I need a function middle that accepts a string and returns the middle character if there are an odd number of characters and the two middle characters if there are a even number of characters in the string in C++ for a program I am writing unfortunately I cannot find anything pre-made for an example thats in c++
std::string middleCharacters(const std::string &str)
{
if (str.length() <= 0) return ""; // For an empty string, return an empty string (customize this as desired)
return str.substr((str.length() - 1) / 2, 2 - str.length() % 2);
}
For proof that this works: http://ideone.com/vId2l
Related
This is the homework spec
// Returns a string where the same characters next each other in
// string n are separated by "22"
//
// Pseudocode Example:
// doubleDouble("goodbye") => "go22odbye"
// doubleDouble("yyuu") => "y22yu22u"
// doubleDouble("aaaa") => "a22a22a22a"
//
string doubleDouble(string n)
{
return ""; // This is not always correct.
}
This is part of a homework set that deals with recursion. I know how to use recursion with an int function but I'm not entirely sure how to approach this problem when a string is passed through. Is it as simple as n.length == n.length() +1 ? and then simply insert "22" ? Alongside this, how would one traverse the string? Thanks!
I would say that the base case be if n turns out to be just a blank space, or if it has a size 0, then simply return the string back, no changes made.
This seems OK to me (untested code)
string doubleDouble(string n)
{
if (n.size() < 2)
return n;
else if (n[0] == n[1])
return n[0] + ("22" + doubleDouble(n.substr(1)));
else
return n[0] + doubleDouble(n.substr(1));
}
As you can see you recurse on substrings of the original string.
Undoubtedly there are more efficient ways to do this (that minimise the string copying) but I'll leave that for you.
PS you need to enable C++11 to get all the required versions of + for strings.
string doubleDouble(string n)
{
if(n.length() == 2){
if(n[0]==n[1])
return n.insert(1,"22");
else
return n;
}
else if(n.length() == 1)
return n;
else
if(n.length() % 2 == 1 || n[n.length()/2]!=n[n.length()/2+1])
return doubleDouble(n.substr(0,n.length() / 2)) + doubleDouble(n.substr((n.length() / 2),n.length()));
else if(n[n.length()/2]==n[n.length()/2+1])
return doubleDouble(n.substr(0,n.length() / 2)) +"22"+ doubleDouble(n.substr((n.length() / 2),n.length()));
}
My solution uses 'Divide-and-conquer' paradigm to separate the string in 2 halfs again and again until it finds 1 or 2 characters only. The solution above is simpler. I tried an academic style.
I am in the beginnings of learning C++ and I am wondering if there is a way to assert that a substring can be created from a String, given a range. My String will vary in size each iteration. I am trying to create six substrings from that original String. With this variation in size, I am sometimes trying to access indexes of the String that do not exist for that particular iteration.
For example, if my String in iteration 1 is 11 characters
My first substring is from 3 characters - valid
My second substring is the next 3 characters - valid
My third substring is the next 5 characters - valid
My fourth substring is the next 4 characters - not valid - crashes program
My fifth substring - not valid, out of range
My sixth substring - not valid, out of range
I am wondering if there is a small check I can do to assert the length is valid. It's worth noting, I suppose, that I have not set any default values to these substrings. They are declared as:
string subS1
string subS2
string subS3
...
...
string subS6
Would setting all 6 substrings to null upon declaration alleviate this issue and for any valid substring, the value will just be overwritten?
Thanks in advance
subS1 = str.substr(0, 3); // Could be valid range
subS2 = str.substr(3, 3); // Could be valid range
subS3 = str.substr(6, 5); // Could be valid range
subS4 = str.substr(11, 4); // Could be valid range
subS5 = str.substr(15, 4); // Could be valid range
subS6 = str.substr(19); // from the nineteenth character to the end
Algorithm--->
step 1: Get the length of string in current iteration in variable size.
step 2: Write this code in itertaion.
int i=0;
i= str.substr(start,end).length();
if( i>size) then,
std::cout<<"index exceeded";
Either check the size of str before extracting the string, or rely on std::string::substr's len parameter:
Number of characters to include in the substring (if the string is
shorter, as many characters as possible are used). A value of
string::npos indicates all characters until the end of the string.
Example:
#include <iostream>
#include <string>
int main ()
{
std::string str="Alexander the Great";
std::string str2 = str.substr (16, 25);
std::cout << str2 << '\n'; // eat
return 0;
}
It won't crash, it will just use as many characters as possible.
This however:
std::string str2 = str.substr (20, 25);
should crash, so do it like this in this case:
std::string str2 = ""; // initialise to empty string, that is the NULL you are saying I guess
if(str.size() > 20)
str2 = str.substr (20, 25);
// 'str2' will be an empty string
I am creating a small game where the user will have hints(Characters of a string) to guess the word of a string. I have the code to see each individual character of the string, but is it possible that I can see those characters printed out randomly?
string str("TEST");
for (int i = 0; i < str.size(); i++){
cout <<" "<< str[i];
output:T E S T
desired sample output: E T S T
Use random_shuffle on the string:
random_shuffle(str.begin(), str.end());
Edits:
C++11 onwards use:
auto engine = std::default_random_engine{};
shuffle ( begin(str), end(str), engine );
Use the following code to generate the letters randomly.
const int stl = str.size();
int stl2 = stl;
while (stl2 >= 0)
{
int r = rand() % stl;
if (str[r] != '0')
{
cout<<" "<<str[r];
str[r] = '0';
stl2--;
}
}
This code basically generates the random number based on the size of the String and then prints the character placed at that particular position of the string.
