alternative solutions to this programming problem? all languages welcome

Question

Create a method that:

Given a string, look for a mirror image (backwards) string at both the beginning and end of the given string. In other words, zero or more characters at the very beginning of the given string, and at the very end of the string in reverse order (possibly overlapping). For example, the string "abXYZba" has the mirror end "ab".

Examples

mirrorEnds("abXYZba") → "ab" 
mirrorEnds("abca") → "a"  
mirrorEnds("aba") → "aba"

Java solution:

public String mirrorEnds(String string) 
{
  boolean matches = true;
  StringBuilder mirror = new StringBuilder();
  int i = 0;
  int length = pString.length();

  while (matches && i < length) 
  {
    if (pString.charAt(i) == pString.charAt(length-i-1))
       mirror.append(pString.charAt(i));
    else
       matches = false;

    i++; 
   }

   String str = mirror.toString(); 
   return str;
}

Answer 1

Ruby

class String
  def mirror_ends
    chars.zip(chars.reverse_each).take_while {|a, b| a == b }.map(&:first).join
  end
end

Testsuite:

require 'test/unit'
class TestMirrorEnds < Test::Unit::TestCase
  def test_that_abXYZba_mirror_ends_is_ab
    assert_equal 'ab', 'abXYZba'.mirror_ends
  end
  def test_that_abca_mirror_ends_is_a
    assert_equal 'a', 'abca'.mirror_ends
  end
  def test_that_aba_mirror_ends_is_aba
    assert_equal 'aba', 'aba'.mirror_ends
  end
end

Answer 2

C++

#include <iostream>
#include <string>
#include <algorithm>
std::string mirrorEnds(const std::string& in)
{
    return std::string(in.begin(),
                       mismatch(in.begin(), in.end(), in.rbegin()).first);
}
int main()
{
    std::cout << mirrorEnds("abXYZba") << '\n'
              << mirrorEnds("abca") << '\n'
              << mirrorEnds("aba") << '\n';
}

EDIT: Since people are asking: in C++, any sequence can be represented as a pair of iterators, and any pair of iterators can become a sequence again. Here, the string in is viewed through the pair of iterators in.begin()...in.end() which present a view of the sequence going forward, and the same string in is viewed through a pair of iterators in.rbegin()...in.rend(), which represent a view of the sequence going backwards. std::mismatch() compares the two sequences that are given to it, element by element, until it finds a mismatch, which it returns as an iterator. Then the pair of iterators in.begin()...<return of mismatch> are viewed as a string again, and that string is returned.

Answer 3

Java

public static String mirrorEnds(String s) {
    int i = 0;
    while (i < s.length() && s.charAt(i) == s.charAt(s.length() - 1 - i))
        i++;
    return s.substring(0, i);
}

Answer 4

Scala

def mirrorEnds(s: String) =
  s.zip(s.reverse).takeWhile {p => p._1 == p._2}.map {_._1}.mkString

Answer 5

Python:

from itertools import takewhile
def mirrorEnds(s):
    return ''.join(a for a, b in takewhile(lambda p: p[0] == p[1], zip(s, reversed(s))))

Answer 6

Common Lisp

(defun mirror-ends (string)
  (loop for i upfrom 0
        for j downfrom (1- (length string))
        while (and (<= i j) (char= (aref string i) (aref string j)))
        finally (return (subseq string 0 i))))

Answer 7

C#

public static string MirrorEnds(this string s)
{
    return String.Join(
        "", from p in s.Zip(s.Reverse(), (a, b) => new { a, b }).
            TakeWhile(p => p.a == p.b) select p.a
    );
}

Answer 8

Tcl

The clearest way to do this I can think of is to convert the string into a list and compare with its reverse:

proc mirrorEnds {txt} {
    set x [split $txt {}]
    set y [lreverse $x]

    set mirror ""

    foreach a $x b $y {
        if {$a == $b} {
            append mirror $a
        } else {
            break
        }
    }

    return $mirror
}

Alternatively, you can do the standard iterate-from-both-directions thing. But it's more verbose:

proc mirrorEnds {txt} {
    set mirror ""

    for {set i 0} {$i < [string length $txt]} {incr i} {
        if {[string index $txt $i] == [string index $txt end-$i]} {
            append mirror [string index $txt $i]
        } else {
            break
        }
    }

    return $mirror
}

Answer 9

C#

Func<string, string> mirrorEnds =
    s => new string(s.Reverse().TakeWhile((c, i) => c == s[i]).ToArray());

For completeness, here was my initial take on it:

Func<string, string> mirrorEnds =
    s => new string(s.Zip(s.Reverse(), (a, b) => new { a, b })
                     .TakeWhile(c => c.a == c.b)
                     .Select(c => c.a).ToArray());

I changed it once I saw how close it was to Jorg's solution.

