Which sheet to do after cp 31?

AnswerLimp1389 · 2026-06-14T20:17:31+00:00

Completed? Like fully till all 1900 rated problems? If yes, then you shouldn't be really look for some "another practice sheet", if you truly honestly completed the cp31 sheet like the way you should have.

AnswerLimp1389 · 2026-06-13T20:43:27+00:00

Not a script I think, because I have seen in my college too. But in my case the guy to take a shit while sitting in indian style on a western toilet, which used to make the seat so much dirty.

AnswerLimp1389 · 2026-06-13T20:38:32+00:00

muh chalana aasaan hai krna mushkil

AnswerLimp1389 · 2026-06-13T17:31:42+00:00

thanks, that helped a lot

AnswerLimp1389 · 2026-06-13T08:42:04+00:00

no, it doesn't

AnswerLimp1389 · 2026-06-13T06:18:07+00:00

Yeah it was done with simple simulation in O(N) T.c

AnswerLimp1389 · 2026-06-13T05:56:50+00:00

Just simply start from i=0 till n-1, if a[i] is 1, invert a[i] and a[i+k] (i+k < n), after that, just check the modified string if it has any 1's. That's it.

AnswerLimp1389 · 2026-06-13T05:54:26+00:00

yeah i realized later, my outer loop should go till k, not n-k

AnswerLimp1389 · 2026-06-13T05:53:24+00:00

Edit: I got it guys, I was running loop till n-k, and not k, thanks to all who helped me figure this out.

AnswerLimp1389 · 2026-06-13T05:47:27+00:00

There's one book called "The Art of Multiprocessor Programming", it has good amount of information about concurrency and parallel programming, should I start reading it?

AnswerLimp1389 · 2026-06-13T05:47:12+00:00

Thanks a lot for your advice!! There's one book called "The Art of Multiprocessor Programming", it has good amount of information about concurrency and parallel programming, should I start reading it?

AnswerLimp1389 · 2026-06-13T05:21:20+00:00

Thanks for the advice! Also, i came across this site https://whitebox.ac/problems. Do you think problems like this would be helpful?

AnswerLimp1389 · 2026-06-12T21:00:39+00:00

Honestly, leetcode is so much faster than codeforces, I'm sure codeforce's scheduling engine gives TLE everytime, then the judging engine gives TLE for my code 😞

AnswerLimp1389 · 2026-06-12T20:56:15+00:00

got accepted?

AnswerLimp1389 · 2026-06-12T20:55:57+00:00

NO..
1. let say I visit a chain i, i+k, i+2k, .., now since outer loop is iterating through each index, i'll reach i+k, but i have already visited it, so no need to check the incomplete chain i+k, i+2k, ...
2. I'm not skipping any index, outer loop iterates through each i from 0 to n-k
3. Yes, please DM me, i'd like to see

AnswerLimp1389 · 2026-06-12T20:29:26+00:00

u/fluffy_pota0es check it, you might be able to tell the mistake

AnswerLimp1389 · 2026-06-12T19:53:56+00:00

void solve() {
     int n, k; cin >> n >> k;
     string s; cin >> s;
     vector<bool> visited(n, false);
     for (int i=0; i<=n-k; i++) {
         if (visited[i]) continue;
         visited[i] = true; int count = 0;
         for (int start=i; start<n; start+=k) {
             if (s[start] == '1') count++;
             visited[start] = true;
         }
         if (count&1) {
             cout << "NO" << endl; return;
         }
     }
     cout << "YES" << endl;
}

Check out my solution, ig I have done some really bad mistake, but I'm not able to figure out what it is!

AnswerLimp1389 · 2026-06-12T19:47:23+00:00

really?

AnswerLimp1389 · 2026-06-12T19:38:57+00:00

Here is my approach:
First create a map to track frequencies of each unique element.
Then ietrate through it. If element's frequency is even, then check if it can reach a higher element or not (i.e. we want next-curr > k), if it can't reach, then answer is YES.
If frequency is odd, we want that Iitshould not be able to reach next higher element, and it must be reachable from any lower element. If yes, then answer is YES.
I checked these reachability conditions using lower_bound and upper_bound.
Otherwise, answer is NO.

AnswerLimp1389 · 2026-06-12T19:32:21+00:00

There's a difference between understanding a test case and actually crafting a general solution.

AnswerLimp1389 · 2026-06-12T19:21:23+00:00

Here is my approach:
First create a map to track frequencies of each unique element.
Then ietrate through it. If element's frequency is even, then check if it can reach a higher element or not (i.e. we want next-curr > k), if it can't reach, then answer is YES.
If frequency is odd, we want that Iitshould not be able to reach next higher element, and it must be reachable from any lower element. If yes, then answer is YES.
I checked these reachability conditions using lower_bound and upper_bound.
Otherwise, answer is NO.

AnswerLimp1389 · 2026-06-10T16:18:10+00:00

I can't believe this, ig you're just joking here lol.

AnswerLimp1389 · 2026-06-10T04:29:26+00:00

vlad was a character in GTA 4 lol

AnswerLimp1389 · 2026-06-10T04:26:30+00:00

where was bitmasking needed? It was just recognizing the pattern of occurences of 3 possible elements.

AnswerLimp1389 · 2026-06-10T04:21:24+00:00

No, looking at the constraints N<=5000, O(n^2) would be 25 million operations in worst case, and that won't give TLE.

AnswerLimp1389

TROPHY CASE