Problem Description

You are given an m x n grid. A robot starts at the top-left corner of the grid (0, 0) and wants to reach the bottom-right corner (m - 1, n - 1). The robot can move either right or down at any point in time.

The grid contains a value coins[i][j] in each cell:

• If coins[i][j] >= 0, the robot gains that many coins.
• If coins[i][j] < 0, the robot encounters a robber, and the robber steals the absolute value of coins[i][j] coins.

The robot has a special ability to neutralize robbers in at most 2 cells on its path, preventing them from stealing coins in those cells.

Note: The robot's total coins can be negative.

Return the maximum profit the robot can gain on the route.

Intuition

To solve the problem, consider the robot's objective to maximize coin collection while traversing from the top-left to the bottom-right corner of the grid. The robot can negate the effect of robbers in up to two cells, which complicates the decision-making process. The dynamic programming approach is useful here, utilizing recursive depth-first search (DFS) with memoization to efficiently explore possible paths.

We define dfs(i, j, k) as a function representing the maximum coins collected starting from cell (i, j) with k opportunities left to neutralize robbers. The robot can only move right or down, so dfs(i, j, k) depends on possible maximum coins from positions (i+1, j) and (i, j+1).

• If i >= m or j >= n, the robot is out of bounds, leading to an invalid path. Return a very small value to indicate this.

• If i = m - 1 and j = n - 1, the robot reaches the grid's bottom-right corner. Consider conversion of the robber if possible. Return the maximum between coins[i][j] and 0 if conversions are left, else return coins[i][j].

• For cells with robbers (coins[i][j] < 0), decide whether to neutralize the robber or not:

  • If neutralizing is possible (k > 0), explore both with and without neutralization to determine the maximum benefit.

• Use memoization to store results of computed states (i, j, k) to avoid redundant computations and improve efficiency.

The described approach efficiently explores the maximum profit path considering the special ability to bypass robbers in select positions.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses a Dynamic Programming (DP) approach with a recursive depth-first search (DFS) to explore all potential paths the robot can take across the grid, tracking the maximum profit obtainable. Here's a detailed walkthrough of the implementation:

1. Define the Recursive Function: We define a function dfs(i, j, k) where i and j represent the current row and column in the grid, respectively, and k is the number of special opportunities remaining to neutralize one bandit cell cost. The function computes the maximum coins obtainable from the current cell onwards.

2. Base Cases:

   • If the robot moves out of bounds (i >= m or j >= n), return a very small number (negative infinity) to signify an invalid path.
   • If the robot reaches the destination cell (m - 1, n - 1), return max(0, coins[i][j]) if neutralization opportunities are available (k > 0), else return coins[i][j].

3. Recursive Exploration:

   • Compute the potential maximum profit from moving right (dfs(i, j+1, k)) or down (dfs(i+1, j, k)).
   • Include the coins value of the current cell coins[i][j] in the computation.

4. Handling Bandit Cells:

   • If the current cell has a negative coins value (indicating a bandit), consider two scenarios:
     • Not using a neutralization opportunity: Continue the computation normally.
     • Using a neutralization opportunity (if k > 0): Compute the maximum profit from both possible moves with k - 1 opportunities remaining.

5. Memoization: Use Python's @cache decorator to remember results of computed states, avoiding repeated calculations for already explored paths with the same conditions.

Here's the implementation encapsulating the above logic:

class Solution:
    def maximumAmount(self, coins: List[List[int]]) -> int:
        @cache
        def dfs(i: int, j: int, k: int) -> int:
            if i >= m or j >= n:
                return -inf
            if i == m - 1 and j == n - 1:
                return max(coins[i][j], 0) if k else coins[i][j]
            ans = coins[i][j] + max(dfs(i + 1, j, k), dfs(i, j + 1, k))
            if coins[i][j] < 0 and k:
                ans = max(ans, dfs(i + 1, j, k - 1), dfs(i, j + 1, k - 1))
            return ans

        m, n = len(coins), len(coins[0])
        return dfs(0, 0, 2)

This solution carefully balances choices at each step and uses memoization to efficiently find the path yielding the highest profit.

Example Walkthrough

Let's consider a simple 3x3 grid as an example:

coins = [
    [0, 1, -2],
    [1, -3, 2],
    [-1, 4, 5]
]

In this grid, the robot starts at the top-left corner (0, 0) and wants to reach the bottom-right corner (2, 2). The robot can neutralize the negative impact of robbers up to two times. The goal is to maximize the total coins collected.

1. Start at (0, 0):

   • Begin at (0, 0) with 2 neutralization opportunities.
   • The cell (0, 0) has 0 coins, so we move onward.

2. Move to (0, 1):

   • Choose to move right to (0, 1) which contains 1 coin.
   • Total coins are now 1.

3. Move to (0, 2):

   • From (0, 1), move to (0, 2) encountering a robber with -2 coins.
   • Decide to neutralize the robber, utilizing one special ability (k = 1 left).
   • Total remains 1 coin, as the robber's effect is negated.

4. Move to (1, 2):

   • From (0, 2), move down to (1, 2) gaining 2 coins.
   • Total coins become 3.

5. Move to (2, 2):

   • Move to (2, 2), the destination, with 5 coins.
   • Total coins become 8.
   • The path is reached with one special ability unused (k = 1).

Using dynamic programming and memoization, the function maximizes the coins by considering alternative paths and neutralization choices. After evaluating possible paths and using the special abilities optimally, the final profit is determined. In this example, the maximum profit the robot could gain on the route is 8 coins.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(m * n * k), where m is the number of rows and n is the number of columns in the coins array, and k is the number of conversion opportunities, which is 2 in this problem.

The space complexity of the code is also O(m * n * k) due to the memoization used in the depth-first search, storing results for each state (i, j, k).

Learn more about how to find time and space complexity quickly.