hackercupai/hackercup huggingface dataset:
The key observation required is that the common part of Mr. Krabs and Plankton's paths will always be a single sub-row or sub-column of the input grid. We can use this observation to construct a dynamic programming approach.

Let's denote \(dp(r, c, d)\) as the minimum possible number of candies Mr. Krabs can collect if:
- Plankton started his route from house \((r, c)\),
- Plankton's route's next segment's direction is \(d\) (\(0\) for vertical, \(1\) for horizontal),
- Mr. Krabs's path intersects Plankton's path, and
- Mr. Krabs strives to maximize the number of candies he gets.

We see that the answer to the problem is equal to \(\min(dp(1, C, 0), dp(1, C, 1))\). To compute each \(dp(r, c, d)\), we will need to consider all valid lengths of the next path's segment. For each length, consider two scenarios and pick the maximum of those:
1. Mr. Krabs's path intersects the current segment.
2. Mr. Krabs's path will intersect one of the next segments.

For case 1, we can precompute helper tables \(\text{max\_up\_left}(r, c)\) and \(\text{max\_down\_right}(r, c)\) for the maximum number of pieces of candy that can be collected on the untouched grid. These computations are a classic dynamic programming problem, which can be done in \(O(R* C)\) time. The implementation requires considering all possible ways of crossing the current segment and finding the most profitable way for Mr. Krabs.

For case 2, we should simply check the value of \(dp(r', c', 1 - d)\), where \((r', c')\) is the first cell of the next segment.

In total, there are \(\mathcal{O}(R*C)\) DP states. Since we need to consider all possible valid segment lengths for each state, the overall complexity of the solution is \(\mathcal{O}(R*C*(R + C))\).