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.
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.
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.
Related
I am trying to make a program that turns a string into encryption by going ten letters ahead of each letter. https://gyazo.com/86f9d708c2f02cf2d70dbc1cd9fa9a06 I am doing part 2. When I input "helloworld" something like 0x45 something comes up. Please help! This is due soon!
I am tried messing around with the for loops but it didn't help.
#include <iostream>
using namespace std;
int main()
{
//Input Message
cout << "Enter a message" << endl;
string message;
getline(cin, message);
//Convert Message to Numbers
int numMess[message.length()];
for (int i = 0; i<message.length(); i++) {
numMess[i] = (int)message[i];
}
cout << numMess << endl;
//Encrypt Number Message by adding ten to each one
int encryptNumMess[message.length()];
for (int a = 0; a < message.length(); a++){
encryptNumMess[a] = numMess[a] + 10;
if (encryptNumMess[a] > 122) {
encryptNumMess[a] = 97;
}
}
cout << encryptNumMess << endl;
//Convert Encrypted Number Message to letters
string encryption[message.length()];
for (int b = 0; b<message.length(); b++) {
encryption[b] = (char)encryptNumMess[b];
}
cout << encryption << endl;
return 0;
}
I expect when I type "helloworld" the final product will be "rovvygybvn"
If you are willing to scrap the hand-coded loops, you can use the STL algorithms such as std::transform to accomplish this:
But first, there are a few things you should do:
Don't use magic numbers such as 122, 97, etc. Instead use the actual character constants, i.e a, b, etc. However if we assume ASCII, where the alphabetic character codes are contiguous, your particular program could simply use a constant string to denote the alphabet, and then use simple indexing to pick out the character.
const char *alphabet = "abcdefghijklmnopqrstuvwxyz";
Then to get the letter a, a simple subtraction is all that's required to get the index:
char ch = 'b';
int index = ch - 'a'; // same as 'b' - 'a' == 98 - 97 == 1
std::cout << alphabet[index]; // will print 'b'
Given this, the next thing is to figure out what character is reached if you add 10 to the value, and if greater than 26, wrap around to the beginning of the alphabet. This can be done using modulus (remainder after division)
char ch = 'x';
int index = (ch - 'a' + 10) % 26; // Same as ('x' - 'a' + 10) % 26 == (120 - 97 + 10) % 26 == 33 % 26 == 7
std::cout << alphabet[index]; // will print 'h'
The next thing is to figure out the opposite, where given an encrypted character, you have to find the unencrypted character by subtracting 10. Here this wraps the opposite way, so a little more work needs to be done (not shown, but code sample reflects what is done).
Putting this all together, and using std::transform and lambdas, we get the following small program:
#include <iostream>
#include <algorithm>
#include <string>
#include <iterator>
#include <cmath>
int main()
{
//Input Message
const char *alphabet="abcdefghijklmnopqrstuvwxyz";
std::string message = "helloworld";
std::string result;
// set the encrypted string using the formula above and std::transform
std::transform(message.begin(), message.end(), std::back_inserter(result),
[&](char ch) { return alphabet[(ch - 'a' + 10) % 26]; });
std::cout << "Encrypted: " << result << '\n';
// convert back to unencrypted using the above formula and std::transform
std::string result2;
std::transform(result.begin(), result.end(), std::back_inserter(result2),
[&](char ch)
{ int index = ch - 'a' - 10; index = index < 0?26 - (abs(index) % 26):index % 26; return alphabet[index];});
std::cout << "Unencrypted: " << result2;
}
Output:
Encrypted: rovvygybvn
Unencrypted: helloworld
This code works for encrypt, if you want to decrypt you should chande newAlphabet and oldAlphabet
I comment in the code that which newAlphabet and oldAlphabet are for encrypt and which are for decrypt
#include <windows.h>
#include <stdio.h>
#include <string>
#include <iostream>
using namespace std;
int main()
{
// For Encrypt
string newAlphabet = "abcdefghijklmnopqrstuvwxyz";
string oldAlphabet = "klmnopqrstuvwxyzabcdefghij";
// For Decrypt
//string newAlphabet = "klmnopqrstuvwxyzabcdefghij";
//string oldAlphabet = "abcdefghijklmnopqrstuvwxyz";
string input = "";
string output = "";
getline(cin, input);
int inputLen = input.size();
if (oldAlphabet.size() != newAlphabet.size())
return false;
for (int i = 0; i < inputLen; ++i)
{
int oldCharIndex = oldAlphabet.find(tolower(input[i]));
if (oldCharIndex >= 0)
output += isupper(input[i]) ? toupper(newAlphabet[oldCharIndex]) : newAlphabet[oldCharIndex];
else
output += input[i];
}
cout << output << endl;
return 0;
}
As others have already mentioned int numMess[message.length()]; is not valid c++.
