CUTBOARD - Editorial

Problem Link:

Practice

Contest

Setter: Alei Reyes

Tester: Triveni Mahatha

Editorialist: Triveni Mahatha


Difficulty:

CAKEWALK

Prerequisites:

None

Problem:

Given N x M chessboard. Make cuts on some of the edges but don’t cut the board into pieces. How many maximum such cuts can you make?

Explanation:

Note that we can make cuts between every two consecutive row of cells. There will be N - 1. Such rows to be cut. Within a row we can make M - 1 cuts. This way of cutting will make the board look like an extended E - shape structure. Refer to the figure for more understanding -

So number of cuts is (N - 1) \times (M - 1)

See code below -

    int T = readInt();
    while(T --> 0) {
       int n = readInt();
       int m = readInt();
       int ans = (n - 1) * (m - 1);
       printInt(ans);
    }

Image - Insert Image Here

SOLUTION:

Setter

Tester

Time Complexity:

O(1), per test case.

Space Complexity:

O(1)

6 Likes

Imagine it as a graph with N*M nodes. In each row we have M-1 edges. Similarly in each column we have N-1 edges. Total we have N*(M-1)+M*(N-1) edges. And we have to make a spanning tree from this graph which which will have total edges equal to it’s number of nodes-1 which is N*M-1 edges.
Difference is N*(M-1)+M*(N-1)-(N*M-1)
which finally evaluates to same expression (N-1)*(M-1) as above.

12 Likes

I think too many example cases were given in this question. One can easily solve the question without even reading it and looking on the test cases. I did not read the question carefully and found the pattern by examining the example cases. Tried it and got AC-ed.

Seems like overkill-imagination for a cakewalk question xD. BTW- fixed editing.

4 Likes

How can you prove this is the maximum number of steps possible? What is we cut in an other way?
i think we have to prove that, in any way of cutting the edges, the maximum possible steps is (n-1)*(m-1).

1 Like