CHEFSUBA - Editorial

chefsuba
editorial
long-contest
may17

#17

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


#18

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


#19

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.


#20

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


#21

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


#22

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


#23

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


#24

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


#25

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

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


#26

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).


#27

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


#28

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; }


#29

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


#30

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


#31

what will the size of sum array?


#32

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


#33

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:


#34

Now its done. Still gives TLE in 3-4 cases.
Code: https://www.codechef.com/viewsolution/13789884


#35

My code is giving NZEC in certain cases. Can anyone please help why is this happening?

class CHEFSUBA
{
public static void main(String[] args) throws IOException
{
boolean readFromFile = false;
boolean outputToFile = false;

	CustomInputOutputClass cio = new CustomInputOutputClass(readFromFile, outputToFile);

	int[] temp = cio.readIntegerArrayFromSingleLine();
	int n = temp[0];
	int k = temp[1];
	int p = temp[2];
	int[] origArr = cio.readIntegerArrayFromSingleLine();
	String queryString = cio.readString();
	
	ArrayList<Integer> answerList = new ArrayList<Integer>();
	int length = origArr.length;
	int[] arr = new int[2 * length];
	int arrLength = 2 * length;
	int[] maxArray = new int[arrLength - k + 1];

	for(int i = 0; i < 2 * length; i++)
	{
		arr* = origArr[i % length];
	}
	for(int i = 0; i < length; i++)
	{
		int t = arr*;
		arr* = arr[2 * length - 1 - i];
		arr[2 * length - 1 - i] = t;
	}

	int count = 0;
	for(int i = 0; i < k; i++)
	{
		count = count + arr*;
	}
	maxArray[0] = count;

	for(int i = 1; i < arrLength - k + 1; i++)
	{
		maxArray* = maxArray[i - 1] - arr[i - 1] + arr[i + k - 1];
	}

	/*
	for(int x: maxArray)
		System.out.print(x + "	");
	System.out.println();
	*/

	Deque<Integer> deque = new LinkedList<Integer>();
	
	for(int i = 0; i < length - k + 1; i++)
	{
		if(deque.isEmpty())
			deque.addFirst(i);
		else
		{
			if(maxArray* < maxArray[deque.getLast()])
				deque.addLast(i);
			else
			{
				while(!deque.isEmpty() && maxArray[deque.getLast()] <= maxArray*)
					deque.removeLast();
				deque.addLast(i);
			}
		}
	} 
	
	//for(int x: deque)
	//	System.out.print(x + "	"); 			

	answerList.add(maxArray[deque.getFirst()]);

	count = 0;
	for(int i = length - k + 1; i < maxArray.length; i++)
	{
		if(deque.getFirst() == count)
			deque.removeFirst();

		if(deque.isEmpty())
			deque.addFirst(i);
		else
		{
			if(maxArray* < maxArray[deque.getLast()])
				deque.addLast(i);
			else
			{
				while(!deque.isEmpty() && maxArray[deque.getLast()] <= maxArray*)
					deque.removeLast();
				deque.addLast(i);
			}
		}
		answerList.add(maxArray[deque.getFirst()]);
		count++;
	}

	//for(int x: answerList)
	//	System.out.print(x + "	");

	count = 0;
	for(int i = 0; i < queryString.length(); i++)
	{
		char query = queryString.charAt(i);
		if(query == '?')
			System.out.println(answerList.get(count));
		else if(query == '!')
			count++;
		
	}
	cio.cleanup();
}

}


#36

My solution is showing TLE for few cases. Please help.

https://www.codechef.com/viewsolution/13978376