# BOXINBOX Editorial

BOX IN BOX EDITORIAL.

SETTER : @aayushindwan
Editorialist : @jprm2

Difficulty : Easy

EXPLANATION

In this problem It was clearly visible that for each frame we have to select the minimumvalue of that frame so if N is even then only N/2 frames are possible and if N is odd then only N/2 +1 frames are visible and so always we have to sum up (N+1)/2 minimum values 1 from each frame and finally print the answers. So this is greedy approach when we choose always minimum value from the frames.

Now the question is how you will iterate frames wise now this might have many different approaches.What I preferred was choosing left=1, right=N and up=N and down=N values
of the currrent iterator and moving from current left to right with this fixed up and down iterators for covering the rows and in reverse way moving from

up to down with this fixed right and left iterator to move over the columns. Now the minimum of both these iterations updated simultaneously be minr , minc.
and finally updating answer with ans+=min(minr,minc); and finally printing the answer over all such (N+1)/2 frames and summing up the answer also in each iteration reducing the size of the matrix by incrementing left by1 , decrementing right by1 , incrementing up by 1 and decrementing down by 1.

Editorialistâ€™s Solution:

Editorialists Solutions.