Most efficient way to find out if two words have the same letters within them

If I understand what you're asking, you can make a copy of both strings, sort both of the copies and see if the copies are equal.

I'm not completely sure but here is an idea,

Let A be some string of characters
Let B be some string of characters

To check if A and B has the same character you can do the following,

• radixSort(A) # using the ascii values of each letter in A -- O(n)


• radixSort(B) # using the ascii values of each letter in B -- O(n)


• compare each letter by letter in A and B

Not sure if thats what your trying to achieve.

You could use something similar to the counting sort algorithm. Count how many times each ASCII character appears in each string - if the counts are equal, then both strings contain the same characters.
e.g.

The problem with the prime number approach is you have a limited amount of characters you can check per number of bits available to store the multiplied primes. Assuming a 32-bit unsigned integer and 26 letters, 'z' would equal '101'. The string "zzzzzz" won't fit in 32-bits. In reality you only get like 5 characters that way. Even going up to 128 bits you only get up to 20 characters.

I'd go with SiCrane's or Hodgman's ideas.

Quote:

Original post by Hodgman You could use something similar to the counting sort algorithm. Count how many times each ASCII character appears in each string - if the counts are equal, then both strings contain the same characters. e.g.*** Source Snippet Removed ***

this should be pretty efficient.

You can check first if the lengths of the strings are equal and if they are, you can use the same loop to count the occurrences.

@nobodynews: Good catch there, I didn't even notice that error.
@Hodgman: I like that idea. I'm gonna go with your approach on this problem.

Thank you for all the replies. Learn something new everyday :)

Here's an idea: Quickcomp. (Quickeq? I can't think of a good name.) Like quicksort, but you sort the two arrays in parallel, and early-out if the partition sizes differ.

def partition(A):  if len(A) == 1: return  pivot = A[0]  cursor = 0  for i in range(len(A)):    if A <= pivot:      A[cursor],A = A,A[cursor]      cursor += 1  return cursordef quickcomp(A, B):  if len(A) != len(B): return false  if len(A) <= 1:     if len(A) == 0: return true    else: return (A[0] == B[0])  a_pivloc = partition(A)  b_pivloc = partition(B)  if not quickcomp(A[:a_pivloc-1], B[:a_pivloc-1]): return false  if not quickcomp(A[a_pivloc+1:], B[a_pivloc+1:]): return false  return true

Without thinking about it too hard, it seems like this should perform in average time O(n) if the sets differ, and O(n log n) if they are the same. My intuition is that that's the best you can do without non-comparison-based stuff.

If your goal is to find all possible words based on scrambled word by searching some sort of database of words, presented solution taken as is could be to slow.
In such a case I would recommend preprocessing your words database first in the following manner:

1. You take each word from the database, sort it first, and next copy it to other database, remembering original word and sorted one. This new database should be sorted by sorted names (sounds strange ;))
For example, if you have words: 'cat', 'dog', 'god', 'good', 'ape', that would be:
act->cat
aep->ape
dog->dog, god
doog->good

It is better to perform this step previously and next work on preprocessed database, as when comparing scrambled name you will not have to sort each word from database during comparison, so it will be much faster.
If you fear it will take too long, consider that it has to be done only once and if you won't do this such way, you will have to perform this task during each search.

2. When searching for true word from the scrambled one, this task will be just simple searching word from the database, what could be obtained either by some sort of logarithmic search, or by using hash-arrays, or by constructing tree from the database.

Good luck
Pomnico

Edit:
By database I don't mean SQL necessary.

Quote:

Original post by Sneftel Here's an idea: Quickcomp. (Quickeq? I can't think of a good name.) Like quicksort, but you sort the two arrays in parallel, and early-out if the partition sizes differ. def partition(A): if len(A) == 1: return pivot = A[0] cursor = 0 for i in range(len(A)): if A <= pivot: A[cursor],A = A,A[cursor] cursor += 1 return cursordef quickcomp(A, B): if len(A) != len(B): return false if len(A) <= 1: if len(A) == 0: return true else: return (A[0] == B[0]) a_pivloc = partition(A) b_pivloc = partition(B) if not quickcomp(A[:a_pivloc-1], B[:a_pivloc-1]): return false if not quickcomp(A[a_pivloc+1:], B[a_pivloc+1:]): return false return true Without thinking about it too hard, it seems like this should perform in average time O(n) if the sets differ, and O(n log n) if they are the same. My intuition is that that's the best you can do without non-comparison-based stuff.

That's a cool algorithm idea. I decided to implement it myself and one problem I ran into is that you need to ensure that the partitioning step picks the same item for both a and b. If it picks a different pivot then a different number of items will be less than it.

SiCrane mentioned my idea.

