Related
I was asked this question in an interview.
Given an array of characters, find the shortest word in a dictionary that contains all the characters. Also, propose an implementation for the dictionary that would optimize this function call.
for e.g. char[] chars = { 'R' , 'C' }. The result should be the word "CAR".
I could not come up with anything that would run reasonably quickly. I thought of pre-processing the dictionary by building a hash table to retrieve all words of a particular length. Then I could only think of retrieving all words in the increasing order of length and checking if the required characters were present in any of those ( maybe by using a bitmask . )
This is a common software interview question, and its solution is this: sort the dictionary itself by length and sort each value alphabetically. When given the characters, sort them and find the needed letters.
First sort the dictionary in ascending order of length.
For each letter, construct a bit map of the locations in the dictionary of the words containing that letter. Each bit map will be long, but there will not be many.
For each search, take the intersection of the bitmaps for the letters in the array. The first one bit in the result will be at the index corresponding to the location in the dictionary of the shortest word containing all the letters.
The other answers are better, but I realized this is entirely precomputable.
For each word
sort the letters and remove duplicates
The sequence of letters can be viewed as a bitmask, A=0bit, B=1bit...Z=26bit. Set the bits of a mask A according to the letters in this word.
For each combination of set bits in the mask A, make a subset mask B
If there is already a word associated with this mask B
and this word is shorter, replace the associated word with this one
otherwise try next B
If there is no word associated with mask B
Associate this word with mask B.
This would take a huge amount of setup time, and the subsequent association storage would be in the vicinity of 1.7GB, but you'd be able to find the shortest word containing a superset of the letters in O(1) time guaranteed.
The obvious preprocessing is to sort all words in the dictionary by their length and alphabetical re-ordering: "word" under "dorw", for example. Then you can use general search algorithms (e.g., regex) to search for the letters you need. An efficient (DFA) search requires only one pass over the dictionary in the worst case, and much less if the first match is short.
Here is a solution in C#:
using System.Collections.Generic;
using System.Linq;
public class ShortestWordFinder
{
public ShortestWordFinder(IEnumerable<string> dictionary)
{
this.dictionary = dictionary;
}
public string ShortestWordContaining(IEnumerable<char> chars)
{
var wordsContaining = dictionary.Where(s =>
{
foreach (var c in chars)
{
if (!s.Contains(c))
{
return false;
}
s = s.Remove(s.IndexOf(c), 1);
}
return true;
}).ToList();
if (!wordsContaining.Any())
{
return null;
}
var minLength = wordsContaining.Min(word => word.Length);
return wordsContaining.First(word => word.Length == minLength);
}
private readonly IEnumerable<string> dictionary;
}
Simple test:
using System.Diagnostics;
using Xunit;
public class ShortestWordFinderTests
{
[Fact]
public void Works()
{
var words = new[] { "dog", "moose", "gargoyle" };
var finder = new ShortestWordFinder(words);
Trace.WriteLine(finder.ShortestWordContaining("o"));
Trace.WriteLine(finder.ShortestWordContaining("oo"));
Trace.WriteLine(finder.ShortestWordContaining("oy"));
Trace.WriteLine(finder.ShortestWordContaining("eyg"));
Trace.WriteLine(finder.ShortestWordContaining("go"));
Assert.Null(finder.ShortestWordContaining("ooo"));
}
}
Pre processing
a. Sort words into alphabetic char arrays. Retain mapping from sorted to original word
b. Split dictionary by word length as you suggest
c. Sort entries in each word length set alphabetically
On function call
Sort char array alphabetically
Start with group of same length as array
Loop through entries testing for characters until first letter of entry lexicographically greater than first in your char array then break. If match then return original word (see a above for mapping)
Back to 2 for next longest word group
Interesting extensions. Multiple words might map to same entry in a. Which one (s) should you return...
Given a string array of variable length, print the lengths of each element in the array.
