Whats wrong in Beautiful Garland

general
,
1

https://www.codechef.com/viewsolution/21673749

i have applied the concept that if length of garland is 1 then garland is beautiful else i hae counted number of R and G and if they are equal and their sum is even and then i have counted the mismatched pair and printed the result accordingly.

Can anyone please help me to figure out whats wrong…??

2

When searching for pairs, you didn’t check for a pair at the start-to-end join. The garland is circular.

You could fix it with:

       prev = s[len-1];
       for(int i=0;i < len;i++){
            if(s[i]==prev){
                ct++;
                if(s[i]=='R') rr=1;
                else gg=1;
            }
            prev = s[i];
        }

(or something like that to include first==last… not checked)

Aside: The second part of your test

(r==g && (r+g)%2==0)

is redundant; if r==g then of course r+g is even.

1 Like
3

its still showing WA

4

i have read the editorial and have done exactly same but still getting WA

5

@joffan is absolutely right. @rk_221b you didn’t check for the start and end joint. Even if you did it later, it isn’t correct. Here is the small modification to your code which gives AC.

6

Thanks @joffan and @yadnesh_viper for pointing out my mistake.