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Strings

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  • melroy_tellis + 6 comments
    #include <bits/stdc++.h>

using namespace std;

int main()
{
    string a, b;
    cin >> a >> b;
    
    cout << a.length() << ' ' << b.length() << endl;
    
    cout << a + b << endl;
    
    swap( a[0], b[0] );
    cout << a << ' ' <<  b << endl;
    
    return 0;
}
    • ghadageseema + 1 comment

      all code is wrong

      • melroy_tellis + 0 comments

        You're mistaken.

    • CodeGenie + 1 comment

      awesome!! but can u please explain it a bit...

      • warrior_ + 0 comments

        My best suggesion to u that , please first understand about the few below functions 1)str.size() 2)swap(str[0],str1[0]) 3)str3=str1+str2 then u can try

    • PiCTosi + 3 comments

      Upvoted for (after ignoring #include <bits/stdc++.h>!)

      swap( a[0], b[0] );

      I personally did it this way:

      cout << b.at(0) << a.substr(1) << ' ';
cout << a.at(0) << b.substr(1) << endl;

      I know that mine would throw out_of_range but what happens, with your solution, if a or b is empty?

      By the way, for anyone wondering: string::size() and string::length() are exactly the same function! Which is pretty uncommon ...

      • Tolani + 3 comments

        Similar to yours , I did

        #include <iostream>
#include <string>
using namespace std;

int main() {
   // Complete the program
    string a;
    string b;
    
    cin >>a;
    cin >> b;
    cout << a.size() <<" "<<b.size()<<endl;
    cout << a + b<< endl;
    cout <<b.front() + a.substr(1, a.size()-1) << " "<< a.front() + b.substr(1, b.size()-1); 
    return 0;
}
        • harshitjain2626 + 1 comment

          hey can anyone tell me what does => a.substr(1,a.size()-1) do??

          • mgomezmunoz92 + 1 comment

            The function substr(x, y) gives us a new string that starts at position x and has a length of y.

            Let suppose that we have a string: a = "house", And we want to take all the string except the first letter, that is "ouse".

            Then we have to start at position 1 ('o') and go until the end of the string. We know that the string "house" has a length of a.size(), therefore the string "ouse" will have a length of a.size()-1.

            • harshitjain2626 + 1 comment

              Thanks!!

              • Valkyrie55 + 0 comments

                that's something new thanks

        • endurance1 + 0 comments

          thats cool

        • tanmayrshinde00 + 1 comment

          Hey can you tell me why you did a.substr(1,a.size()-1) rather than a.substr(a[1],a.size()-1)

          • jluisg_23 + 0 comments

            The method substr() of the class string works as in his prototype:

            std::string::substr(size_t pos, size_t len) const;

            where, the first argument that it takes is the position of the fisrt character of the string, but not the character itself. Whereas the second argument indicates the number of characters to include in the subtring.

            Reference: www.cplusplus.com/reference/string/string/substr/

      • drkato + 0 comments

        I did it this way also, and like it because you aren't altering your local variable, which means you can still use it in its original form later on if you want to. Of course, I also check for empty strings before performing the operation as well.

      • abhi_sc30 + 0 comments
        [deleted]
    • him825422 + 0 comments

      I want to do this code without using function like length() and swap(). pls help

    • rohanjeet + 0 comments
      #include <iostream>
#include <string>
using namespace std;

int main() {
   // Complete the program
    string a,b;
    cin>>a>>b;
    cout<<a.size() <<" " << b.size() <<endl;
    cout<< a+b << endl;
    a[0]^= b[0]^= a[0]^= b[0];
    cout<<a<<" "<<b;
    return 0;
}
    • amark720 + 0 comments
      [deleted]

  • goetz + 1 comment

    Looks like the compiler is not properly configured to handle the input. Try this with getline() and you will find an extra \n character is captured. I don't care enough to verify (cin works just fine), but this is probably a Window/Linux/OSX issue with end-of-line (\r\n, \n, \r respectively). Its like the input was created in Windows and the compiler is on a Linux machine.

    Still, I could be wrong about this.