For example, given:
string[] ex = {"abc", "adf", "df", "ergd", "adfdfd");
The output should be:
2 3 4 6
One possibility I'm considering is to use a linked list to save each string length, and sort while inserting and finally display the results.
Any other suggestions for efficient solutions to this problem?
Whenever you want to maintain a collection of distinct things (ie: filter out duplicates), you probably want a set.
There are many different data structures for storing sets. Some of these, like search trees, will also "sort" the values for you. You could try using one of the many forms of binary search trees.
What you are doing now (or the given answer) is called the insertion sort. It basically compare the length of the string-to-insert from the inserted strings. After then, when printing, teh length of string-to-print (at current pointer) will be compared to the length of the string before it and after it, if has the same length, do not print!
Another approach is, the bubble sort, it will sort two strings at a time, sort them, then move to next string...
The printing is the most important part in your program, regardless of what sorting algorithm you use, it doesn't matter.
Here's an algorithm for bubble sort and printing process, it's VB so just convert it...
Dim YourString(4) As String
YourString(0) = "12345" 'Will not be printed
YourString(1) = "12345" 'Will not be printed
YourString(2) = "123" 'Will be printed
YourString(3) = "1234" 'Will be printed
Dim RoundLimit As Integer = YourString.Length - 2
'Outer loop for how many times we will sort the whole array...
For CycleCounter = 0 To RoundLimit
Dim CompareCounter As Integer
'Inner loop to compare strings...
For CompareCounter = 0 To RoundLimit - CycleCounter - 1
'Compare lengths... If the first is greater, sort! Note: this is ascending
If YourString(CompareCounter).Length > YourString(CompareCounter + 1).Length Then
'Sorting process...
Dim TempString = YourString(CompareCounter)
YourString(CompareCounter) = YourString(CompareCounter + 1)
YourString(CompareCounter + 1) = TempString
End If
Next
Next
'Cycles = Array length - 2 , so we have 2 cycles here
'First Cycle!!!
'"12345","12345","123","1234" Compare 1: index 0 and 1 no changes
'"12345","123","12345","1234" Compare 2: index 1 and 2 changed
'"12345","123","1234","12345" Compare 3: index 2 and 3 changed
'Second Cycle!!!
'"123","12345","1234","12345" Compare 1: index 0 and 1 changed
'"123","1234","12345","12345" Compare 2: index 1 and 2 changed
'"123","1234","12345","12345" Compare 3: index 2 and 3 no changes
'No more cycle!
'Now print it! Or use messagebox...
Dim CompareLimit As Integer = YourString.Length - 2
For CycleCounter = 0 To CompareLimit
'If length is equal to next string or the preceeding string, do not print...
If ((CycleCounter - 1) <> -1) Then 'Check if index exist
If YourString(CycleCounter).Length = YourString(CycleCounter - 1).Length Then
Continue For 'The length is not unique, exit compare, go to next iteration...
End If
End If
If ((CycleCounter + 1) <> YourString.Length - 1) Then 'Check if index exist
If YourString(CycleCounter).Length = YourString(CycleCounter + 1).Length Then
Continue For 'The length is not unique, exit compare, go to next iteration...
End If
End If
'All test passed, the length is unique, show a dialog!
MsgBox(YourString(CycleCounter))
Next
The question as stated doesn't say anything about sorting or removing duplicates from the results. It is only the given output that implies the sorting and duplicate removal. It doesn't say anything about optimisation for speed or space or writing for maintainability.
So there really isn't enough information for a "best" solution.
If you want a solution that will work in most languages you probably should stick with an array. Put the lengths in a new array, sort it, then print in a loop that remembers that last value to skip duplicates. I wouldn't want to use a language that couldn't cope with that.
If a language is specified you might be able to take advantage of set or associate array type data structures to handle the duplicates and/or sorting automatically. E.g., in Java you could pick a collection class that automatically ignores duplicates and sorts, and you could structure your code such that a one line change to use a different class would let you keep duplicates, or not sort. If you are using C# you could probably write the whole thing as a one-line LINQ statement...
