Embarking on a daily coding challenge journey is an excellent way to sharpen problem-solving skills and delve deeper into complex algorithms. Today, we’re tackling a classic dynamic programming problem often referred to as “Cherry Pickup I” or, in this context, “Chocolate Pickup,” from GeeksforGeeks. This problem tests your ability to think about paths in a grid and optimize for maximum collection under specific constraints.

The “Chocolate Pickup” Challenge: Problem Description

Imagine a square grid of size n x n. Each cell (i, j) in this grid either contains a certain number of chocolates (a non-negative integer) or is blocked, indicated by a value of -1.

Your task is to find the maximum number of chocolates a robot can collect by completing a round trip:
1. Forward Journey: Starting from the top-left cell (0, 0), the robot moves to the bottom-right cell (n-1, n-1). During this phase, it can only move right (i, j+1) or down (i+1, j).
2. Return Journey: From the bottom-right cell (n-1, n-1), the robot returns to the top-left cell (0, 0). During this phase, it can only move left (i, j-1) or up (i-1, j).

Crucial Rules:
* If a cell (i, j) contains chocolates, the robot collects them, and the cell becomes empty (its value becomes 0) for any subsequent visits. This means chocolates can only be collected once from any given cell.
* If a cell is blocked (-1), it cannot be visited.
* If there’s no valid path for either the forward or the return journey, the total chocolates collected should be 0.

Constraints:
* 1 <= n <= 50 (Grid size)
* -1 <= mat[i][j] <= 100 (Chocolates in a cell or blocked)

Example Scenario:
Consider a grid where paths might overlap. If an optimal path involves collecting 7 chocolates, it’s typically achieved by one robot taking a forward path and another taking a path that, when reversed, constitutes the optimal return journey. The key is to ensure shared cells are only counted once. If an example path results in 0, it means no valid route exists due to blocked cells.

Transforming the Problem for an Optimal Solution

The “forward and back” journey can be elegantly rephrased for a dynamic programming approach. Instead of one robot making a round trip, we can consider two robots simultaneously moving from the starting cell (0, 0) to the destination cell (n-1, n-1).

• Robot 1 follows the path from (0, 0) to (n-1, n-1).
• Robot 2 also follows a path from (0, 0) to (n-1, n-1).

The rules for movement (down or right) and chocolate collection (cell becomes empty after visit) still apply. If both robots arrive at the same cell, the chocolates are collected only once. This transformation simplifies the problem into finding two optimal paths that maximize total chocolates collected, with the constraint of single collection per cell.

Dynamic Programming to the Rescue

This problem exhibits optimal substructure (the optimal solution for a larger problem can be constructed from optimal solutions of subproblems) and overlapping subproblems (the same subproblems are solved repeatedly). These characteristics make Dynamic Programming an ideal candidate. We can use memoization (top-down DP) to store the results of subproblems and avoid redundant calculations.

The state for our dynamic programming solution will be (r1, c1, r2, c2), representing the current positions of Robot 1 (r1, c1) and Robot 2 (r2, c2) as they both move towards (n-1, n-1).

Understanding the Solution Code

Let’s break down the provided Python solution:

class Solution:
    def chocolatePickup(self, mat):
        n = len(mat)
        from functools import lru_cache

        @lru_cache(None)
        def solve(r1, c1, r2, c2):
            # Base Cases for invalid paths
            if r1 >= n or c1 >= n or r2 >= n or c2 >= n:
                return float('-inf')  # Out of bounds
            if mat[r1][c1] == -1 or mat[r2][c2] == -1:
                return float('-inf')  # Blocked cell

            # Base Case: If both robots reach the destination (n-1, n-1)
            # This logic implicitly handles a robot reaching the end and collecting its final chocolate.
            if r1 == n - 1 and c1 == n - 1:
                return mat[r1][c1]
            if r2 == n - 1 and c2 == n - 1:
                return mat[r2][c2]

            # Calculate chocolates collected at current positions
            current_chocolates = 0
            if r1 == r2 and c1 == c2:
                # Both robots at the same cell, collect once
                current_chocolates += mat[r1][c1]
            else:
                # Robots at different cells, collect from both
                current_chocolates += mat[r1][c1] + mat[r2][c2]

            # Explore all possible next moves for both robots
            # Each robot can move Down (D) or Right (R)
            # Possible combinations: (D,D), (R,R), (D,R), (R,D)
            max_next_chocolates = max(
                solve(r1 + 1, c1, r2 + 1, c2),  # Robot 1 Down, Robot 2 Down
                solve(r1, c1 + 1, r2, c2 + 1),  # Robot 1 Right, Robot 2 Right
                solve(r1 + 1, c1, r2, c2 + 1),  # Robot 1 Down, Robot 2 Right
                solve(r1, c1 + 1, r2 + 1, c2)   # Robot 1 Right, Robot 2 Down
            )

            # Add current chocolates to the maximum possible future chocolates
            return current_chocolates + max_next_chocolates

        # Start the recursive calls from (0,0) for both robots
        ans = solve(0, 0, 0, 0)
        # Return 0 if no valid path exists (ans would be float('-inf'))
        return max(0, ans)

Explanation of Key Parts:

1. @lru_cache(None): This decorator from functools automatically memoizes the results of the solve function calls. If solve is called with the same arguments multiple times, it returns the cached result instead of recomputing, drastically improving performance.
2. solve(r1, c1, r2, c2): This is our recursive function representing the DP state.
   • Invalid Path Base Cases:
     • if r1 >= n or c1 >= n or r2 >= n or c2 >= n: If either robot moves out of the grid boundaries, this path is invalid, and we return float('-inf') to ensure it’s not chosen.
     • if mat[r1][c1] == -1 or mat[r2][c2] == -1: If either robot lands on a blocked cell, the path is invalid, also returning float('-inf').
   • Destination Base Cases:
     • if r1 == n - 1 and c1 == n - 1: return mat[r1][c1]
     • if r2 == n - 1 and c2 == n - 1: return mat[r2][c2]
       These lines handle the scenario where one of the robots reaches the destination (n-1, n-1). At this point, it collects its chocolate, and its future contribution from that point for further moves is implicitly handled as moves beyond (n-1, n-1) would lead to out-of-bounds.
   • Chocolate Collection Logic:
     • if r1 == r2 and c1 == c2: If both robots are at the same cell, chocolates are counted only once (mat[r1][c1]).
     • else: If they are at different cells, chocolates from both cells are added (mat[r1][c1] + mat[r2][c2]).
   • Recursive Step: max_next_chocolates = max(...) explores all four possible combinations of moves for the two robots. Each robot can move one step down (r+1) or one step right (c+1). The function then recursively calls solve for each of these next states to find the maximum possible chocolates from that point onwards.
   • The total chocolates for the current state are current_chocolates + max_next_chocolates.
3. Initial Call and Result: ans = solve(0, 0, 0, 0) starts the process with both robots at the top-left corner. Finally, return max(0, ans) ensures that if no valid path exists (meaning ans would be float('-inf')), the function correctly returns 0 as per the problem statement.

Conclusion

The “Chocolate Pickup” problem is a fantastic illustration of how dynamic programming can be used to solve complex pathfinding problems involving multiple agents or round trips. By transforming the problem into two concurrent forward journeys and meticulously defining the DP state and transitions, we can efficiently compute the maximum possible chocolate collection. This approach, combined with memoization, prevents redundant computations and provides an optimal solution. Keep practicing these challenging problems to elevate your algorithmic thinking!