If it works for you, you're using compiler extension which you really shouldn't rely on. The correct way would be:
std::vector <int> numMess(message.length());
Look up the std::vector reference for more info.
Next, int encryptNumMess[100]; creates a C array style array. encryptNumMess is the base pointer to the array. when you try std::cout << encryptNumMess it'll output the pointer value, NOT the array. You'll need a for loop for doing that, like so :
for(int i = 0; i < 100; ++i)
std::cout << encryptNumMess[i] << " ";
std::cout << endl;
The above also works when you convert this to a vector like we did with numMess whereas in that case, std::cout << encryptNumMess wouldn't even compile.
Thirdly, string encryption[100] creates an array of 100 strings! Not a string of size 100. To do that:
std::string foo(message.length(), '\0');
We have to specify what character to fill the string with. Thus us '\0'.
And now, for the string, to output it, you may use std::cout << foo.
Lastly, since arithmetic is allowed on char, the entire program may be shortened to just this
#include <iostream>
int main()
{
// Input Message
std::cout << "Enter a message" << std::endl;
std::string message, encryption;
getline(std::cin, message);
// Resize encryption string to the desired length
encryption.resize(message.length());
// Do the encryption
for(size_t i = 0; i < message.length(); ++i) {
encryption[i] = message[i] + 10;
if (encryption[i] > 122) {
encryption[i] = 97;
}
}
// Output the string
std::cout << encryption << std::endl;
return 0;
}
Of course, your encryption algorithm is still not correct as per instructions, but I'll leave that for you to figure out. I believe #PaulMcKenzie has already told you most of how to fix it, and also to not use magic numbers.
I am trying to write a code that will search userInput for the word "darn" and if it is found, print out "Censored". if it is not found, it will just print out the userInput. It works in some cases, but not others. If the userInput is "That darn cat!", it will print out "Censored". However, if the userInput is "Dang, that was scary!", it also prints out "Censored". I am trying to use find() to search for the string literal "darn " (the space is because it should be able to determine between the word "darn" and words like "darning". I am not worrying about punctuation after "darn"). However, it seems as though find() is not doing what I would like. Is there another way I could search for a string literal? I tried using substr() but I couldn't figure out what the index and the len should be.
#include <iostream>
#include <string>
using namespace std;
int main() {
string userInput;
userInput = "That darn cat.";
if (userInput.find("darn ") > 0){
cout << "Censored" << endl;
}
else {
cout << userInput << endl;
} //userText.substr(0, 7)
return 0;
}
The problem here is your condition. std::string::find returns a object of std::string::size_type which is an unsigned integer type. That means it can never be less than 0 which means
if (userInput.find("darn ") > 0)
will always be true unless userInput starts with "darn ". Because of this if find doesn't find anything then it returns std::string::npos. What you need to do is compare against that like
if (userInput.find("darn ") != std::string::npos)
Do note that userInput.find("darn ") will not work in all cases. If userInput is just "darn" or "Darn" then it won't match. The space needs to be handled as a separate element. For example:
std::string::size_type position = userInput.find("darn");
if (position != std::string::npos) {
// now you can check which character is at userInput[position + 4]
}
std::search and std::string::replace were made for this:
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main() {
string userInput;
userInput = "That darn cat is a varmint.";
static const string bad_words [] = {
"darn",
"varmint"
};
for(auto&& bad : bad_words)
{
const auto size = distance(bad.begin(), bad.end());
auto i = userInput.begin();
while ((i = std::search(i, userInput.end(), bad.begin(), bad.end())) != userInput.end())
{
// modify this part to allow more or fewer leading letters from the offending words
constexpr std::size_t leading_letters = 1;
// never allow the whole word to appear - hide at least the last letter
auto leading = std::min(leading_letters, std::size_t(size - 1));
auto replacement = std::string(i, i + leading) + std::string(size - leading, '*');
userInput.replace(i, i + size, replacement.begin(), replacement.end());
i += size;
}
}
cout << userInput << endl;
return 0;
}
expected output:
That d*** cat is a v******.
