# Big O and algorithm to decide string A contains string B?

Discussion in 'C++' started by usgog@yahoo.com, Mar 24, 2005.

1. ### Guest

I need to implement a function to return True/false whether String A
contains String B. For example, String A = "This is a test"; String B =
"is is". So it will return TRUE if String A includes two "i" and two
"s". The function should also handle if String A and B have huge
values, like two big dictionary.

What's the best approach to achieve this with the best performance?
what's the Big O then?

I am thinking to put A and B into two hashtable, like key = "i" and
value = "2" and so on. Then compare these two hashtable. But how to

, Mar 24, 2005

2. ### Mark PGuest

wrote:
> I need to implement a function to return True/false whether String A
> contains String B. For example, String A = "This is a test"; String B =
> "is is". So it will return TRUE if String A includes two "i" and two
> "s". The function should also handle if String A and B have huge
> values, like two big dictionary.
>
> What's the best approach to achieve this with the best performance?
> what's the Big O then?
>
> I am thinking to put A and B into two hashtable, like key = "i" and
> value = "2" and so on. Then compare these two hashtable. But how to
>

One simple possibility is to iterate through each position i within
string A and see if the substring A(i,i+length(B)) = B. Java and C++
have facilities to make this quite simple. The worst case runtime is
O(length(A)*length(B)) under reasonable assumptions about how string
comparisons are performed.

Perhaps one can do better if one is clever.

Mark P, Mar 24, 2005

3. ### James AguilarGuest

"Mark P" <> wrote in message
news:HFu0e.14995\$...
>
> One simple possibility is to iterate through each position i within string
> A and see if the substring A(i,i+length(B)) = B. Java and C++ have
> facilities to make this quite simple. The worst case runtime is
> O(length(A)*length(B)) under reasonable assumptions about how string
> comparisons are performed.

Way too slow if your pattern is 1k chars long and your text is 5 million
chars long (one small gene).

> Perhaps one can do better if one is clever.

Hmmm . . . the Manber-Meyers suffix array algorithm is O(n + log m) where n
is the size of the pattern and m is the length of the text (this is after
suffix array construction, which is . . . O(n), I think, for suffix tree
(O(n)) then tree -> array (O(n))). This algorithm is independent of
alphabet size. I can't find a version of it on the internet, and by no
means will I explain it to you.

The phrase "DO YOUR OWN HOMEWORK" comes to mind as well. Please.

- JFA1

James Aguilar, Mar 24, 2005
4. ### Siemel NaranGuest

<> wrote in message

> I need to implement a function to return True/false whether String A
> contains String B. For example, String A = "This is a test"; String B =
> "is is". So it will return TRUE if String A includes two "i" and two
> "s". The function should also handle if String A and B have huge
> values, like two big dictionary.

Because String B = "is is" (note the space), should the function return true
if String A contains two "i", two "s", one space?

The first step would be to parse String B so that we know to expect 2 "i", 2
"s", 1 " ".

What's the number of distinct chars? Is it A-Z and a-z for a total of 52?
If the number of distinct chars is small, say 256 distinct chars, create an
array, such as unsigned expect[256], to represent the number of chars to
expect. expect['i'] would equal 2, expect['s'] would equal 2, expect[' ']
would equal 1, and all other expect elements would be zero. If the number
of distinct chars is large, then you could use a map<char_type, unsigned> or
hashtable.

Now step through every char in String A. Let the char in question be char
c. Decrement expect[c] by one, but if it is zero then don't decrement it.
Now check if all the elements in expect are zero. As an optimization, you
only need to do this check if you decremented expect[c].

To check if all the elements in expect are zero, you could for example
create another array zero, such as unsigned zero[256] = {0}, and use memcmp
to compare expect to zero. There are other ways, maybe platform specific
ways that might be faster.

Remember to deal with the special case that String B is the empty string, in
which case you can probably immediately return true.

> What's the best approach to achieve this with the best performance?
> what's the Big O then?

My algorithm is O(length(String A)) + O(length(String B)).

> I am thinking to put A and B into two hashtable, like key = "i" and
> value = "2" and so on. Then compare these two hashtable. But how to

In principle it is doable, but putting all the chars of String A into a
hashtable might be rather expensive.

To compare two hashtables, you could iterate through the elements in the
first hash table, using a for_each(hash1.begin(), hash1.end()) structure, or
even for (Iter iter = hash1.begin(); iter != hash1.end(); ++iter). For each
element in hash1, look up the corresponding value in hash2, for example
hash2[iter->key]. Then check if the values are equal, for example
iter->value == hash2[iter->key].

Siemel Naran, Mar 24, 2005
5. ### Guest

Thanks for ur input! This topic is open to discuss the algorithm
actually.