Well another way to test if the letters of word2 are the ones in word1 you can do the following:

1. Have an array A[26]. Initialize it to 0

2. For every letter of word1, set A[letter] = 1

3. set sum=0

4. For every letter of word2 check if A[letter_from_word2]==1. If yes sum++

if sum equals the length of word2 then its true

The counting method is pretty common and effecient.
However,I'm particulary fond of this method :


Example
1)"TomMarvoloRiddle" "IamLordVoldemort" -> "TomMarvoloRiddle" "TamLordVoldemorI"
2)"omMarvoloRiddle" "amLordVoldemorI" -> "omMarvoloRiddle" "TmLardVoldemorI"
3)"mMarvoloRiddle" "mLardVoldemorI" -> "mMarvoloRiddle" "mLordVoldemorI"
4)"MarvoloRiddle" "LardVoldemorI" -> "MarvoloRiddle" "mardVoldeLorI"
5)"arvoloRiddle" "ardVoldeLorI" -> "arvoloRiddle" "ardVoldeLorI"
6)"rvoloRiddle" "rdVoldeLorI" -> "arvoloRiddle" "ardVoldeLorI"
7)"voloRiddle" "dVoldeLorI" -> "voloRiddle" "VdoldeLorI"
8)"oloRiddle" "doldeLorI" ->"oloRiddle" "odldeLorI"
9)"loRiddle" "dldeLorI" -> "loRiddle" "lddeLorI"
10)"oRiddle" "ddeLorI" -> "oRiddle" "odeLdrI"
11)"Riddle" "deLdrI" -> "Riddle" "reLddI"
12)"iddle" "eLddI" -> "iddle" "iLdde"
13)"ddle" "Ldde" -> "ddle" "dlde"
14)"dle" "lde" -> "dle" "dle"
15)"le" "de" -> "le" "le"
16)"e" "e" -> "e" "e"
"" "" -> True

Quote:

Original post by iMalc one problem I ran into is that you need to ensure that the partitioning step picks the same item for both a and b. If it picks a different pivot then a different number of items will be less than it.

Wow, I can't believe I didn't think of that. Nice catch.

def partition(A, pivot):  if len(A) == 1: return  cursor = 0  for i in range(len(A)):    if A <= pivot:      A[cursor],A = A,A[cursor]      cursor += 1  return cursordef quickcomp(A, B):  if len(A) != len(B): return false  if len(A) <= 1:     if len(A) == 0: return true    else: return (A[0] == B[0])  pivot = A[0]  a_pivloc = partition(A, pivot)  b_pivloc = partition(B, pivot)  if not quickcomp(A[:a_pivloc-1], B[:a_pivloc-1]): return false  if not quickcomp(A[a_pivloc+1:], B[a_pivloc+1:]): return false  return true

I was going to pitch my solution, but the prob is, i am already a manager.

Ah, the joys of technical interviews... And people wonder why 99% can't solve FizBuzz.

The best way for just two strings is the counting sort suggestion that was mentioned. It is O(n) in time and O(1) in memory.

If the problem is "find the matching palindromic string in a database" then the problem is harder...doing just the counting sort suggestion on each string in the database sequentially would be O(nm) where n is the max string size and m is the size of the database.
Using a binary search algorithm, this could be significantly reduced to O(n log m).

However, there's a different algorithm for the database approach. Using a Trie you can solve the database lookup in O(n log n) time. To construct the trie, sort each string in the database and add it to the trie. This is O(n log n m) time. Then, to look up whether a string is in the database, sort the string ( O(n log n) ) and traverse the trie (O(n)) to see if it exists.

Quote:

Original post by Steve132 The best way for just two strings is the counting sort suggestion that was mentioned. It is O(n) in time and O(1) in memory.

Counting sort is vastly inferior:
- O(1) space is very big, wchar_t can be a 32-bit number, some encodings of chars can be even longer
- it cannot detect equal characters with different encodings
- Each call must reset the buckets, which will be up to 2^32 writes and will dwarf running time of the rest of algorithm, the n^2 algorithm with perhaps early exit will complete faster for typical strings

The space/initialization problem can be solved using different lookup structure, but those have logn complexity in general, so the end result is n logn.


Unless the "best" way is academia or interview question, which are completely disjoint from real world issues.

For the search-in-a-database problem you could also compute a hash of the char counts in O(n) and then index the database by it in O(1).

EDIT: Take a look at this code:


Of course, having the same commutative_hash doesn't mean that the strings contain the same letters, but it is a necessary condition and collisions should be exceedingly rare. For many applications this might be enough. In the case of a database it would at least dramatically narrow the set of words you have to compare against.



[Edited by - alvaro on December 3, 2010 1:43:42 PM]

you can constantly shuffle the second string and check if it is equal to the first
if it is there you go
if it is not try again

after some hours if you still could not detected their equality than you can be %99.99 sure that they do not include same letters

this is not a deterministic algorithm but I think it will work on most cases

Quote:

Original post by Serdar-X you can constantly shuffle the second string and check if it is equal to the first if it is there you go if it is not try again after some hours if you still could not detected their equality than you can be %99.99 sure that they do not include same letters this is not a deterministic algorithm but I think it will work on most cases

I can't figure out if this is a joke or a horrible idea.

Quote:

Original post by alvaro I can't figure out if this is a joke or a horrible idea.

If the shuffles are permutations without repetition, then it's a valid solution.

If shuffling can result in repeated strings, then it's not guaranteed to ever stop. It would be possible to derive the probability out of that, but that is not a deterministic algorithm anymore.


While poor fit for Von Neumann architecture, it's probably fine for quantum computers. And maybe even for genetic programming (the biological kind).