My code is not getting accepted ('infectious letters' on codeforces)

Question

1.00/5 (1 vote)

See more:

#include<bits/stdc++.h>
using namespace std;
void letters(string &s,int n,int &i){
if(i==n) return;
if(s[i]=='a'){
int m=i;
while(s[m-1]!='b' && m>0){
s[m-1]='a';
m--;
}
int j=i;
while(s[j+1]!='b' && j<n){
s[j+1]='a';
j++;
}
}
letters(s,n,++i);
}
int main(){
string s;
cin>>s;
int i=1;
int n=s.length();
letters(s,n,i);
int ans=count(s.begin(),s.end(),'a');
cout<<ans;
}

What I have tried:

Tried to submit this code on codeforces but it is giving partial output as result.
There are  no solutions or editorials related to this.
Please help???

Posted 30-Aug-22 23:32pm

Animesh _1SG20IS008

Updated 31-Aug-22 11:24am

v3

Add a Solution

Comments

CHill60 31-Aug-22 5:32am

"accepted" for what?

Animesh _1SG20IS008 31-Aug-22 5:40am

When tried to submit this code on codeforces it is giving partial output as result.

CPallini 31-Aug-22 6:31am

That means your code is incorrect. You should try to locate the bug either by code inspection or providing ad hoc inputs to verify what is happening.

3 solutions

Add a Solution

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

CPallini · Answer 1 · 2022-08-31T00:48:00

Quote:
while(s[j+1]!='b' && j<n){
s[j+1]='a';
j++;
}

This is a bug: when j is equal to (n-1) item s[n] (out of array bounds) is accessed.

Moreover:

You don't need recursion. At least you don't need recursion the way you have implemented it.
If you are at index i then you propagate the 'a' both backward and forward. Once you finished the forward propagation then you should consider index (i+j), instead of (i+1) (that is inefficient).

OriginalGriff · Answer 2 · 2022-08-30T23:44:00

Solution 1

We can't help: we have no access to whatever standards codeforces may expect your code to meet.

Talk to them: they are the only people who can help you.

Posted 30-Aug-22 23:44pm

OriginalGriff

Patrice T · Answer 3 · 2022-08-31T11:24:00

Problem - 1028923 - Codeforces[^]

Quote:
Please help???

It is amazing to see how your small piece of code is complicated for such a simple problem.
Using recursion is overkill and storing 'a's every where in string is unnecessary.
All you need is keep track of last 'a' and last 'b'.

C++

int letters(string &s){
    int counter= 0;
    int LastA= -2;
    int LastB= -1;
    for (int i= 0; i<s.length(); i++) {
        if (s[i]== 'a') {
            // if last time was a 'a'
            //    update counter from LastA
            //    update LastA
            // else
            //    update counter from LastB
            //    update LastA
        }
        else if (s[i]== 'b') {
            // if last time was a 'a'
            //    update counter from LastA
            //    update LastB
            // else
            //    update LastB
        }
    // At the end, check if you have a run of 'a's to finish
    }
    return counter;
}

This code is only general structure and not intended to give a full solution.

My code is not getting accepted ('infectious letters' on codeforces)

3 solutions

Solution 2

Solution 1

Solution 3

Add your solution here

Preview 0

My code is not getting accepted ('infectious letters' on codeforces)

3 solutions

Solution 2

Solution 1

Solution 3

Add your solution here

Preview 0

Existing Members

...or Join us