Here is a C++ solution:
#include <set>
#include <vector>
#include <string>
#include <iostream>
using namespace std;
int main()
{
string strarr[] = {"abc", "adf", "df", "ergd", "adfsgf"};
vector< string > vstr(strarr, strarr + 5);
set< size_t > s;
for (size_t i = 0; i < vstr.size(); i++)
{
s.insert( vstr[i].size() );
}
for (set<size_t>::iterator ii = s.begin(); ii != s.end(); ii++)
cout << *ii << " ";
cout << endl;
return 0;
}
Output:
$ g++ -o set-str set-str.cpp
$ ./set-str
2 3 4 6
A set is used because (quoting from here):
Sets are a kind of associative container that stores unique elements,
and in which the elements themselves are the keys.
Associative containers are containers especially designed to be
efficient accessing its elements by their key (unlike sequence
containers, which are more efficient accessing elements by their
relative or absolute position).
Internally, the elements in a set are always sorted from lower to
higher following a specific strict weak ordering criterion set on
container construction.
Sets are typically implemented as binary search trees.
And for details on vector see here and here for string.
Depending on the language, the easiest way might be to iterate through the array using a for loop
for (i=0;i<array.length;i++){
print array[i].length;
}
do you need to print them in order?
Why I can't use table.sort to sort tables with associative indexes?
In general, Lua tables are pure associative arrays. There is no "natural" order other than the as a side effect of the particular hash table implementation used in the Lua core. This makes sense because values of any Lua data type (other than nil) can be used as both keys and values; but only strings and numbers have any kind of sensible ordering, and then only between values of like type.
For example, what should the sorted order of this table be:
unsortable = {
answer=42,
true="Beauty",
[function() return 17 end] = function() return 42 end,
[math.pi] = "pi",
[ {} ] = {},
12, 11, 10, 9, 8
}
It has one string key, one boolean key, one function key, one non-integral key, one table key, and five integer keys. Should the function sort ahead of the string? How do you compare the string to a number? Where should the table sort? And what about userdata and thread values which don't happen to appear in this table?
By convention, values indexed by sequential integers beginning with 1 are commonly used as lists. Several functions and common idioms follow this convention, and table.sort is one example. Functions that operate over lists usually ignore any values stored at keys that are not part of the list. Again, table.sort is an example: it sorts only those elements that are stored at keys that are part of the list.
Another example is the # operator. For the above table, #unsortable is 5 because unsortable[5] ~= nil and unsortable[6] == nil. Notice that the value stored at the numeric index math.pi is not counted even though pi is between 3 and 4 because it is not an integer. Furthermore, none of the other non-integer keys are counted either. This means that a simple for loop can iterate over the entire list:
for i in 1,#unsortable do
print(i,unsortable[i])
end
Although that is often written as
for i,v in ipairs(unsortable) do
print(i,v)
end
In short, Lua tables are unordered collections of values, each indexed by a key; but there is a special convention for sequential integer keys beginning at 1.
Edit: For the special case of non-integral keys with a suitable partial ordering, there is a work-around involving a separate index table. The described content of tables keyed by string values is a suitable example for this trick.
First, collect the keys in a new table, in the form of a list. That is, make a table indexed by consecutive integers beginning at 1 with keys as values and sort that. Then, use that index to iterate over the original table in the desired order.
For example, here is foreachinorder(), which uses this technique to iterate over all values of a table, calling a function for each key/value pair, in an order determined by a comparison function.
function foreachinorder(t, f, cmp)
-- first extract a list of the keys from t
local keys = {}
for k,_ in pairs(t) do
keys[#keys+1] = k
end
-- sort the keys according to the function cmp. If cmp
-- is omitted, table.sort() defaults to the < operator
table.sort(keys,cmp)
-- finally, loop over the keys in sorted order, and operate
-- on elements of t
for _,k in ipairs(keys) do
f(k,t[k])
end
end
It constructs an index, sorts it with table.sort(), then loops over each element in the sorted index and calls the function f for each one. The function f is passed the key and value. The sort order is determined by an optional comparison function which is passed to table.sort. It is called with two elements to compare (the keys to the table t in this case) and must return true if the first is less than the second. If omitted, table.sort uses the built-in < operator.
