3D DP with clean and short code
Intuitiondp[i][j][k] means maximum coins robot can get at position (i, j) with status k neutralize left.ApproachComplexity
Time complexity: O(mn)

Space complex
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
dp[i][j][k] means maximum coins robot can get at position (i, j) with status k neutralize left.
# Approach
<!-- Describe your approach to solving the problem. -->

# Complexity
- Time complexity: O(mn)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity: O(mn)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```python3 []
class Solution:
    def maximumAmount(self, coins: List[List[int]]) -> int:

        dp = [[[float('-inf') for i in range(3)] for _ in range(len(coins[0])+1)] for _ in range(len(coins)+1)]
        dp[0][1][0] = 0
        dp[0][1][1] = 0
        dp[0][1][2] = 0
        dp[1][0][0] = 0
        dp[1][0][1] = 0
        dp[1][0][2] = 0

        for i in range(1, len(dp)):
            for j in range(1, len(dp[0])):
                dp[i][j][2] = max(dp[i][j-1][2], dp[i-1][j][2]) + coins[i-1][j-1]   # no use neutralize
                dp[i][j][1] = max(dp[i][j-1][1] + coins[i-1][j-1], dp[i-1][j][1] + coins[i-1][j-1], # no use neutralize
                                    dp[i][j-1][2], dp[i-1][j][2])                                   # use neutralize
                dp[i][j][0] = max(dp[i][j-1][0] + coins[i-1][j-1], dp[i-1][j][0] + coins[i-1][j-1], # no use neutralize
                                    dp[i][j-1][1], dp[i-1][j][1])                                   # use neutralize

            
        return max(dp[-1][-1])
        
        





        
        
```
['Python3']