, Mar 24, 2005
6. ### Guest

I do not quite undertstand the problem. You write that the function
should return true when string A contains string B. But your example
shows that it should return true when string A contains all characters
that are in string B. This is something very different. Which one is

cheers,
Marcin Kalicinski

, Mar 24, 2005
7. ### Marco CassianiGuest

You can use the Rabin-Karrp alghoritm

with average access time O(n+m) in the common case.

<> ha scritto nel messaggio
news:...
> I need to implement a function to return True/false whether String A
> contains String B. For example, String A = "This is a test"; String B =
> "is is". So it will return TRUE if String A includes two "i" and two
> "s". The function should also handle if String A and B have huge
> values, like two big dictionary.
>
> What's the best approach to achieve this with the best performance?
> what's the Big O then?
>
> I am thinking to put A and B into two hashtable, like key = "i" and
> value = "2" and so on. Then compare these two hashtable. But how to
>

Marco Cassiani, Mar 24, 2005
8. ### Gernot FrischGuest

<> schrieb im Newsbeitrag
news:...
>I need to implement a function to return True/false whether String A
> contains String B. For example, String A = "This is a test"; String
> B =
> "is is". So it will return TRUE if String A includes two "i" and two
> "s". The function should also handle if String A and B have huge
> values, like two big dictionary.
>
> What's the best approach to achieve this with the best performance?
> what's the Big O then?
>
> I am thinking to put A and B into two hashtable, like key = "i" and
> value = "2" and so on. Then compare these two hashtable. But how to
>

bool XY(const char* src, const char* seek)
{
std::map<char, int> counter;
std::map<char, int>::iterator it;

// Fill map with number of chars in dest
for(; *seek!='\0'; ++seek)
{
if(is_char_to_be_counted(*seek))
++counter[*seek];
}
// subtract the count of these chars on src
for(; *src!='\0'; ++src)
{
it = counter.find(*src);
if(it!=counter.end() && --it->second<0)
return false; // src has more 'x's than dest
}
// See if any of these has different count
for(it=counter.begin(); it!=counter.end(); ++it)
{
if(it->second) return false; // dest has more 'x's than src
}
return true;
}

this should so the trick quickly. If you really have chars, you can
replace the std::map with a simple array of 256 ints...
-Gernot

Gernot Frisch, Mar 24, 2005
9. ### Guest

Acutally the problem is not about String Matching. The function will
return TRUE if string A contains all characters
that are in string B. Or String A has what String B has. For example,
String B is "issi" and String A is "This is a test", the function will
return TRUE. So what if String A and B have big values, what's the best
algorithm and Big O to achieve this?

I am thinking creating two hashtable. For String B, key=i, value=2;
key=s, value=2. For String A, key=i, value=2; key=s, value=3, etc. Then
compare this two hashtable. So constructing these two hashtable will be
expensive but the Big O will be O(nlogn). Is it good?

, Mar 24, 2005
10. ### WillemGuest

wrote:
) Acutally the problem is not about String Matching. The function will
) return TRUE if string A contains all characters
) that are in string B. Or String A has what String B has. For example,
) String B is "issi" and String A is "This is a test", the function will
) return TRUE. So what if String A and B have big values, what's the best
) algorithm and Big O to achieve this?
)
) I am thinking creating two hashtable. For String B, key=i, value=2;
) key=s, value=2. For String A, key=i, value=2; key=s, value=3, etc. Then
) compare this two hashtable. So constructing these two hashtable will be
) expensive but the Big O will be O(nlogn). Is it good?

Why a hashtable ? There are only 256 different characters, so you
can just make an array with 256 entries and count.

The BigO will be O(n).

SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

Willem, Mar 24, 2005
11. ### Guest

Yes. An array with 256 entries should be enough. But how to create an
array using char as the index? array index is supposed to be int,
right? For example, array[0] = 'a' instead of array['a'] = 0. Sorry, I
am a newbie to algorithm...

, Mar 25, 2005
12. ### Karl Heinz BucheggerGuest

wrote:
>
> Yes. An array with 256 entries should be enough. But how to create an
> array using char as the index? array index is supposed to be int,
> right?

a char is nothing more then a small integer in C and C++. Only
during input and output a char is treated differently.

> For example, array[0] = 'a' instead of array['a'] = 0. Sorry, I
> am a newbie to algorithm...

array['a'] = 0;

is fine. (Remember: a char is nothing more then an small integer. Its
value is the code number of the character on your system).

