (int)a_char - (int)b_char; } // Destructively modifies inputs. char find_letter_sort(char* s, char* t) { int n = strlen(s); qsort(s, n, sizeof(char), compare); qsort(t, n + 1, sizeof(char), compare); while (*s && *s == *t) { ++s; ++t; } return *t; } int main(int argc, char* argv[]) { if (argc != 3) { fprintf(stderr, "Usage: %s ARG1 ARG2\n", argv[0]); return EXIT_FAILURE; } char* s = argv[1]; char* t = argv[2]; char c_table = find_letter_table(s, t); char c_xor = find_letter_xor(s, t); char c_sum_diff = find_letter_sum_diff(s, t); char c_sort = find_letter_sort(s, t); assert(c_table == c_xor); assert(c_table == c_sum_diff); assert(c_table == c_sort); printf("%c\n", c_table); return EXIT_SUCCESS; }Example:
I forgot to free my memory allocations :-(.
Here’s an update.
char find_letter_table(char* s, char* t) { int* s_counts = calloc(26, sizeof(int)); int* t_counts = calloc(26, sizeof(int)); for (char c = *s; c; c = *(++s)) ++(s_counts[c - 'a']); for (char c = *t; c; c = *(++t)) ++(t_counts[c - 'a']); char result = '\0'; for (int i = 0; i < 26; ++i) { if (t_counts[i] > s_counts[i]) { result = i + 'a'; break; } } free(s_counts); free(t_counts); return result; }The fastest solution I could find in Python.
from collections import Counter def diff1(s, t): d = Counter(t) - Counter(s) return next(iter(d.keys()))Prolog solution here (not optimized):
Then just
?- difference("abcdef","bfxdeac",Diff).