The first section will go over the full solution and the last paragraph will touch upon the intended solutions for some subtasks. Let the mean of the entire array be ~\mu~. Notice that the mean of each component of a valid partition must be ~\mu~. Now consider the transformation of ~A_{i,j} = T_{i,j} - \mu~. This reduces the problem to splitting array ~A~ into the maximum number of parts such that each part has a sum of ~0~.

Now let ~P_{j,i}~ be the prefix sum array for ~A_j~. That is,

$$\displaystyle P_{j,i} = \sum_{c=1}^{i} A_{j,c}$$

We will solve the new problem with dynamic programming. Let ~dp[k][i]~ represent the maximum number of parts we can split the first ~i~ columns of the array such that the sum of each part is ~0~ and there is a (possibly empty) component sticking out of the ~k^{\text{th}}~ row (row ~1~ or row ~2~). Thus, for the base case we have that ~dp[k][0] = 0~.

We can incrementally update ~dp[k][i]~ starting from the smallest ~i~. First set ~dp[k][i]~ to be the maximum of

1. ~dp[k][i-1]~ (~A_{k,i}~ is sticking out)
2. ~dp[3-k][i-1]~ (~A_{k,i}~ and ~A_{3-k,i}~ are in one component)
3. ~dp[k][j] + 1~ such that ~P_{3-k,j} = P_{3-k,i}~ and ~j < i~ (Add a "line component" from ~A_{3-k,j+1}~ to ~A_{3-k,i}~)
4. ~dp[3-k][j]+1~ such that ~P_{k,i} + P_{3-k,j} = 0~ (Finish the component by cutting a chunk from the first ~i~ cells on row ~3-k~ and the first ~j~ cells on row ~k~)

Then if ~P_{1,i} +P_{2,i} = 0~, simultaneously update ~dp[1][i]~ and ~dp[2][i]~ as ~max(dp[0][i], dp[1][i]) + 1~ to signify that we split between column ~i~ and ~i + 1~ (or we reach the end of ~i = N~).

The final answer is ~dp[1][N]~. To help understand how this is sufficient, it is helpful to draw out the transitions with multiple cases and notice that the sum of the component that is sticking out for ~dp[k][i]~ is ~P_{k,i}~.

This results in an ~\mathcal{O}(N^2)~ algorithm as seen in cases ~3~ and ~4~. We can make it run in sub-quadratic time by maintaining maps of ~P_{k,i} \rightarrow~ max ~dp[k][i]~ and ~P_{k,i} \rightarrow~ max ~dp[3 - k][i]~ as we compute ~dp[k][i]~.

For the first ~2~ subtasks, we can generate all the possible partitions by hand for Subtask ~1~ or by running a clever brute force for Subtask ~2~. The other subtasks are used to reward suboptimal dynamic programming solutions such as using the sum of the component as a state, using the prefix of the first row and the second row simultaneously as a state or with suboptimal transitions.