CHEFSUBA - Editorial

editorial
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21

Woah, you’re all right, actually the site went under maintenance when I submitted previously and it didn’t show the WA in the task. Now, I made the change and its working (all AC’s). Thanks guys. Lol … hours of debugging and missed this one.

22

https://www.codechef.com/viewsolution/13531464
I think it is better Approach ! :stuck_out_tongue:

23

I did this Question using sliding windows by precompute values and deque to get the maximum value

24

O(n) solution can be done without deques, using frequency table to get the max. Since values are 0 and 1 only, the max value changes by at most 1.

Solution:
https://www.codechef.com/viewsolution/13519755

2 Likes
25

Here is my solution: https://www.codechef.com/viewsolution/13634699 as same as author solution but I don’t know why my solution got wrong. Please tell me why I got wrong. Thank you.

26

One question guys!
Is the complexity O(n) for the entire subtask or O(n) for each querie?

27

Can someone tell me whats wrong with my code. I would be very greatful.
Submission : https://www.codechef.com/viewsolution/13640002

28

This Solution is giving TLE. I have used Segment Tree/ Range Maximum Query. Please share your insight.

29

A deque approach with window sliding and taking care of k>n condition and array sizes declaration gives an AC.
here is my easy to understand solution easy solution

30

I used trie https://www.codechef.com/viewsolution/13584501

31

Please check this solution @chaitya_62

32

i am using segment trees as said above . i am getting WA. please help me

https://www.codechef.com/viewsolution/13739007

33

can anyone explain this part
else the answer is the maximum sum we can obtain by starting the window from 1 to (n−k+1).

34

Someone please tell me why my code is giving SIGABRT
https://www.codechef.com/viewsolution/13742388

35

using namespace std;

include

define _ iosbase::syncwith_stdio(false);cin.tie(NULL);

define D(x) cout<< #x " = "<<(x)<<endl

int main() { _

int n;cin>>n;
int acum = 0;
int ans = 0;
int t;
for (int i = 0; i < n; ++i) {
cin>>t;
if (t == 0) {
ans = max(ans, acum);
acum = 0;
} else {
acum++;
}
}

ans = max(ans, acum);
cout<<ans<<endl;

return 0;
}

36

I have solved the same problem without using dequeue, segment tree or any sparse table.
I have done it just by checking the contribution of the shifted 1 during the shift if last digit is 1 into the new forward window formed using that new 1 and removing its contribution from the last window formed earlier using that 1 digit. Similar is in case of a 0. To maintain the counts of 1’s in the array used a multiset to store the counts of continuous 1’s in the array after the shift. Answer will be the largest value in multiset whenever query occurs.
Do check once!

https://www.codechef.com/viewsolution/13527132

Thanks

37

Can someone help me with my code, I’m getting WA in 2 test cases - one from small and one from large
I’ve checked the condition for k>n using k = min(k,n).

https://www.codechef.com/viewsolution/13745994

38

what will the size of sum array?

39

Why do I get a SIGSEV error on exactly one of the testcases? All the others are AC. I used two sliding windows.

Code here

40

Can anyone reply with side cases and all the test cases you prepared for this problem. I need to run the code on those as its giving a WA.
If anyone is willing to look at the code here is the link: https://www.codechef.com/viewsolution/13789471
The code is simple with no use of special data structures, easy to check.
Thank you in advance :slight_smile: