Why does the erase() function works properly only for the first time?

Question

I am trying to erase particular characters using the erase() function from a string but its not working.

The question says you have to remove a substring which is either "AB" or "BB". When the substring gets deleted from the string the remaining parts of the string get concatenated and the process continues...

Here is the code:

#include <bits/stdc++.h>

#define ios ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define int long long int

using namespace std;

int32_t main()
{

    ios;
    int t;
    cin>>t;

    while(t--)
    {
        string s;
        cin>>s;
        int i;
        for(i=0;i<s.length();i++)
        {
            if(s[i]=='A'&&s[i+1]=='B')
            {
                s.erase(i,i+1);
                cout<<s<<"\n";
                i=-1; // I want to start from begining therefore initializing i=-1 after i++ it becomes i=0;
            }                                        
            else if(s[i]=='B'&&s[i+1]=='B')
            {
                s.erase(i,i+1);
                cout<<s<<"\n";
                i=-1;
            }
        }
        //cout<<s.length()<<"\n";
    }
    return 0;
}

Input:

1
AABBBABBBB

The output is :

ABBABBBB

BBABBBB

BABBBB

BBBB

BBB

BB

B

But the output should be:

ABBABBBB

BABBBB

BBBB

BB

Am I doing something wrong?

[Don’t use `` at all](https://stackoverflow.com/questions/31816095/). Don’t use `#define int ...`, use a `typedef` or `using` alias instead. Also, your loops are going out of bounds of the string. Use a debugger to figure out why your output is not what you are expecting. — Remy Lebeau, Oct 18 '20 at 09:01
... and done `using namespace std;`. Also, use [RAII](https://en.wikipedia.org/wiki/Resource_acquisition_is_initialization): don't predeclare `int i;`. Where did you base this on? The `#define`s are just plain weird. What's next? `#define false true`? — JHBonarius, Oct 18 '20 at 09:15
`#define int long long int` technically is illegal, because that's a keyword and a standard header is included — Swift - Friday Pie, Oct 18 '20 at 09:27
`#define int long long int` -- Don't do this. There is an `int64_t` type in C++ now. There is absolutely no need for this abomination of a macro. — PaulMcKenzie, Oct 18 '20 at 10:12
`#define int long long int` means your program has undefined behaviour, i.e. the compiler can emit object code that does *absolutely anything* and by the rules of C++ that's valid. — Caleth, Oct 19 '20 at 10:56
@chi plz correct me if i am wrong, s.length() gives the length of the string excluding the null character therefore accessing the last character will be possible only when the condition is i — Bhargav Reddy, Oct 19 '20 at 11:21
@BhargavReddy You are right -- `s[s.length()]` since C++11 is guaranteed to return a null character. Still, you do not really need to check that: I would make the loop to stop earlier. Keep in mind that in your loop you access `s[i+1]` and `i+1` can be equal to `s.length()`, so you do access the null character at the end. — chi, Oct 19 '20 at 11:43

czs108 · Answer 1 · 2020-10-18T09:13:55.293

1

You can change two s.erase(i, i + 1); statements to s.erase(i, 2);.

1
AABBBABBBB
ABBABBBB
BABBBB
BBBB
BB

And you can use C++ STL.

#include <vector>
#include <iostream>
#include <string>

int main() {

    int t{0};
    std::cin >> t;

    while (t--) {
        std::string s{};
        std::cin >> s;
        for (auto i = 0; i < s.length() && i + 1 < s.length(); i++) {
            if (s[i] == 'A' && s[i + 1] == 'B') {
                s.erase(i, 2);
                std::cout << s << std::endl;
                i = -1;
            } else if (s[i] == 'B' && s[i + 1] == 'B') {
                s.erase(i, 2);
                std::cout << s << std::endl;
                i = -1;
            }
        }
    }

