In the recent snackdown round A qualifier, there was a problem I could not figure out.

Link: https://www.codechef.com/SNCK1A19/problems/AVGMAT

From skimming the solutions and from making my own observations I have realized the following:

• Most solution use some type of DP with two passes storing DP1[i][j] and DP2[i][j], where i and j presumably mean row and column.
• For any given distance D, there are O(D) other nodes D away in manhattan distance
• We can construct ans from dp1 and dp2 where ans[i] stores the amount of pairs of nodes i away from each other in manhattan distance

The editorial is not out, as codechef editorials can be quite slow (some in 2016 still pending).

Can anybody explain the solution here?