It's not hard to notice that if we have x same kind of stones, then we will need at least x boxes. So the answer is maximum frequency of any element in the array.

void __jaiShreeRam() {

getArr

map<ll,ll> mp;

for(auto&i:arr) mp[i]++;

    ll ans = 0;

for(auto&i:mp) ans = max(ans, i.second);

    cout<<ans<<'\n';

}

Since we have to take the shakes from a single shop, so we will move to each shop and will try to find how many maximum shakes we can buy from that particular shop with the amount of money we are having.

No. of shakes we can buy from i'th shop will be : min(a[i], money/price) And we will take the maximum of each shop

void __jaiShreeRam() {

ll money,q; cin>>money>>q;

ll ans = -1, ind_ = 1;

ll shakes = 0;

while(q--){ ll a,m; cin>>a>>m;

ll have = money/m;

ll shakes_have = min(have,a);

if(shakes_have>shakes){
    shakes = shakes_have;
    ans = ind_;
}

ind_++;

}

cout<<ans<<'\n';

}

I was completely exhausted today in my work, Internship e.t.c so my brain was not able to figure out the approach during the contest because I haven't focused on constraints and alphabets. I've also asked the solution for D above but later on I was able to figure out the solution and I have done it in simpler manner than the guy explained above. By the way Thanks to him for explaining.

Let's fix the left point of my substring [l,r] for now. Now think how much do we have to traverse for r ?

There will be at max 26 alphabets, and we want 7 char as max. So In worst case we have to check for at most 26*7 = 182 . So my r will be from l+1 to min(n,l+182). After this max frequency will become 8.

So we will iterate over l and check for r how far we go and getting the substring with at max 7 freq. of any character. To check this we can maintain a vector of size 26.

void __jaiShreeRam() {

ll n; cin>>n;

string s; cin>>s;

ll ans = 0;

loop(left,0,n){
    vl freq(26,0);

    ll right = left;

    bool found = false;

    while(right<n){

     freq[s[right] - 'a']++;

     if(freq[s[right] - 'a']>7){
        break;
    }

    if(freq[s[right] -'a'] == 7){
        found = true;
       // cout<<left+1<<" "<<right+1<<endl;
    }

    ans+=found;

    right++;
}

}

cout<<ans<<"\n";

}

You must be familiar with the data structure SEGMENT TREE.

We can use Segment tree to solve this. And each node of segment tree will store 3 things left(left point of the segment), right(right point of the segment) and a multiset, in which we will store the values of each segment. So that after cancellation we can restore the state of the array.

Now Just process the queries. if query is of type 1 then add the element x by updating the multiset. if query is of type 2 then remove the element and restore the state by removing element from multiset if query if of type 3 then print the maximum by querying max from segment tree.

This Problem is Very interesting. Try by yourself so that you can improve your understanding of segment tree.

I'll share the code also If you want. So please do reply if you are facing any issues in Implementation