String Re-ordering
March 27, 2015
Today’s exercise is a fun little interview question:
You are given a string O that specifies the desired ordering of letters in a target string T. For example, given string O = “eloh” the target string T = “hello” would be re-ordered “elloh” and the target string T = “help” would be re-ordered “pelh” (letters not in the order string are moved to the beginning of the output in some unspecified order).
Your task is to write a program that produces the requested string re-ordering. 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 
A solution using PHP >= 5.3: http://3v4l.org/vnoC1
In Python.
import string # re_order builds a list of characters that are used to sort. o is added at the # end of the list, such that all other characters are at front of the result. # lowercase only def re_order(o, t): alphabet = list(string.ascii_lowercase) for c in o: try: v = alphabet.remove(c) except: pass sortlist = alphabet + list(o) new_list = sorted(t, key=lambda word: [sortlist.index(c) for c in word]) return "".join(new_list) assert re_order('eloh', 'hello') == 'elloh' assert re_order('eloh', 'help') == 'pelh'Haskell:
import Data.List f s o = sortBy (\a b -> compare (e a) (e b)) s where e x = maybe (-1) id (elemIndex x o)In Python the string.find() method returns the index of a character or -1 if a letter is not found in the string.
def reorder(t, o):
return ”.join(sorted(t, key=o.find))
This time with formatting:
An alternative Python solution.
The OrderedCounter counts the number of each character seen in the input, and remembers the order in which they are first seen.
Then each character in the ‘order’ string is moved to the end of the counter.
OrderedCounter.elements() returns each character repeated the number of times it was seen.
from collections import Counter, OrderedDict class OrderedCounter(Counter, OrderedDict): pass def reorder(o, t): oc = OrderedCounter(t) for ch in o: if ch in oc: oc.move_to_end(ch) return ''.join(oc.elements()) reorder('eloh', 'alphabravocharliedeltaecho') # ---> 'aaaaapbrrvccidteeellloohhh'Bucket sort a nice idea. Here’s an in-place solution, scan the first string recording which characters are present, then scan the second string, moving uncounted characters to the front, then scan the first again, copying the right number of characters into the second. I’ve ignored problems with using a char to index an array, (don’t do this at work):
#include <stdio.h> int counts[256]; bool needed[256]; int main(int argc, char *argv[]) { char *s, *t; for (s = argv[1]; *s; s++) { needed[*s] = true; } for (t = s = argv[2]; *s; s++) { if (needed[*s]) { counts[*s]++; } else { *t++ = *s; } } for (s = argv[1]; *s; s++) { while (counts[*s]) { counts[*s]--; *t++ = *s; } } printf("%s\n",argv[2]); }With Python list comprehensions, first gather the characters not in the ordering string and then the ones from the ordering string in their order.
def substr(ostr,tstr):
wstr = [x for x in tstr if x not in ostr]
for x in ostr:
wstr += [y for y in tstr if y == x]
return ”.join(wstr)
if __name__ == ‘__main__’:
ostr = “eloh”
tstr = “hello”
print substr(ostr,tstr)
A perl solution:
use Test::More;
is reorder(“hello”, “eloh”), “elloh”;
is reorder(“help”, “eloh”), “pelh”;
sub reorder {
my ($target, $order) = @_;
return join ”, sort { index($order,$a) index($order,$b) } split //, $target;
}
done_testing;
I have ported the given scheme solution using decorate-sort-undecorate algorithm to java:
import java.util.ArrayList; import java.util.Collections; import java.util.HashMap; import java.util.Iterator; import java.util.Set; public class DecorateSort { public static void main(String[] args) { order("heleo", "eloh"); order("help", "eloh"); } private static void order(String input, String output) { int index; Set<Integer> rank; ArrayList<Integer> listRank; StringBuffer sb = new StringBuffer(); HashMap<Integer, String> rankMap = new HashMap<Integer, String>(); int counter = 0; for(int i = 0;i<input.length();i++) { index = output.indexOf(input.substring(i, i+1)); if(index<0) { counter = counter-1; index = counter; } if(rankMap.get(index) != null) { rankMap.put(index, rankMap.get(index)+input.substring(i, i+1)); } else { rankMap.put(index, input.substring(i, i+1)); } } rank = rankMap.keySet(); listRank = new ArrayList<Integer>(rank); Collections.sort(listRank); Iterator<Integer> it = listRank.iterator(); while(it.hasNext()) { sb.append(rankMap.get(it.next())); } System.out.println(sb.toString()); } }This is a modified version of the solution posted by Krishnan R.
import java.util.*; public class StringReorderingTest{ public static void main(String[] args){ test("eloh", "hello"); test("eloh", "help"); } public static void test(String order, String target){ System.out.println(reorder(order, target)); } public static String reorder(String order, String target){ Map<Character, String> container = new HashMap<>(); for(char letter : target.toCharArray()){ String str = container.get(letter); if(str == null) str = ""; str += letter; container.put(letter, str); } StringBuilder strBuilder = new StringBuilder(); for(char letter : order.toCharArray()){ String str = container.remove(letter); if(str == null) str = ""; strBuilder.append(str); } for(String str : container.values()) strBuilder.insert(0, str); return strBuilder.toString(); } }Bucketsort version:
import java.util.ArrayList; import java.util.Arrays; public class BucketSort { static ArrayList<Character> result = new ArrayList<Character>(); public static void main(String[] args) { order2("hello", "eloh"); System.out.println(); order2("help", "eloh"); } private static void order2(String t, String o) { char[] ts = t.toCharArray(); char[] os = o.toCharArray(); int[] buckets = new int[256]; ArrayList<Character> excess = new ArrayList<Character>(t.length()); ArrayList<Character> result = new ArrayList<Character>(t.length()); for(int i=0;i<256;i++) buckets[i] = -1; for(char c : os) buckets[Character.getNumericValue(c)] = 0; for(char c : ts) { int i = Character.getNumericValue(c); if(buckets[i] < 0) excess.add(c); else buckets[i] = buckets[i]+1; } for(Character c : excess) { result.add(c); } for(char c : os) { char[] x = make_array(buckets[Character.getNumericValue(c)], c); for(char c1 : x) result.add(c1); } StringBuilder builder = new StringBuilder(result.size()); for(Character c : result) { builder.append(c); } System.out.println(builder.toString()); } private static char[] make_array(int os, char c) { char[] output = new char[os]; Arrays.fill(output, c); return output; } }A not particularly efficient but straightforward C++(11) version:
void sort(const std::string& O, std::string& T) { std::sort(T.begin(), T.end(), [&O](char a, char b) { return O.find(a) < O.find(b); }); }Java8:
private static String order(String stringToSort, final String orderString) { return Arrays.stream(stringToSort.split("")) .sorted((o1, o2) -> orderString.indexOf(o1) - orderString.indexOf(o2)) .reduce((t, u) -> t + u) .orElse(null); }