Find an hidden permutation of a string C++
up vote
-1
down vote
favorite
I have two strings, and wanted to check if the second is a permutation of the first (and viceversa of course).
So I found out on cplusplus reference that the is_permutation function of the library algorithm could help me. Indeed, I have the following code:
int main () {
string s1 = "bear";
string s2 = "reab";
if ( is_permutation (s1.begin(), s1.end(), s2.begin()) )
cout << "Found permutation.n";
else cout << "No permutations found.n";
return 0;
}
And this works. But now, for example, let's say I still have the string "bear", and a second random string that inside has a permutation of bear, so something like this:
s1 = "bear";
s2 = "AsdVYTcKIyqbNQreabJUoBn";
As you can see there's still the permutation "reab". How can I actually check if there's an hidden permutation? And eventually, save it on a "s3" different string?
Hope you can help me.
c++ string permutation
asked Nov 10 at 10:36
Froooo
81
add a comment |
up vote
-1
down vote
favorite
I have two strings, and wanted to check if the second is a permutation of the first (and viceversa of course).
So I found out on cplusplus reference that the is_permutation function of the library algorithm could help me. Indeed, I have the following code:
int main () {
string s1 = "bear";
string s2 = "reab";
if ( is_permutation (s1.begin(), s1.end(), s2.begin()) )
cout << "Found permutation.n";
else cout << "No permutations found.n";
return 0;
}
And this works. But now, for example, let's say I still have the string "bear", and a second random string that inside has a permutation of bear, so something like this:
s1 = "bear";
s2 = "AsdVYTcKIyqbNQreabJUoBn";
As you can see there's still the permutation "reab". How can I actually check if there's an hidden permutation? And eventually, save it on a "s3" different string?
Hope you can help me.
c++ string permutation
asked Nov 10 at 10:36
Froooo
81
1
You can check all four (s1. length) char consecutive substrings and use the is_permutation on each substting.
– kingW3
Nov 10 at 10:41
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have two strings, and wanted to check if the second is a permutation of the first (and viceversa of course).
So I found out on cplusplus reference that the is_permutation function of the library algorithm could help me. Indeed, I have the following code:
int main () {
string s1 = "bear";
string s2 = "reab";
if ( is_permutation (s1.begin(), s1.end(), s2.begin()) )
cout << "Found permutation.n";
else cout << "No permutations found.n";
return 0;
}
And this works. But now, for example, let's say I still have the string "bear", and a second random string that inside has a permutation of bear, so something like this:
s1 = "bear";
s2 = "AsdVYTcKIyqbNQreabJUoBn";
As you can see there's still the permutation "reab". How can I actually check if there's an hidden permutation? And eventually, save it on a "s3" different string?
Hope you can help me.
c++ string permutation
asked Nov 10 at 10:36
Froooo
81
I have two strings, and wanted to check if the second is a permutation of the first (and viceversa of course).
So I found out on cplusplus reference that the is_permutation function of the library algorithm could help me. Indeed, I have the following code:
int main () {
string s1 = "bear";
string s2 = "reab";
if ( is_permutation (s1.begin(), s1.end(), s2.begin()) )
cout << "Found permutation.n";
else cout << "No permutations found.n";
return 0;
}
And this works. But now, for example, let's say I still have the string "bear", and a second random string that inside has a permutation of bear, so something like this:
s1 = "bear";
s2 = "AsdVYTcKIyqbNQreabJUoBn";
As you can see there's still the permutation "reab". How can I actually check if there's an hidden permutation? And eventually, save it on a "s3" different string?
Hope you can help me.
c++ string permutation
c++ string permutation
asked Nov 10 at 10:36
Froooo
81
asked Nov 10 at 10:36
Froooo
81
asked Nov 10 at 10:36
Froooo
81
asked Nov 10 at 10:36
Froooo
81
asked Nov 10 at 10:36
Froooo
81
81
1
You can check all four (s1. length) char consecutive substrings and use the is_permutation on each substting.
– kingW3
Nov 10 at 10:41
add a comment |
1
You can check all four (s1. length) char consecutive substrings and use the is_permutation on each substting.
– kingW3
Nov 10 at 10:41
1
1
You can check all four (s1. length) char consecutive substrings and use the is_permutation on each substting.
– kingW3
Nov 10 at 10:41
You can check all four (s1. length) char consecutive substrings and use the is_permutation on each substting.
– kingW3
Nov 10 at 10:41
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
You can use a combination of std::string::substr and is_permutation to achieve this.