Answer 10

Java

Here's one way.

public static String reversi(String s) {
    StringBuffer sb2 = new StringBuffer(s).reverse();

    int i = 0;
    while (i < s.length() && s.charAt(i) == sb2.charAt(i)) i++;
    return i>0 ? s.substring(0, i) : "";
}

Answer 11

Ruby

Inspired by @Sheldon L. Cooper's Java solution:

class String
  def mirror_ends
    chars.take_while.with_index {|c, i| c == self[-i-1] }.join
  end
end

Testsuite:

require 'test/unit'
class TestMirrorEnds < Test::Unit::TestCase
  def test_that_abXYZba_mirror_ends_is_ab
    assert_equal 'ab', 'abXYZba'.mirror_ends
  end
  def test_that_abca_mirror_ends_is_a
    assert_equal 'a', 'abca'.mirror_ends
  end
  def test_that_aba_mirror_ends_is_aba
    assert_equal 'aba', 'aba'.mirror_ends
  end
end

Answer 12

C

  char *mirrorEnds(char *s)
  {
    char *a;
    for (a=s+strlen(s) ; *s && *s==*(a-1) ; s++,a--);
    return a;
  }

Answer 13

Perl

sub mirrorEnds {
  my ($a)=@_;
  my ($i,$b);
  for $i (1..length $a) {
    if (substr($a,0,$i) eq reverse(substr($a,-$i))) {
      $b = substr($a,0,$i);
    }
  }
  $b
}

Answer 14

Haskell (somebody reformat this for me):

mirrorEnds a = map fst $ takeWhile (uncurry (==)) $ zip a (reverse a)

mirrorEnds a = zipWith const a $ takeWhile id $ zipWith (==) a (reverse a)

I actually like the second better.

Answer 15

C#

Here's a more conventional approach:

static string mirrorEnds(string strIn)
{
    string strOut = "";
    for (int i = 0, j = strIn.Length - 1; j >= 0 && strIn[i] == strIn[j]; i++, --j)
        strOut += strIn[i];
    return strOut;
}

And probably the fastest method:

static string mirrorEnds(string strIn)
{
    int i = 0, j = strIn.Length - 1;
    for (; j >= 0 && strIn[i] == strIn[j]; i++, --j) ;
    return strIn.Substring(0, i);
}

Answer 16

PHP

$rev=str_split(strrev($str));
$mirror='';
foreach($rev as $k=>$v) {
    if($str[$k]==$v) {
        $mirror.=$v;
    } else {
        break;
    }
}
echo $mirror;

Answer 17

---

Delphi

function MirrorEnds(s: string): string;
var
  i, halfway: integer;
begin
  Result := '';
  halfway := Round((Length(s) / 2) - (1/2));  //halfway pt, rounded down
  for i := 1 to halfway do
    if s[Length(s)-i+1] = s[i] then
      Result := Result + s[i];
  if Length(Result) = halfway then
    Result := s;
end;

Or if you prefer a more code-golfy method (i.e., sacrifice performance for brevity) then you can do this:

function MirrorEnds(s: string): string;
var
  i: integer;
begin
  Result := '';
  for i := 1 to Length(s) do
    if ReverseString(s)[i] = s[i] then
      Result := Result + s[i]
    else
      Break;
end;

And finally, a brief test on my machine indicates that the first one is in fact about five times faster than the second one (although both are virtually instantaneous in practice):

Input string:"abcdefghijklmnopponmlkjihgfedcba"
# of iterations: 900000

Brief = 4973 ms
Fast = 1057 ms

Answer 18

Java

alternative

Didn't see the StringUtils version yet...

  String mirrorEnds(String s) {
    return StringUtils.getCommonPrefix(new String[] { s, s.reverse() });
  }

Answer 19

Python

alternative

def mirror_ends(string):
    return ''.join(string[i] for i in range(len(string)//2) if string[i] == string[-i-1])

>>> mirror_ends("abxyzba")
'ab'

Answer 20

XQuery

substring($s,1,min(let $c := string-to-codepoints($s)
                   return for $x at $p in $c
            where $x != $c[last() - $p + 1]
            return $p))

Answer 21

GAWK-SED-SHELL Oneliner

echo string | sed 's/./& /g' | gawk '{ for(j=1; j<NF/2; j++) if($j==$(NF-j+1)) print $j; else break}' | tr -d "\n"

where string is the string to process.

Answer 22

My Java Implementation

  public static String mirrorEnds(String in) {
    StringBuilder sb = new StringBuilder();
    char[] chars = in.toCharArray();
    int length = chars.length - 1;

    for (int i = 0; i < length; i++) {
        if (chars[i] == chars[length - i]) {
            sb.append(chars[i]);
        } else {
            break;
        }
    }

    return sb.toString();
}

Answer 23

Scheme

(define (equal-start? a b)
  (and
    (pair? a)
    (pair? b)
    (equal? (car a) (car b))))

(define (take-while-equal-start a b)
  (if (equal-start? a b)
      (cons
        (car a)
        (take-while-equal-start (cdr a) (cdr b)))
      '()))

(define (mirror-ends str)
  (let ((lst (string->list str)))
    (list->string
      (take-while-equal-start lst (reverse lst)))))

> (mirror-ends "abXYZba")
"ab"
> (mirror-ends "abca")
"a"
> (mirror-ends "aba")
"aba"
> (mirror-ends "")
""
> (mirror-ends "a")
"a"