LeetCode #3418 — MEDIUM

Maximum Amount of Money Robot Can Earn

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

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) with 0 coins (total coins = 0).
Move to (0, 1), gaining 1 coin (total coins = 0 + 1 = 1).
Move to (1, 1), where there's a robber stealing 2 coins. The robot uses one neutralization here, avoiding the robbery (total coins = 1).
Move to (1, 2), gaining 3 coins (total coins = 1 + 3 = 4).
Move to (2, 2), gaining 4 coins (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) with 10 coins (total coins = 10).
Move to (0, 1), gaining 10 coins (total coins = 10 + 10 = 20).
Move to (0, 2), gaining another 10 coins (total coins = 20 + 10 = 30).
Move to (1, 2), gaining the final 10 coins (total coins = 30 + 10 = 40).

Constraints:

m == coins.length
n == coins[i].length
1 <= m, n <= 500
-1000 <= coins[i][j] <= 1000

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[[0,1,-1],[1,-2,3],[2,-3,4]]

Example 2

[[10,10,10],[10,10,10]]

Step 02

Core Insight

What unlocks the optimal approach

Use Dynamic Programming.
Let <code>dp[i][j][k]</code> denote the maximum amount of money a robot can earn by starting at cell <code>(i,j)</code> and having neutralized <code>k</code> robbers.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3418: Maximum Amount of Money Robot Can Earn
class Solution {
    private Integer[][][] f;
    private int[][] coins;
    private int m;
    private int n;

    public int maximumAmount(int[][] coins) {
        m = coins.length;
        n = coins[0].length;
        this.coins = coins;
        f = new Integer[m][n][3];
        return dfs(0, 0, 2);
    }

    private int dfs(int i, int j, int k) {
        if (i >= m || j >= n) {
            return Integer.MIN_VALUE / 2;
        }
        if (f[i][j][k] != null) {
            return f[i][j][k];
        }
        if (i == m - 1 && j == n - 1) {
            return k > 0 ? Math.max(0, coins[i][j]) : coins[i][j];
        }
        int ans = coins[i][j] + Math.max(dfs(i + 1, j, k), dfs(i, j + 1, k));
        if (coins[i][j] < 0 && k > 0) {
            ans = Math.max(ans, Math.max(dfs(i + 1, j, k - 1), dfs(i, j + 1, k - 1)));
        }
        return f[i][j][k] = ans;
    }
}

// Accepted solution for LeetCode #3418: Maximum Amount of Money Robot Can Earn
func maximumAmount(coins [][]int) int {
	m, n := len(coins), len(coins[0])
	f := make([][][]int, m)
	for i := range f {
		f[i] = make([][]int, n)
		for j := range f[i] {
			f[i][j] = make([]int, 3)
			for k := range f[i][j] {
				f[i][j][k] = math.MinInt32
			}
		}
	}
	var dfs func(i, j, k int) int
	dfs = func(i, j, k int) int {
		if i >= m || j >= n {
			return math.MinInt32 / 2
		}
		if f[i][j][k] != math.MinInt32 {
			return f[i][j][k]
		}
		if i == m-1 && j == n-1 {
			if k > 0 {
				return max(0, coins[i][j])
			}
			return coins[i][j]
		}
		ans := coins[i][j] + max(dfs(i+1, j, k), dfs(i, j+1, k))
		if coins[i][j] < 0 && k > 0 {
			ans = max(ans, max(dfs(i+1, j, k-1), dfs(i, j+1, k-1)))
		}
		f[i][j][k] = ans
		return ans
	}
	return dfs(0, 0, 2)
}

# Accepted solution for LeetCode #3418: Maximum Amount of Money Robot Can Earn
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)

// Accepted solution for LeetCode #3418: Maximum Amount of Money Robot Can Earn
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3418: Maximum Amount of Money Robot Can Earn
// class Solution {
//     private Integer[][][] f;
//     private int[][] coins;
//     private int m;
//     private int n;
// 
//     public int maximumAmount(int[][] coins) {
//         m = coins.length;
//         n = coins[0].length;
//         this.coins = coins;
//         f = new Integer[m][n][3];
//         return dfs(0, 0, 2);
//     }
// 
//     private int dfs(int i, int j, int k) {
//         if (i >= m || j >= n) {
//             return Integer.MIN_VALUE / 2;
//         }
//         if (f[i][j][k] != null) {
//             return f[i][j][k];
//         }
//         if (i == m - 1 && j == n - 1) {
//             return k > 0 ? Math.max(0, coins[i][j]) : coins[i][j];
//         }
//         int ans = coins[i][j] + Math.max(dfs(i + 1, j, k), dfs(i, j + 1, k));
//         if (coins[i][j] < 0 && k > 0) {
//             ans = Math.max(ans, Math.max(dfs(i + 1, j, k - 1), dfs(i, j + 1, k - 1)));
//         }
//         return f[i][j][k] = ans;
//     }
// }

// Accepted solution for LeetCode #3418: Maximum Amount of Money Robot Can Earn
function maximumAmount(coins: number[][]): number {
    const [m, n] = [coins.length, coins[0].length];
    const f = Array.from({ length: m }, () =>
        Array.from({ length: n }, () => Array(3).fill(-Infinity)),
    );
    const dfs = (i: number, j: number, k: number): number => {
        if (i >= m || j >= n) {
            return -Infinity;
        }
        if (f[i][j][k] !== -Infinity) {
            return f[i][j][k];
        }
        if (i === m - 1 && j === n - 1) {
            return k > 0 ? Math.max(0, coins[i][j]) : coins[i][j];
        }
        let ans = coins[i][j] + Math.max(dfs(i + 1, j, k), dfs(i, j + 1, k));
        if (coins[i][j] < 0 && k > 0) {
            ans = Math.max(ans, dfs(i + 1, j, k - 1), dfs(i, j + 1, k - 1));
        }
        return (f[i][j][k] = ans);
    };
    return dfs(0, 0, 2);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n × k)

Space

O(m × n × k)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.