solved
IntuitionThe problem requires the robot to collect the maximum coins while potentially neutralizing up to two robbers. The key challenge is deciding when to use
# Intuition
The problem requires the robot to collect the maximum coins while potentially neutralizing up to two robbers. The key challenge is deciding when to use the neutralizations to avoid the largest losses. Dynamic Programming (DP) is suitable here as it allows tracking the maximum coins collected at each cell while considering the remaining neutralizations.

# Approach
We use a 3D DP array dp[i][j][k], where i and j represent the current cell, and k (0, 1, or 2) tracks the remaining neutralizations. For each cell:

If the cell has coins (coins[i][j] >= 0), add them to the current total without changing k.

If the cell has a robber (coins[i][j] < 0), choose between:

Taking the loss (subtracting its absolute value) and keeping k unchanged.

Neutralizing it (adding 0 coins) and reducing k by 1 (if k > 0).

The DP transitions consider paths from the top and left cells, updating the maximum coins for each possible k.

# Complexity
- Time complexity:
O(m * n)
The algorithm iterates through each cell of the grid once, and for each cell, it checks two directions (top and left) and three possible values of k. This results in O(m * n * 3 * 2) operations, which simplifies to O(m * n).

- Space complexity:
O(m * n)
The 3D DP array uses O(m * n * 3) space. Since k is bounded by a constant (2), the space complexity simplifies to O(m * n).

# Code
```python3 []
from typing import List
from math import inf

class Solution:
    def maximumAmount(self, coins: List[List[int]]) -> int:
        m = len(coins)
        n = len(coins[0]) if m else 0
        if m == 0 or n == 0:
            return 0
        
        # Initialize DP table with -infinity
        dp = [[[-inf] * 3 for _ in range(n)] for __ in range(m)]
        
        # Handle the starting cell (0, 0)
        current_coin = coins[0][0]
        if current_coin >= 0:
            dp[0][0][2] = current_coin
        else:
            # Option 1: Do not neutralize, k remains 2
            dp[0][0][2] = current_coin
            # Option 2: Neutralize, k becomes 1, coins added 0
            dp[0][0][1] = 0
        
        # Iterate through each cell in row-major order
        for i in range(m):
            for j in range(n):
                if i == 0 and j == 0:
                    continue  # already initialized
                
                # Check both possible directions (up and left)
                for di, dj in [(-1, 0), (0, -1)]:
                    pi, pj = i + di, j + dj
                    if pi < 0 or pj < 0:
                        continue  # invalid previous cell
                    
                    # Iterate through all possible k values for the previous cell
                    for prev_k in range(3):
                        prev_val = dp[pi][pj][prev_k]
                        if prev_val == -inf:
                            continue  # no valid path through this previous state
                        
                        current_coin_val = coins[i][j]
                        
                        if current_coin_val >= 0:
                            # Add the current coin value, no change to k
                            new_val = prev_val + current_coin_val
                            new_k = prev_k
                            if new_val > dp[i][j][new_k]:
                                dp[i][j][new_k] = new_val
                        else:
                            # Option 1: Do not neutralize, subtract the absolute value
                            new_val1 = prev_val + current_coin_val
                            new_k1 = prev_k
                            if new_val1 > dp[i][j][new_k1]:
                                dp[i][j][new_k1] = new_val1
                            
                            # Option 2: Neutralize if possible
                            if prev_k >= 1:
                                new_val2 = prev_val + 0
                                new_k2 = prev_k - 1
                                if new_val2 > dp[i][j][new_k2]:
                                    dp[i][j][new_k2] = new_val2
        
        # The result is the maximum value in the bottom-right cell for all k
        max_result = max(dp[m-1][n-1][k] for k in range(3))
        return max_result if max_result != -inf else 0
```
['Python3']