and you kept your promise!!

I kept my promise, I can die with a clear conscience.

Legendary Grandmaster with 1435 Rating??

)

All the best bro

questions ❌

problems ✅

Christmas was on the 25th, it's the 28th now :)

(unless you're talking about Orthodox Christmas on the 7th of January :D)

agreed

Yes

looking forward to your success!

LMFAO!! What a noob

looks pretty real to me

Its real, i clicked the photo

human

Really? Can you send code

How do you get it accepted with 170ms? I tried your code, and got 2000+ms. Where is the difference from?

SecondThread orz

Thanks a lot! It was really nice to see a different view on the tasks

This video is unlisted btw.

might be, solved it using ordered sets, although i believe the author had a simpler solution for F of a Div 4!, btw what was your reasoning behind G?

You can compress the coordinates, sort the intervals and then use fenwick or segment tree to calculate the answer.

Yeah, it's just nlogn inversion count using merge sort

think about the prefix sums of the odd-index and even-index arrays. What can you say about the prefix sum values at the left and right endpoints of a good subarray?

Let $$$\mathrm{sum_{even}}$$$ denote the sum of values on even indices on our current subarray, and define $$$\mathrm{sum_{odd}}$$$ similarly.

$$$\mathrm{sum_{even}} = \mathrm{sum_{odd}}$$$

$$$\mathrm{sum_{even}} - \mathrm{sum_{odd}} = 0$$$

Now, multiply all values on odd indices by $$$-1$$$ in the original array. The condition now becomes

$$$\mathrm{sum_{even}} - (-1) \cdot \mathrm{sum_{odd}} = 0$$$

$$$\mathrm{sum_{even}} + \mathrm{sum_{odd}} = 0$$$

The left side is just the sum of the current subarray. Thus, we need to find a subarray with sum $$$0$$$ $$$\Leftrightarrow$$$ you need to find two equal prefix sums.

a -> pref sum of odd indidces b -> pref sum of even indices for range l...r to be valid ar — al = br — bl -> (ar — br = bl — al) now you know how do it. it's similar to 2 sum problem

you can solve it with a set , make every even index element negative then loop on the array and insert the prefix sum till i in the set , and find if you have the same prefix already , if true then its a YES .

Trying same thing from hours but don't know why it's not working... I hope some explain flaw in this approach

distance is $$$O(n)$$$

Time complexity of the distance method is O(N), which means your final complexity is O(N^2)

(Co-ordinate compression + segment Tree) will work too.

We just have to found the subarray (with certain sum x) whose sum of odd and even places is same. what we can do is if we multiply -1(either of the odd or even position) now we just have to find the contiguous subarray with 0 sum that can be done by maintaining a prefix sum and a set(basic hashing)...

You can take help by this video..

For F, you can see that 2 intervals contribute to the answer only if 1 is completely into another. Noy you can compresd all the coordinates, sort the intervals by ascending xleft coordinate and use segment or fenwick tree to calculate the answer. Tou can see my code for more details.

Problem F can be soved using ordered set.239369576

Bro really overthinking B

Tell me one thing, did you copy C? Someone for whom traversing 2D array is tough, had template in C (but not in A and B where it might be useful) which was anyways not required.

read this blog

    if(i + 3 >= n){
       fin += s[i], fin += s[i + 1], fin += s[i + 2];

       cop.push_back(fin);
       i += 2;
       cc++;
       break;
    }

Here, what if i + 2 >= n

point(4,5) meet with (3,9) at 2nd sec ->count=1; point(2,6) meet with (4,5) at 3rd sec ->count=2; point(1,8) meet with (4,5) and (2,6) at 3rd,5th sec ->count=4; point (-2,100) meet with every other point at 7th,8th,10th,11th,12thsec ->count=4+5=9;

You must find all $$$a_i > a_j$$$ such that $$$b_i < b_j$$$.

The $$$(a_i, b_i)$$$ and $$$(a_j, b_j)$$$ pairs that meet this criteria are as follows:

• 1 8 greets -2 100;
• 2 6 greets -2 100 and 1 8;
• 3 9 greets -2 100;
• 4 5 greets -2 100, 1 8, 2 6 and 3 9;
• 7 10 greets -2 100.

So in total there are 9 greetings.

why

He can only use bikes from the cities he has travelled through, i.e. only the bikes from the cities that lie on the path(or walk) from city 1 to city C.

1 -> 2 -> 3 -> 2 -> 4 -> 5

1 -> 2 time = 10 2 -> 3 time = 10 + 2 3 -> 2 time 10 + 2 + 1 2 -> 4 time = 10 + 2 + 1 + 5 4 -> 5 time = 10 + 2 + 1 + 5 + 1 total time = 19

I'll denote the fastest bike that we currently have by $$$b$$$.

The steps are as follows:

• Start in city 1.
• Buy the bike in city 1, now $$$b = 5$$$.
• Go to city 2, it takes $$$5 \cdot 2 = 10$$$ minutes.
• Buy the bike in city 2, now $$$b = 2$$$.
• Go to city 3, it takes $$$2 \cdot 1 = 2$$$ minutes.
• Buy the bike in city 3, now $$$b = 1$$$.
• Go to city 2, it takes $$$1 \cdot 1 = 1$$$ minute.
• Go to city 4, it takes $$$1 \cdot 5 = 5$$$ minutes.
• Go to city 5, it takes $$$1 \cdot 1 = 1$$$ minute.

Elapsed time: $$$10 + 2 + 1 + 5 + 1 = 19$$$.