I would like to parse through two vectors of strings and find the strings that match each other and the ones that do not.
Example of what I want get:
input vector 1 would look like: [string1, string2, string3]
input vector 2 would look like: [string2, string3, string4]
Ideal output:
string1: No Match
string2: Match
string3: Match
string4: No Match
At the moment I use this code:
vector<string> function(vector<string> sequences, vector<string> second_sequences){
for(vector<string>::size_type i = 0; i != sequences.size(); i++) {
for(vector<string>::size_type j = 0; j != second_sequences.size(); j++){
if (sequences[i] == second_sequences[j]){
cout << "Match: " << sequences[i];
}else{
cout << "No Match: " << sequences[i];
cout << "No Match: " << second_sequences[j];
}
}
}
}
It works great for the ones that match, but iterates over everything so many times,
and the ones that do not match get printed a large number of times.
How can I improve this?
The best code is the code that you did not have to write.
If you take a (STL) map container it will take care for you of sorting and memorizing the different strings you encounter.
So let the container works for us.
I propose a small code quickly written. You need for this syntax to enable at least the C++ 2011 option of your compiler ( -std=c++11 on gcc for example ). The syntax that should be used before C++11 is much more verbose (but should be known from a scholar point of view ).
You have only a single loop.
This is only a hint for you ( my code does not take into account that in the second vector string4 could be present more than once, but I let you arrange it to your exact needs)
#include <iostream>
#include <vector>
#include <string>
#include <map>
using namespace std;
vector<string> v1 { "string1","string2","string3"};
vector<string> v2 { "string2","string3","string4"};
//ordered map will take care of "alphabetical" ordering
//The key are the strings
//the value is a counter ( or could be any object of your own
//containing more information )
map<string,int> my_map;
int main()
{
cout << "Hello world!" << endl;
//The first vector feeds the map before comparison with
//The second vector
for ( const auto & cstr_ref:v1)
my_map[cstr_ref] = 0;
//We will look into the second vector ( it could also be the third,
//the fourth... )
for ( const auto & cstr_ref:v2)
{
auto iterpair = my_map.equal_range(cstr_ref);
if ( my_map.end() != iterpair.first )
{
//if the element already exist we increment the counter
iterpair.first->second += 1;
}
else
{
//otherwise we put the string inside the map
my_map[cstr_ref] = 0;
}
}
for ( const auto & map_iter: my_map)
{
if ( 0 < map_iter.second )
{
cout << "Match :";
}
else
{
cout << "No Match :" ;
}
cout << map_iter.first << endl;
}
return 0;
}
Output:
No Match :string1
Match :string2
Match :string3
No Match :string4
std::sort(std::begin(v1), std::end(v1));
std::sort(std::begin(v2), std::end(v2));
std::vector<std::string> common_elements;
std::set_intersection(std::begin(v1), std::end(v1)
, std::begin(v2), std::end(v2)
, std::back_inserter(common_elements));
for(auto const& s : common_elements)
{
std::cout<<s<<std::endl;
}
I just needed to reverse a string, so I declared a new string variable and iteratively copied the elements. Now i want to print the reversed string through cout << reversed; but this is printing nothing. I can print it through a for loop through reverse[i] until the size but is there any better way?
#include <iostream>
#include <string>
using namespace std;
int main()
{
string original = "hello";
string reverse;
int i, j = 0, size = original.length();
for (i = size - 1; i >= 0; i--) // original's last as reversed's first
{
reverse[j] = original[i];
j++;
}
reverse[j] = '\0'; //last value as null
cout << "original string = " << original << endl;
cout << "reversed string = " << reverse << endl;
system("pause");
return 0;
}
A better solution is to use std::reverse:
std::string original = "whatever";
std::string rev = original;
std::reverse(rev.begin(), rev.end());
You can use reverse iterators to instantiate the reversed string from the original:
string original = "hello";
string reverse(original.rbegin(), original.rend());
Then
std::cout << "reversed string = " << reverse << std::endl;
Note: avoid using namespace std;. There is an algorithm called std::reverse, whose name you could be inadvertently pulling into the global namespace. And you do have a variable with that name.
