Thanks for your help!

Now, if you could please explain the problem statement for Q5.

DP I think not quite sure if some Greedy solution is there or not but DP solution did pass

Were test cases for D weak? I couldn't figure out the linear solution so decided to simply submit my $$$O(n^2)$$$ DP solution just for the sake of practice and it surpisingly didn't give me TLE (though there was a bug in my implementation which I couldn't fix in time ): ).

In D, Suppose there are 6 C's, then consider the path -> -> -> <- <- <- (-> means i will go forward and <- i will go backward). Calculate the answer for that path (call it rel), now I will calculate answer for a different path RELATIVE to this path. Call the difference delta. Then answer for arbitrary path = rel + delta. I want to maximize delta (rel is constant). Observe that if I flip -> arrow at the ith index It changes delta by +2*(suffsub (i) — suffadd(i) ) (Suffadd(i) is the no of '+' in string in the indexes from [ i to N ] , Suffsub(i) is the same for '-' ).

Also observe that If I flip -> into <- then I also have to flip some other <- into ->. (To maintain the fact that I have to reach starrting pos at the end). So max value of delta is some j highest values of -> flip and <- flip.

Assume you have $$$2 \cdot x$$$ times $$$C$$$. Now you have to replace $$$x$$$ $$$C$$$ with $$$1$$$ and remaining $$$x$$$ $$$C$$$ with $$$-1$$$.

Suppose you are at some $$$i$$$ such that $$$s[i]='+' $$$, what will you add to the answer for this $$$+$$$?

The answer is number of times $$$1$$$ occured before position $$$i$$$ — number of times $$$-1$$$ occured before position $$$i$$$.

Now think other way round.

Suppose you at position $$$i$$$ such that $$$s[i]=C$$$, and you have $$$x$$$ $$$'+'$$$ to the right of $$$i$$$ and $$$y$$$ $$$'-' $$$ to the right of $$$i$$$. Now you add $$$x-y$$$ to the answer if you have $$$1$$$ at this position, otherwise you add $$$-(x-y)$$$ to the answer.

Thus you take another array $$$b$$$ such that

$$$\bullet$$$ $$$b[i]=0$$$ if $$$s[i]=C$$$

$$$\bullet$$$ $$$b[i]=1$$$ if $$$s[i]=+$$$

$$$\bullet$$$ $$$b[i]=-1$$$ if $$$s[i]=-$$$

You find suffix sum of this array, and append $$$suff[i]$$$ in some vector(say track), if $$$b[i]=0$$$.

Now sort track, and subtract first $$$x$$$ elements from the answer, and add last $$$x$$$ elements to the answer.

void solve(){         
    ll n,ans=0,now=0; cin>>n;  
    vector<ll> track;
    string s; cin>>s;
    reverse(all(s));
    for(auto it:s){  
        if(it=='+'){  
            now++;  
        }
        else if(it=='-'){
            now--;
        }
        else{
            track.pb(now); 
        }
    } 
    sort(all(track));
    n=track.size();
    for(ll i=0;i<n/2;i++){
        ans+=track[n-i-1]-track[i]; 
    }
    cout<<ans;
    return;          
}