LeetCode 3418: Maximum Amount of Money Robot Can Earn Solution

Master LeetCode problem 3418 (Maximum Amount of Money Robot Can Earn), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3418. Maximum Amount of Money Robot Can Earn

Problem Explanation

Explanation:

  • We can solve this problem using dynamic programming.
  • We will create a 2D dp array to store the maximum profit the robot can earn at each cell.
  • At each cell, the robot can either come from the cell above or the cell on the left.
  • We will consider the case where the robot neutralizes a robber in the previous cell by subtracting the absolute value of the robber's coins.
  • The robot can neutralize at most 2 robbers on its path, so we will keep track of the last two maximum values.
  • Finally, the bottom-right cell will contain the maximum profit the robot can earn.

Time Complexity:

The time complexity of this solution is O(m * n) where m and n are the dimensions of the grid.

Space Complexity:

The space complexity is also O(m * n) for the dp array.

Solution Code

class Solution {
    public int getMaxCoins(int[][] coins) {
        int m = coins.length;
        int n = coins[0].length;
        
        int[][] dp = new int[m][n];
        
        dp[0][0] = coins[0][0];
        
        for (int i = 1; i < m; i++) {
            dp[i][0] = dp[i-1][0] + coins[i][0];
        }
        
        for (int j = 1; j < n; j++) {
            dp[0][j] = dp[0][j-1] + coins[0][j];
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]) + coins[i][j];
                if (coins[i][j] < 0) {
                    dp[i][j] = Math.max(dp[i][j], Math.max(dp[i-1][j], dp[i][j-1]) - Math.abs(coins[i][j]));
                }
            }
        }
        
        return dp[m-1][n-1];
    }
}

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