CHEFBM - Editorial



Author: Dmytro Berezin
Tester: Sergey Kulik
Editorialist: Lalit Kundu






There is a N * M matrix, where A[i][j] is j for all i,j. There are P queries of form (i,j), where we do A[i][j]++.
For each row i,
if there exists a j, such that A[i][j] > A[i][j+1], print -1.
else print sum[i=M to 2]{A[i][j]-A[i][j-1]}.
1 ≤ N,M,P ≤ 105


For each i, we also keep note of values (i,j-1) and (i,j+1), if (i,j) is given in input. Traversing over the queries, we can see which values change and print answer accordingly.


We have to process for each i independently. Let’s say for an i, we have NN queries of form (j,k) which means: increase A[i][j] by k. For such a query we also store (j-1,0) and (j+1,0), because these two indexes are also effected.
If there were no changes, our answer will be m-1. Now, we traverse over all queries in increasing order of j and look at index j+1, if the value at j+1 is less than value at j, we know that answer is -1 for this row. Else, we add the (A[i][j+1]-A[i][j]) to the answer and subtract -1 from the answer. We can use map to perform this process efficiently. See (setter/tester/editorialist)'s solution for implementation details.


Author’s solution
Setter’s solution
Editorialist’s solution


" For such a query we also store (j-1,0) and (j+1,0), because these two indexes are also effected."

Can anyone expand on this line? Where do we store that?

for which test case does this code fails what is wrong in it

for input

1 4 5
1 3
1 2
1 2
1 1
1 1

it returns -1

initial matrix


after additions


so the result should be 1…


can you help me as well

@ashish424 see this… …u r printing extra numbers!!!


I tried many times but always got WA . Both the above use case is correct for me .
If any one can give me test case that caused WA , it will be a great help.


try this case…

4 1 1
1 1

ans should be:-


urs is…


Since the explanation states “We can use map to perform this process efficiently”, I assume it means that we have to add an entry in our map which shows that (j-1) and (j+1) are not increased. Or more plainly, j+1 and j-1 are increased by 0.
Can someone please check if this is giving right answers on your test cases? It can calculate results even if input of a row changes after changes in other rows, for example

4 4 6

3 2

2 3

3 2

2 4

1 2


Can the approach be this:
storing count of the incremented index(the coloumn), adding to it the index values. Sort this, if its same as the initial string, ans will be m-1 otherwise -1. For eg.

4 4 6
2 2
3 2
3 2
4 3
4 4
4 3

then for the third row, count(this array will be 1-indexed) will be 0 2 0 0, adding the index values to it, it will be 1 4 3 4. Now if i’ll sort this, i’ll not get the same array and hence answer should be -1. I implemented this here: and its most obvious to give TLE. BUt is there any way I can optimize this??

i corrected that …but it is still showing wrong answer

Why does this ans give TLE:
It works fine on my computer.Please help.

Where shall i get the input file and output file?

sorting is a very expensive operation. Initial cost is M-1. Consider this. If first element is increased, cost goes down by 1. If last element is increased, cost goes up by 1. If any other element is increased, as long as it is smaller than next, cost is unaffected. Hope you get the rest of idea, like checking if it was previously smaller and now larger and all that too.

Why does this ans give WA: It works fine on my computer.Please help.

@darkshadows : this kind of discussion (specially commenter’s reply) clearly shows that there is a lack of effort in editorial making. Similar responses have been in the last long challenge when you were the editorialist (@xwllos0 saved your day by his solution Please don’t slack off on your work, it is an important reason for codechef to attract competitive programmers.


You mean test cases? Those are not available…
its working fine for every test cases…but getting wrong answer…can any one please help

I did that way and got AC also. But i was just thinking if some way of the above kind(using sort) exists.