How to replace specific part of a string with another text?

Question

I'm solving the challenge on Reddit here

And I can't find out how to replace a specific text of a string with another one. I managed to find out how to check if that specific text exists, but I'm having problems replacing it with another text or completely removing it from the string (as the challenge requires that).

Here's my code so far:

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

int law (string a);

int main() {

    string a;
    cin >> a;

    cout << law(a) << endl;

    return 0;
}

int law(string a){
    if (a.find("NOT") != string::npos)
        return a.replace(a.begin(), a.end(), ' ');
    if(a.find("NOT") != string::npos && ((a.find("AND") == string::npos) || (a.find("OR") == string::npos)))
       return a.erase(remove(a.begin(), a.end(), "NOT"), a.end());

}

Why do you need text replacing algorithms to apply De Morgan's Law transformation to a given boolean expression? — Manu343726
– Manu343726, Commented Nov 15, 2013 at 19:22

Manu343726 · Accepted Answer · 2013-11-16 13:03:12Z

3

You could search the substring position, then assign each character of to the original string:

void replace( std::string& input_string , const std::string& searched_string , const std::string& replace_string )
{
    std::size_t replace_pos = input_string.find( searched_string );

    if( replace_pos != std::string::npos )
    {
        std::copy( std::begin( replace_string ) ,
                   std::end( replace_string ) ,
                   std::begin( input_string ) + replace_pos
                 );

        //Or with a simple for loop (Which is what std::copy does):
        for( std::size_t i = 0 ; i < replace_string.size() ; ++i )
            input_string[i + replace_pos}] = replace_string[i];
    }
}

The point of this solution is that there is no reason to use expensive string concatenation at all to replace text.

edited Nov 16, 2013 at 13:03

answered Nov 15, 2013 at 19:20

Manu343726

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2 Comments

Brian Gradin Over a year ago

What if the length of the text you are replacing != the length of the text you are putting in?

Manu343726 Over a year ago

@BrianGradin I had supposed that the replacement text always fits in the original searched string. Of course if thats not true this solution does not work and you need to split and reconcat the string. My point was that not always you should use expensive concatenations.

Brian Gradin · Accepted Answer · 2013-11-15 18:26:29Z

2

Think of the resulting string having three parts:

The beginning part of the original string
The replacement text
The ending part of the original string

You can get the beginning and ending part of the original string by using string.substr(). The code will look something like this:

string originalString; // given
string partToReplace; // given
string replacementText; // given

int positionOfText = originalString.find(partToReplace);

string resultString = originalString.substr(0, positionOfText);
resultString += replacementText;
resultString += originalString.substr(positionOfText + partToReplace.length());

answered Nov 15, 2013 at 18:26

Brian Gradin

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