// Example program
#include <iostream>
#include <string>
#include <algorithm>
using std::string;
using std::cout;
int main () {
string s1 = "bear";
string s2 = "AsdVYTcKIyqbNQJUoBnreab";
size_t i;
for( i = 0; i <= s2.size() - s1.size(); i++)
{
string s3 = s2.substr(i, s1.size());
if ( is_permutation (s1.begin(), s1.end(), s3.begin()))
{
cout << "Found permutation.n";
break;
}
else
{
continue;
}
}
if(i > s2.size() - s1.size())
cout << "No permutations found.n";
return 0;
}
See live demo here.
answered Nov 10 at 11:00
P.W
10.3k2742
You are welcome :) I made a slight change to the code to include the corner case when the permutation appears at the end of the string as well.
– P.W
Nov 10 at 11:28
add a comment |
up vote
0
down vote
as kingW3 already pointed out in the comments on how one might do it.
string s1 = "bear";
string s2 = "AsdVYTcKIyqbNQreabJUoBn";
string key = "";
for (int i = 0; i < s2.length()+1 - s1.length(); i++)
{
key = s2.substr(i, s1.length());
if (is_permutation(s1.begin(), s1.end(), key.begin()))
cout << "Found permutation.n";
else cout << "No permutations found.n";
}
return 0;
Edit: Please note that the for loop condition has to be writtenm with either +1 or -1 in order to get the last character from your second string.
i < s2.length()+1 - s1.length()
or
i < s2.length() - (s1.length()-1)
hope it helps.
answered Nov 10 at 11:01
ats
694
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You can use a combination of std::string::substr and is_permutation to achieve this.
// Example program
#include <iostream>
#include <string>
#include <algorithm>
using std::string;
using std::cout;
int main () {
string s1 = "bear";
string s2 = "AsdVYTcKIyqbNQJUoBnreab";
size_t i;
for( i = 0; i <= s2.size() - s1.size(); i++)
{
string s3 = s2.substr(i, s1.size());
if ( is_permutation (s1.begin(), s1.end(), s3.begin()))
{
cout << "Found permutation.n";
break;
}
else
{
continue;
}
}
if(i > s2.size() - s1.size())
cout << "No permutations found.n";
return 0;
}
See live demo here.
answered Nov 10 at 11:00
P.W
10.3k2742
You are welcome :) I made a slight change to the code to include the corner case when the permutation appears at the end of the string as well.
– P.W
Nov 10 at 11:28
add a comment |
up vote
0
down vote
accepted
You can use a combination of std::string::substr and is_permutation to achieve this.
// Example program
#include <iostream>
#include <string>
#include <algorithm>
using std::string;
using std::cout;
int main () {
string s1 = "bear";
string s2 = "AsdVYTcKIyqbNQJUoBnreab";
size_t i;
for( i = 0; i <= s2.size() - s1.size(); i++)
{
string s3 = s2.substr(i, s1.size());
if ( is_permutation (s1.begin(), s1.end(), s3.begin()))
{
cout << "Found permutation.n";
break;
}
else
{
continue;
}
}
if(i > s2.size() - s1.size())
cout << "No permutations found.n";
return 0;
}
See live demo here.
answered Nov 10 at 11:00
P.W
10.3k2742
You are welcome :) I made a slight change to the code to include the corner case when the permutation appears at the end of the string as well.
– P.W
Nov 10 at 11:28
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You can use a combination of std::string::substr and is_permutation to achieve this.
// Example program
#include <iostream>
#include <string>
#include <algorithm>
using std::string;
using std::cout;
int main () {
string s1 = "bear";
string s2 = "AsdVYTcKIyqbNQJUoBnreab";
size_t i;
for( i = 0; i <= s2.size() - s1.size(); i++)
{
string s3 = s2.substr(i, s1.size());
if ( is_permutation (s1.begin(), s1.end(), s3.begin()))
{
cout << "Found permutation.n";
break;
}
else
{
continue;
}
}
if(i > s2.size() - s1.size())
cout << "No permutations found.n";
return 0;
}
See live demo here.
answered Nov 10 at 11:00
P.W
10.3k2742
You can use a combination of std::string::substr and is_permutation to achieve this.