--
Karl Heinz Buchegger

Karl Heinz Buchegger, Mar 25, 2005
13. ### msaltersGuest

wrote:
> Yes. An array with 256 entries should be enough. But how to create an
> array using char as the index? array index is supposed to be int,
> right? For example, array[0] = 'a' instead of array['a'] = 0. Sorry,

I
> am a newbie to algorithm...

char is an integral type, like short, int and long. They'll convert
without problems. i.e.

int frequency[ 1<<CHAR_BIT ] = {0}; // typically 256 entries
std::string A = foo();
for( int i=0;i!=A.size(); ++i )
++frequency[ A ];

std::string B = bar();
for( int i=0;i!=B.size(); ++i )
if( frequency[ B ]-- == 0 )
std::cout << "B not in A";

Obvious O(A.size+B.size) and you can't do better
in general.

HTH,
Michiel Salters

msalters, Mar 25, 2005
14. ### Roger WillcocksGuest

"msalters" <> wrote in message
news:...
>
> wrote:
> > Yes. An array with 256 entries should be enough. But how to create an
> > array using char as the index? array index is supposed to be int,
> > right? For example, array[0] = 'a' instead of array['a'] = 0. Sorry,

> I
> > am a newbie to algorithm...

>
> char is an integral type, like short, int and long. They'll convert
> without problems. i.e.
>
> int frequency[ 1<<CHAR_BIT ] = {0}; // typically 256 entries
> std::string A = foo();
> for( int i=0;i!=A.size(); ++i )
> ++frequency[ A ];
>
> std::string B = bar();
> for( int i=0;i!=B.size(); ++i )
> if( frequency[ B ]-- == 0 )
> std::cout << "B not in A";
>
> Obvious O(A.size+B.size) and you can't do better
> in general.
>
> HTH,
> Michiel Salters
>

std:string is represented as 'char's which can be signed or unsigned
depending on the platform, so you probably want e.g. ++frequency[A &
0xff].

--
Roger

Roger Willcocks, Mar 25, 2005
15. ### Pointless HarlowsGuest

"Willem" <> wrote in message
news:...
> wrote:
> ) Acutally the problem is not about String Matching. The function will
> ) return TRUE if string A contains all characters
> ) that are in string B. Or String A has what String B has. For example,
> ) String B is "issi" and String A is "This is a test", the function will
> ) return TRUE. So what if String A and B have big values, what's the best
> ) algorithm and Big O to achieve this?
> )
> ) I am thinking creating two hashtable. For String B, key=i, value=2;
> ) key=s, value=2. For String A, key=i, value=2; key=s, value=3, etc. Then
> ) compare this two hashtable. So constructing these two hashtable will be
> ) expensive but the Big O will be O(nlogn). Is it good?
>
> Why a hashtable ? There are only 256 different characters, so you
> can just make an array with 256 entries and count.
>
> The BigO will be O(n).
>
>
> SaSW, Willem
> --
> Disclaimer: I am in no way responsible for any of the statements
> made in the above text. For all I know I might be
> drugged or something..
> No I'm not paranoid. You all think I'm paranoid, don't you !
> #EOT

You cross post this in a Jav newsgroup and have the cheek to suggest that
there are only 256 characters ! That's why I abandoned C and it's
derivatives... which simply assumed that nobody in China, Japan, Korea etc
would ever need to use a computer ;-)

Pointless Harlows, Apr 9, 2005
16. ### Charles RichmondGuest

Pointless Harlows wrote:
>
> "Willem" <> wrote in message
> news:...
> > wrote:
> > ) Acutally the problem is not about String Matching. The function will
> > ) return TRUE if string A contains all characters
> > ) that are in string B. Or String A has what String B has. For example,
> > ) String B is "issi" and String A is "This is a test", the function will
> > ) return TRUE. So what if String A and B have big values, what's the best
> > ) algorithm and Big O to achieve this?
> > )
> > ) I am thinking creating two hashtable. For String B, key=i, value=2;
> > ) key=s, value=2. For String A, key=i, value=2; key=s, value=3, etc. Then
> > ) compare this two hashtable. So constructing these two hashtable will be
> > ) expensive but the Big O will be O(nlogn). Is it good?
> >
> > Why a hashtable ? There are only 256 different characters, so you
> > can just make an array with 256 entries and count.
> >
> > The BigO will be O(n).
> >
> >
> > SaSW, Willem
> > --
> > Disclaimer: I am in no way responsible for any of the statements
> > made in the above text. For all I know I might be
> > drugged or something..
> > No I'm not paranoid. You all think I'm paranoid, don't you !
> > #EOT

>
> You cross post this in a Jav newsgroup and have the cheek to suggest that
> there are only 256 characters ! That's why I abandoned C and it's
> derivatives... which simply assumed that nobody in China, Japan, Korea etc
> would ever need to use a computer ;-)
>

No, C just assumes that people in China, Japan, Korea, and India will use
computers in English!!! ;-)

--
+----------------------------------------------------------------+
| Charles and Francis Richmond It is moral cowardice to leave |
| undone what one perceives right |
| richmond at plano dot net to do. -- Confucius |

+----------------------------------------------------------------+

Charles Richmond, Apr 10, 2005