For example, given the following table:
t1 = {
a = 1,
b = 2,
c = 3,
}
then foreachinorder(t1,print) prints:
a 1
b 2
c 3
and foreachinorder(t1,print,function(a,b) return a>b end) prints:
c 3
b 2
a 1
You can only sort tables with consecutive integer keys starting at 1, i.e., lists. If you have another table of key-value pairs, you can make a list of pairs and sort that:
function sortpairs(t, lt)
local u = { }
for k, v in pairs(t) do table.insert(u, { key = k, value = v }) end
table.sort(u, lt)
return u
end
Of course this is useful only if you provide a custom ordering (lt) which expects as arguments key/value pairs.
This issue is discussed at greater length in a related question about sorting Lua tables.
Because they don't have any order in the first place. It's like trying to sort a garbage bag full of bananas.
I have a string, and another text file which contains a list of strings.
We call 2 strings "brotherhood strings" when they're exactly the same after sorting alphabetically.
For example, "abc" and "cba" will be sorted into "abc" and "abc", so the original two are brotherhood. But "abc" and "aaa" are not.
So, is there an efficient way to pick out all brotherhood strings from the text file, according to the one string provided?
For example, we have "abc" and a text file which writes like this:
abc
cba
acb
lalala
then "abc", "cba", "acb" are the answers.
Of course, "sort & compare" is a nice try, but by "efficient", i mean if there is a way, we can determine a candidate string is or not brotherhood of the original one after one pass processing.
This is the most efficient way, i think. After all, you can not tell out the answer without even reading candidate strings. For sorting, most of the time, we need to do more than 1 pass to the candidate string. So, hash table might be a good solution, but i've no idea what hash function to choose.
Most efficient algorithm I can think of:
Set up a hash table for the original string. Let each letter be the key, and the number of times the letter appears in the string be the value. Call this hash table inputStringTable
Parse the input string, and each time you see a character, increment the value of the hash entry by one
for each string in the file
create a new hash table. Call this one brotherStringTable.
for each character in the string, add one to a new hash table. If brotherStringTable[character] > inputStringTable[character], this string is not a brother (one character shows up too many times)
once string is parsed, compare each inputStringTable value with the corresponding brotherStringTable value. If one is different, then this string is not a brother string. If all match, then the string is a brother string.
This will be O(nk), where n is the length of the input string (any strings longer than the input string can be discarded immediately) and k is the number of strings in the file. Any sort based algorithm will be O(nk lg n), so in certain cases, this algorithm is faster than a sort based algorithm.
Sorting each string, then comparing it, works out to something like O(N*(k+log S)), where N is the number of strings, k is the search key length, and S is the average string length.
It seems like counting the occurrences of each character might be a possible way to go here (assuming the strings are of a reasonable length). That gives you O(k+N*S). Whether that's actually faster than the sort & compare is obviously going to depend on the values of k, N, and S.
I think that in practice, the cache-thrashing effect of re-writing all the strings in the sorting case will kill performance, compared to any algorithm that doesn't modify the strings...
iterate, sort, compare. that shouldn't be too hard, right?
Let's assume your alphabet is from 'a' to 'z' and you can index an array based on the characters. Then, for each element in a 26 element array, you store the number of times that letter appears in the input string.
Then you go through the set of strings you're searching, and iterate through the characters in each string. You can decrement the count associated with each letter in (a copy of) the array of counts from the key string.
If you finish your loop through the candidate string without having to stop, and you have seen the same number of characters as there were in the input string, it's a match.
This allows you to skip the sorts in favor of a constant-time array copy and a single iteration through each string.
EDIT: Upon further reflection, this is effectively sorting the characters of the first string using a bucket sort.