    return EXIT_SUCCESS;
}

edited Oct 18 '20 at 09:13

answered Oct 18 '20 at 09:05

czs108

151
4

Yep. The OP wants to erase *two* characters, so the range has to have a width of 2, not 1. – David Schwartz Oct 18 '20 at 09:06
`i < s.length() && i + 1 < s.length()` <-- uhm... don't. And where are you using the STL = standard template library? You `include ` but never use it. `return EXIT_SUCCESS;` is C. `std::endl` flushes the buffer unnecessarily: use `'\n'` instead. No need to initialize `t` and `s`. `i = -1` goes back to the start of the string every time -> `i = std::max(-1, i-1);`. – JHBonarius Oct 18 '20 at 09:29
@JHBonarius Sorry `include ` is unnecessary. I mean use C++ STL header rather ``. About `t` and `s`, in my company, initializing every variable is a good style. You can see this https://dl.acm.org/doi/10.1145/3319535.3345654 and https://cwe.mitre.org/data/definitions/908.html – czs108 Oct 18 '20 at 09:32
That it's your company choosing guideline doesn't mean it's a good. Yes, you don't want issues with using uninitialized variables. But you also don't want unnecessary initialization. – JHBonarius Oct 18 '20 at 09:51
@DavidSchwartz Does erase(i,i+1) means I am deleting only one character ? If is it so then how its deleting two characters in the very first operation ? – Bhargav Reddy Oct 19 '20 at 11:30
@BhargavReddy It deletes only the first character in the very first operation. – David Schwartz Oct 19 '20 at 14:28
@DavidSchwartz but Its deleting "AB" at index 1 and 2 (0 is the 1st index). You can see it yourself. – Bhargav Reddy Oct 19 '20 at 14:37
@BhargavReddy The B is still there. – David Schwartz Oct 19 '20 at 14:58

score 1 · Answer 2 · answered Oct 19 '20 at 11:20

Am I doing something wrong?

Yes, almost everything.

#include <iostream>
#include <string>
#include <regex>

int main()
{
    std::regex re("AB|BB");

    int t;
    std::cin >> t;

    while(t--)
    {
        std::string s;
        std::cin >> s;
        std::string prev;

        // do one replacement at a time, until there are no changes
        do
        {
            std::cout << s << '\n';
            prev = s;
            s = std::regex_replace(s, re, "", std::regex_constants::format_first_only);
        } while (s != prev);
    }
    return 0;
}

Aniket Kariya · Answer 3 · 2020-10-18T09:59:48.730

-1

It's just a simple variation of bubble sort.

#include <iostream>
#include <string>

using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int t;
    cin >> t;

    while(t > 0 && t--)
    {
        string s;
        cin >> s;

        for(size_t i = 0; i < s.length(); i++) {
            for(size_t j = 0; j < s.length() - 1; j++) {
                if((s[j]=='A' || s[j] == 'B') && s[j+1]=='B')
                {
                    s.erase(j,2);
                    cout<<s<<"\n";
                    break;
                }      
            }
        }
    }
    return 0;
}

edited Oct 18 '20 at 09:59

answered Oct 18 '20 at 09:24

Aniket Kariya

1,471
2
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26

@JHBonarius Hope it's better now. – Aniket Kariya Oct 18 '20 at 09:38
i have added the question plz change your code accordingly. – Bhargav Reddy Oct 19 '20 at 10:50
@BhargavReddy There's not much change that I've made to your code. You can just copy the nestled for loops from this code and paste in place of yours and it should still work. But you should follow others advise. What you're doing.. It's a bad code practice. – Aniket Kariya Oct 19 '20 at 10:56
Is (j,2) equal to (j,j+2) ?? – Bhargav Reddy Oct 19 '20 at 11:12
@JHBonarius I am sorry sir, I am still a student who is doing competitive programming and therefore all this bad code i have learned is just to save time. – Bhargav Reddy Oct 19 '20 at 11:16
Why are you apologizing to me? – JHBonarius Oct 19 '20 at 11:18
@BhargavReddy Speed is important when it comes to competitive programming. But remember one thing, CP is for improving your coding skills, your logical reasoning, it might help you become better at coding, but you should always follow best practices. because in the real world, both matters. your logical skills as well as your code quality. – Aniket Kariya Oct 19 '20 at 14:03
@AniketKariya will do sir. – Bhargav Reddy Oct 19 '20 at 14:35

Why does the erase() function works properly only for the first time?

3 Answers3