To avoid the reprinting of already printed character, I have converted the character printed to "0", so next time same position number is generated, it will check if the character is "0" or not.
If you need to preserve the original string, then you may copy the string to another variable and use it in the code.
Note: It is assumed that string will contain only alphabetic characters and so to prevent repetition, "0" is used. If your string may contain numbers, you may use a different character for comparison purpose
I'm solving the following problem:
The assignment is to create and return a string object that consists of digits in an int that is sent in through the function's parameter; so the expected output of the function call string pattern(int n) would be "1\n22\n..n\n".
In case you're interested, here is the URL (You need to be signed in to view) to the full assignment, a CodeWars Kata
This is one of the tests (with my return included):
Test-case input: pattern(2)
Expected:
1
22
Actual: "OUTPUT"
//string header file and namespace are already included for you
string pattern(int n){
string out = "OUTPUT";
for (int i = 1; i <= n; ++i){
string temp = "";
temp.insert(0, i, i);
out += temp;
}
return out;
}
The code is self-explanatory and I'm sure there are multiple ways of making it run quicker and more efficiently.
My question is two-fold. Why doesn't my loop start (even though my expression should hold true (1 <= 2) for above case)?
And how does my code hold in the grand scheme of things? Am I breaking any best-practices?
The overload of std::string::insert() that you are using takes three arguments:
index
count
character
You are using i as both count and character. However, the function expects the character to be of char type. In your case, your i is interpreted as a character with the code of 1 and 2, which are basically spaces (well, not really, but whatever). So your output really looks like OUTPUT___ where ___ are three spaces.
If you look at the ascii table, you will notice that digits 0123...9 have indexes from 48 to 57, so to get an index of a particular number, you can do i + 48, or i + '0' (where '0' is the index of 0, which is 48). Finally, you can do it all in the constructor:
string temp(i, i + '0');
The loop works - but does nothing visible. You insert the character-code 1 - not the character '1'; use:
temp.insert(0, i, '0'+i);
the insert method is not called right:
temp.insert(0, i, i); --->
temp.insert(0, i, i+'0');
Please tell me if I am understanding the the substr member function correctly?
result = result.substr(0, pos) + result.substr(pos + 1);
It takes the string from pos, 0 until (but not including), remove[i]
and then + result.substr(pos + 1); concatenates the rest of the string, except but not including the string / char in remove?
string removeLetters2(string text, string remove)
{
int pos;
string result = text;
for (int i = 0; i < remove.length(); i++)
{
while (true)
{
pos = result.find(remove[i]);
if (pos == string::npos)
{
break;
}
else
{
result = result.substr(0, pos) +
result.substr(pos + 1);
}
}
}
return result;
}
In short, you are asking if
result = result.substr(0, pos) +
result.substr(pos + 1);
removes the character at position pos, right?
Short Answer:
Yes.
Longer Answer:
The two-argument call takes the start index and the length (the one argument call goes to the end of string).
It helps to imagine the string like this:
F o o / B a r
0 1 2 3 4 5 6 <- indices
Now remove /:
F o o / B a r
0 1 2 3 4 5 6 <- indices
1 2 3 | <- 1st length
| 1 2 3 <- 2nd length
result = result.substr(0, 3) <- from index 0 with length 3
+ result.substr(4); <- from index 4 to end
As a programmer, always be aware of the difference between distance/index and length.
Better: If index is known:
Your code creates two new, temporary strings, which are then concatenated into a third temporary string, which is then copied to result.
It would be better to ask string to erase (wink wink) in place:
result.erase(pos,1);
// or by iterator
string::iterator it = ....;
result.erase(it,it+1);
This leaves more optimization freedom to the string implementer, who may choose to just move all characters after pos by one to the left. This could, in a specialized scenario, be implemented with a single assignment, a single loop, and within the loop with the x86 swap instruction.
Better: If characters to be deleted are known:
Or, but I am not sure if this gives better performance, but it may give better code, the algorithm remove_if:
#include <algorithm>
// this would remove all slashes, question marks and dots
....
std::string foobar = "ab/d?...";
std::remove_if (foobar.begin(), foobar.end(), [](char c) {
return c=='/' || c=='?' || '.';
});
remove_if accepts any function object.
If there is just one character, it gets easier:
// this would remove all slashes
std::remove (foobar.begin(), foobar.end(), '/');
Although the answer to your question is "yes", there is a better way to go about what you are trying to do. Use string::erase, like this:
result.erase(pos, 1);
This API is designed for removal of characters from the string; it achieves the same result much more efficiently.
Yes, this function removes all letters in remove from text.
since you seem to delete more than one type of character have a look at remove_if from <algorithm> with a special predicate too, although the response of dasblinkenlignt is the good one