    • EliahKagan + 0 comments

      This is apparently still not fixed -- which is unfortunate, since std::getline is the self-documenting way to read a full line of input. (Changing https://paste.ubuntu.com/15327941/ to https://paste.ubuntu.com/15327947/ went from all tests failing to all tests passing -- both work equally well when tested locally.)

  • Ethan13310 + 0 comments

    Here's my answer:

    #include <iostream>
#include <string>
#include <utility> // std::swap

int main()
{
    std::string a{};
    std::string b{};
    std::cin >> a >> b;
    
    std::cout << a.size() << ' ' << b.size() << std::endl;
    std::cout << a + b << std::endl;
    // Don't forget to check if both strings are not null
    if (a.size() > 0 && b.size() > 0)
    {
        std::swap(a[0], b[0]);
        std::cout << a << ' ' << b << std::endl;
    }
    return 0;
}

  • rock29520 + 2 comments

    I hope this will Help you to think something new.don't copy paste. }

    #include <iostream>
#include <string>
using namespace std;

int main() {
   string a,b;
    string tmp;
    cin>>a;
    cin>>b;
    cout<<a.size()<<" "<<b.size()<<endl<<(a+b)<<endl;
    tmp[0]=a[0];
    tmp[1]=b[0];
    a[0]=tmp[1];
    b[0]=tmp[0];
    cout<<a<<" ";
    cout<<b;
  
    return 0;
}
    • rahulgupta9336 + 0 comments

      thanx

    • iamishaan + 1 comment

      you can just one temp variable, no need of two.

      • ak8414034 + 0 comments

        string a,b,c;

        cin>>a>>b; 
cout<<a.size()<<" "<<b.size()<<endl;
c=a+b;
cout<<c<<endl;
char temp;
temp=b[0];
b[0]=a[0];
a[0]=temp;
cout<<a<<" "<<b;

  • A_Goerlitz + 2 comments

    love to use XOR swapping:

    a[0] = a[0] ^ b[0];
b[0] = b[0] ^ a[0];
a[0] = a[0] ^ b[0];

    :-)

    • imraanpattan + 0 comments

      nice swapping trick.keep it up

    • rohanjeet + 0 comments
      a[0]^= b[0]^= a[0]^= b[0];

  • michaelhazani + 0 comments

    Didn't know the useful str.swap(), ended up with the slightly more verbose:

    #include <iostream>
#include <string>
using namespace std;

int main() {
   // Complete the program    
   string a, b;
   cin >> a >> b;
   cout << a.size() << " " << b.size() << "\n" << a + b << "\n";
   cout << b[0] + a.substr(1) << " " << a[0] + b.substr(1);    
   return 0;
}

  • Thirumagaldhivya + 0 comments

    This is an alternative,without using swap function.

    #include <iostream>
#include <string>
using namespace std;

int main() {
    
    int t;
    string s1,s2;
    cin>>s1>>s2;
    cout<<s1.size()<<" "<<s2.size()<<endl;
    cout<<s1+s2<<endl;
    t=s1[0];
    s1[0]=s2[0];
    s2[0]=t;
    cout<<s1<<" "<<s2;
    return 0;
}

  • abdelazizwmuhmd + 0 comments
    int main()
{
	char save;
	string a, b;
	cin >> a;
	cin >> b;
	cout << a.length() << " " << b.length() << endl;
	cout << a + b << endl;
	save = a[0];
	a[0] = b[0];
    b[0] = save;
	cout << a << " " << b << endl;


    return 0;
}

  • alfonso_delamor1 + 0 comments
        cout << a.size() << " " << b.size() << endl;
    cout << a << b << endl;
    cout << b[0] << a.substr(1,a.size() - 1) << " " 
         << a[0] << b.substr(1,b.size() - 1);

  • rupeshshukla + 0 comments

    without any extra effort.

    int main()
{
	// Complete the program
    string stfirst,stSecond;
    cin>>stfirst>>stSecond;
    cout<<stfirst.size()<<" "<<stSecond.size();  
    cout<<endl<<stfirst + stSecond;
    char ch = stSecond[0];
    stSecond[0] = stfirst[0];
    stfirst[0] =ch;
    cout<<endl<<stfirst <<" " <<stSecond;
    return 0;
}
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