Class Solution
java.lang.Object
g3401_3500.s3418_maximum_amount_of_money_robot_can_earn.Solution
3418 - Maximum Amount of Money Robot Can Earn.
Medium
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 ofcoins[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.
Example 1:
Input: coins = [[0,1,-1],[1,-2,3],[2,-3,4]]
Output: 8
Explanation:
An optimal path for maximum coins is:
- Start at
(0, 0)with0coins (total coins =0). - Move to
(0, 1), gaining1coin (total coins =0 + 1 = 1). - Move to
(1, 1), where there’s a robber stealing2coins. The robot uses one neutralization here, avoiding the robbery (total coins =1). - Move to
(1, 2), gaining3coins (total coins =1 + 3 = 4). - Move to
(2, 2), gaining4coins (total coins =4 + 4 = 8).
Example 2:
Input: coins = [[10,10,10],[10,10,10]]
Output: 40
Explanation:
An optimal path for maximum coins is:
- Start at
(0, 0)with10coins (total coins =10). - Move to
(0, 1), gaining10coins (total coins =10 + 10 = 20). - Move to
(0, 2), gaining another10coins (total coins =20 + 10 = 30). - Move to
(1, 2), gaining the final10coins (total coins =30 + 10 = 40).
Constraints:
m == coins.lengthn == coins[i].length1 <= m, n <= 500-1000 <= coins[i][j] <= 1000
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Constructor Summary
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Method Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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maximumAmount
public int maximumAmount(int[][] coins)
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