// Example program
#include <iostream>
#include <string>
#include <algorithm>
using std::string;
using std::cout;
int main () {
string s1 = "bear";
string s2 = "AsdVYTcKIyqbNQJUoBnreab";
size_t i;
for( i = 0; i <= s2.size() - s1.size(); i++)
{
string s3 = s2.substr(i, s1.size());
if ( is_permutation (s1.begin(), s1.end(), s3.begin()))
{
cout << "Found permutation.n";
break;
}
else
{
continue;
}
}
if(i > s2.size() - s1.size())
cout << "No permutations found.n";
return 0;
}
See live demo here.
answered Nov 10 at 11:00
P.W
10.3k2742
edited Nov 10 at 11:27
answered Nov 10 at 11:00
P.W
10.3k2742
answered Nov 10 at 11:00
P.W
10.3k2742
answered Nov 10 at 11:00
P.W
10.3k2742
10.3k2742
You are welcome :) I made a slight change to the code to include the corner case when the permutation appears at the end of the string as well.
– P.W
Nov 10 at 11:28
add a comment |
You are welcome :) I made a slight change to the code to include the corner case when the permutation appears at the end of the string as well.
– P.W
Nov 10 at 11:28
You are welcome :) I made a slight change to the code to include the corner case when the permutation appears at the end of the string as well.
– P.W
Nov 10 at 11:28
You are welcome :) I made a slight change to the code to include the corner case when the permutation appears at the end of the string as well.
– P.W
Nov 10 at 11:28
add a comment |
up vote
0
down vote
as kingW3 already pointed out in the comments on how one might do it.
string s1 = "bear";
string s2 = "AsdVYTcKIyqbNQreabJUoBn";
string key = "";
for (int i = 0; i < s2.length()+1 - s1.length(); i++)
{
key = s2.substr(i, s1.length());
if (is_permutation(s1.begin(), s1.end(), key.begin()))
cout << "Found permutation.n";
else cout << "No permutations found.n";
}
return 0;
Edit: Please note that the for loop condition has to be writtenm with either +1 or -1 in order to get the last character from your second string.
i < s2.length()+1 - s1.length()
or
i < s2.length() - (s1.length()-1)
hope it helps.
answered Nov 10 at 11:01
ats
694
add a comment |
up vote
0
down vote
as kingW3 already pointed out in the comments on how one might do it.
string s1 = "bear";
string s2 = "AsdVYTcKIyqbNQreabJUoBn";
string key = "";
for (int i = 0; i < s2.length()+1 - s1.length(); i++)
{
key = s2.substr(i, s1.length());
if (is_permutation(s1.begin(), s1.end(), key.begin()))
cout << "Found permutation.n";
else cout << "No permutations found.n";
}
return 0;
Edit: Please note that the for loop condition has to be writtenm with either +1 or -1 in order to get the last character from your second string.
i < s2.length()+1 - s1.length()
or
i < s2.length() - (s1.length()-1)
hope it helps.
answered Nov 10 at 11:01
ats
694
add a comment |
up vote
0
down vote
up vote
0
down vote
as kingW3 already pointed out in the comments on how one might do it.
string s1 = "bear";
string s2 = "AsdVYTcKIyqbNQreabJUoBn";
string key = "";
for (int i = 0; i < s2.length()+1 - s1.length(); i++)
{
key = s2.substr(i, s1.length());
if (is_permutation(s1.begin(), s1.end(), key.begin()))
cout << "Found permutation.n";
else cout << "No permutations found.n";
}
return 0;
Edit: Please note that the for loop condition has to be writtenm with either +1 or -1 in order to get the last character from your second string.
i < s2.length()+1 - s1.length()
or
i < s2.length() - (s1.length()-1)
hope it helps.
answered Nov 10 at 11:01
ats
694
as kingW3 already pointed out in the comments on how one might do it.
string s1 = "bear";
string s2 = "AsdVYTcKIyqbNQreabJUoBn";
string key = "";
for (int i = 0; i < s2.length()+1 - s1.length(); i++)
{
key = s2.substr(i, s1.length());
if (is_permutation(s1.begin(), s1.end(), key.begin()))
cout << "Found permutation.n";
else cout << "No permutations found.n";
}
return 0;
Edit: Please note that the for loop condition has to be writtenm with either +1 or -1 in order to get the last character from your second string.
i < s2.length()+1 - s1.length()
or
i < s2.length() - (s1.length()-1)
hope it helps.
answered Nov 10 at 11:01
ats
694
edited Nov 10 at 11:21
answered Nov 10 at 11:01
ats
694
answered Nov 10 at 11:01
ats
694
answered Nov 10 at 11:01
ats
694
694
add a comment |
add a comment |
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1
You can check all four (s1. length) char consecutive substrings and use the is_permutation on each substting.
– kingW3
Nov 10 at 10:41