I think what will help you is the test if two strings are anagrams. Here is how you can do it. I am assuming the string can contain 256 ascii characters for now.
#define NUM_ALPHABETS 256
int alphabets[NUM_ALPHABETS];
bool isAnagram(char *src, char *dest) {
len1 = strlen(src);
len2 = strlen(dest);
if (len1 != len2)
return false;
memset(alphabets, 0, sizeof(alphabets));
for (i = 0; i < len1; i++)
alphabets[src[i]]++;
for (i = 0; i < len2; i++) {
alphabets[dest[i]]--;
if (alphabets[dest[i]] < 0)
return false;
}
return true;
}
This will run in O(mn) if you have 'm' strings in the file of average length 'n'
Sort your query string
Iterate through the Collection, doing the following:
Sort current string
Compare against query string
If it matches, this is a "brotherhood" match, save it/index/whatever you want
That's pretty much it. If you're doing lots of searching, presorting all of your collection will make the routine a lot faster (at the cost of extra memory). If you are doing this even more, you could pre-sort and save a dictionary (or some hashed collection) based off the first character, etc, to find matches much faster.
It's fairly obvious that each brotherhood string will have the same histogram of letters as the original. It is trivial to construct such a histogram, and fairly efficient to test whether the input string has the same histogram as the test string ( you have to increment or decrement counters for twice the length of the input string ).
The steps would be:
construct histogram of test string ( zero an array int histogram[128] and increment position for each character in test string )
for each input string
for each character in input string c, test whether histogram[c] is zero. If it is, it is a non-match and restore the histogram.
decrement histogram[c]
to restore the histogram, traverse the input string back to its start incrementing rather than decrementing
At most, it requires two increments/decrements of an array for each character in the input.
The most efficient answer will depend on the contents of the file. Any algorithm we come up with will have complexity proportional to N (number of words in file) and L (average length of the strings) and possibly V (variety in the length of strings)
If this were a real world situation, I would start with KISS and not try to overcomplicate it. Checking the length of the target string is simple but could help avoid lots of nlogn sort operations.
target = sort_characters("target string")
count = 0
foreach (word in inputfile){
if target.len == word.len && target == sort_characters(word){
count++
}
}
I would recommend:
for each string in text file :
compare size with "source string" (size of brotherhood strings should be equal)
compare hashes (CRC or default framework hash should be good)
in case of equity, do a finer compare with string sorted.
It's not the fastest algorithm but it will work for any alphabet/encoding.
Here's another method, which works if you have a relatively small set of possible "letters" in the strings, or good support for large integers. Basically consists of writing a position-independent hash function...
Assign a different prime number for each letter:
prime['a']=2;
prime['b']=3;
prime['c']=5;
Write a function that runs through a string, repeatedly multiplying the prime associated with each letter into a running product
long long key(char *string)
{
long long product=1;
while (*string++) {
product *= prime[*string];
}
return product;
}
This function will return a guaranteed-unique integer for any set of letters, independent of the order that they appear in the string. Once you've got the value for the "key", you can go through the list of strings to match, and perform the same operation.
Time complexity of this is O(N), of course. You can even re-generate the (sorted) search string by factoring the key. The disadvantage, of course, is that the keys do get large pretty quickly if you have a large alphabet.
Here's an implementation. It creates a dict of the letters of the master, and a string version of the same as string comparisons will be done at C++ speed. When creating a dict of the letters in a trial string, it checks against the master dict in order to fail at the first possible moment - if it finds a letter not in the original, or more of that letter than the original, it will fail. You could replace the strings with integer-based hashes (as per one answer regarding base 26) if that proves quicker. Currently the hash for comparison looks like a3c2b1 for abacca.