See this working demo.
The major problem with your code is that reverse is empty, so using any index leads to you indexing out of bounds and undefined behavior.
As a side-note, you don't have to terminate std::string objects, they are automatically terminated.
you should notice that string reverse doesn't have a size , so when you do something like this
for (i = size - 1; i >= 0; i--) // original's last as reversed's first
{
reverse[j] = original[i];
j++;
}
Here, you are accessing an index that is not found in string reverse,as it's size is 0 . You should take care of something like that.
There are plenty of answers that your received about this question , I just wanted to make you notice this mistake
instead , you can make
reverse+=original[i];
You don't even need to create a new string to hold the reversed string. Just iterator over the original string.
void print_reversed(const std::string& str)
{
std::copy(str.crbegin(), str.crend(), std::ostream_iterator<char>(std::cout));
}
I want to search a large string for all the locations of a string.
The two other answers are correct but they are very slow and have O(N^2) complexity. But there is the Knuth-Morris-Pratt algorithm, which finds all substrings in O(N) complexity.
Edit:
Also, there is another algorithm: the so-called "Z-function" with O(N) complexity, but I couldn't find an English source for this algorithm (maybe because there is also another more famous one with same name - the Z-function of Rieman), so I will just put its code here and explain what it does.
void calc_z (string &s, vector<int> & z)
{
int len = s.size();
z.resize (len);
int l = 0, r = 0;
for (int i=1; i<len; ++i)
if (z[i-l]+i <= r)
z[i] = z[i-l];
else
{
l = i;
if (i > r) r = i;
for (z[i] = r-i; r<len; ++r, ++z[i])
if (s[r] != s[z[i]])
break;
--r;
}
}
int main()
{
string main_string = "some string where we want to find substring or sub of string or just sub";
string substring = "sub";
string working_string = substring + main_string;
vector<int> z;
calc_z(working_string, z);
//after this z[i] is maximal length of prefix of working_string
//which is equal to string which starting from i-th position of
//working_string. So the positions where z[i] >= substring.size()
//are positions of substrings.
for(int i = substring.size(); i < working_string.size(); ++i)
if(z[i] >=substring.size())
cout << i - substring.size() << endl; //to get position in main_string
}
Using std::string::find. You can do something like:
std::string::size_type start_pos = 0;
while( std::string::npos !=
( start_pos = mystring.find( my_sub_string, start_pos ) ) )
{
// do something with start_pos or store it in a container
++start_pos;
}
EDIT: Doh! Thanks for the remark, Nawaz! Better?
I'll add for completeness, there is another approach that is possible with std::search, works like std::string::find, difference is that you work with iterators, something like:
std::string::iterator it(str.begin()), end(str.end());
std::string::iterator s_it(search_str.begin()), s_end(search_str.end());
it = std::search(it, end, s_it, s_end);
while(it != end)
{
// do something with this position..
// a tiny optimisation could be to buffer the result of the std::distance - heyho..
it = std::search(std::advance(it, std::distance(s_it, s_end)), end, s_it, s_end);
}
I find that this sometimes outperforms std::string::find, esp. if you represent your string as a vector<char>.
Simply use std::string::find() which returns the position at which the substring was found, or std::string::npos if none was found.
Here is the documentation.
An here is the example taken from this documentation:
// string::find
#include <iostream>
#include <string>
using namespace std;
int main ()
{
string str ("There are two needles in this haystack with needles.");
string str2 ("needle");
size_t found;
// different member versions of find in the same order as above:
found=str.find(str2);
if (found!=string::npos)
cout << "first 'needle' found at: " << int(found) << endl;
found=str.find("needles are small",found+1,6);
if (found!=string::npos)
cout << "second 'needle' found at: " << int(found) << endl;
found=str.find("haystack");
if (found!=string::npos)
cout << "'haystack' also found at: " << int(found) << endl;
found=str.find('.');
if (found!=string::npos)
cout << "Period found at: " << int(found) << endl;
// let's replace the first needle:
str.replace(str.find(str2),str2.length(),"preposition");
cout << str << endl;
return 0;
}