This should work out O(N log( min(M,K) )) for N strings of length M and a reference string of length K, and requires the minimum number of lookups of the trial string.
master = "abc"
wordset = "def cba accb aepojpaohge abd bac ajghe aegage abc".split()
def dictmaster(str):
charmap = {}
for char in str:
if char not in charmap:
charmap[char]=1
else:
charmap[char] += 1
return charmap
def dicttrial(str,mastermap):
trialmap = {}
for char in str:
if char in mastermap:
# check if this means there are more incidences
# than in the master
if char not in trialmap:
trialmap[char]=1
else:
trialmap[char] += 1
else:
return False
return trialmap
def dicttostring(hash):
if hash==False:
return False
str = ""
for char in hash:
str += char + `hash[char]`
return str
def testtrial(str,master,mastermap,masterhashstring):
if len(master) != len(str):
return False
trialhashstring=dicttostring(dicttrial(str,mastermap))
if (trialhashstring==False) or (trialhashstring != masterhashstring):
return False
else:
return True
mastermap = dictmaster(master)
masterhashstring = dicttostring(mastermap)
for word in wordset:
if testtrial(word,master,mastermap,masterhashstring):
print word+"\n"
Being used to the standard way of sorting strings, I was surprised when I noticed that Windows sorts files by their names in a kind of advanced way. Let me give you an example:
Track1.mp3
Track2.mp3
Track10.mp3
Track20.mp3
I think that those names are compared (during sorting) based on letters and by numbers separately.
On the other hand, the following is the same list sorted in a standard way:
Track1.mp3
Track10.mp3
Track2.mp3
Track20.mp3
I would like to create a comparing alogorithm in Delphi that would let me sort strings in the same way. At first I thought it would be enough to compare consecutive characters of two strings while they are letters. When a digit would be found at some position of both the strings, I would read all digits following them to form a number and then compare the numbers.
To give you an example, I'll compare "Track10" and "Track2" strings this way:
1) read characters while they are equal and while they are letters: "Track", "Track"
2) if a digit is found, read all following digits: "10", "2"
2a) if they are equal, go to 1 or else finish
Ten is greater than two, so "Track10" is greater than "Track2"
It had seemed that everything would be all right until I noticed, during my tests, that Windows considered "Track010" lower than "Track10", while I thought the first one was greater as it was longer (not mentioning that according to my algorithm both the strings would be equal, which is wrong).
Could you provide me with the idea how exactly Windows sorts files by names or maybe you have a ready-to-use algorithm (in any programming language) that I could base on?
Thanks a lot!
Mariusz
Jeff wrote up an article about this on Coding Horror. This is called natural sorting, where you effectively treat a group of digits as a single "character". There are implementations out there in every language under the sun, but strangely it's not usually built-in to most languages' standard libraries.
The mother of all sorts:
ls '*.mp3' | sort --version-sort
The absolute easiest way, I found, was isolate the string you want, so in the OP's case, Path.GetFileNameWithoutExtension(), remove the non-digits, convert to int, and sort. Using LINQ and some extension methods, it's a one-liner. In my case, I was going on directories:
Directory.GetDirectories(#"a:\b\c").OrderBy(x => x.RemoveNonDigits().ToIntOrZero())
Where RemoveNonDigits and ToIntOrZero are extensions methods:
public static string RemoveNonDigits(this string value) {
return Regex.Replace(value, "[^0-9]", string.Empty);
}
public static int ToIntOrZero(this string toConvert) {
try {
if (toConvert == null || toConvert.Trim() == string.Empty) return 0;
return int.Parse(toConvert);
} catch (Exception) {
return 0;
}
}
The extension methods are common tools I use everywhere. YMMV.
Here's a Python approach:
import re
def tryint(s):
"""
Return an int if possible, or `s` unchanged.
"""
try:
return int(s)
except ValueError:
return s
def alphanum_key(s):
"""
Turn a string into a list of string and number chunks.
>>> alphanum_key("z23a")
["z", 23, "a"]
"""
return [ tryint(c) for c in re.split('([0-9]+)', s) ]
def human_sort(l):
"""
Sort a list in the way that humans expect.
"""
l.sort(key=alphanum_key)
And a blog post with more detail: https://nedbatchelder.com/blog/200712/human_sorting.html