Maximum Amount of Money Robot Can Earn - Practice Coding Problems

Maximum Amount of Money Robot Can Earn - Problem

🤖 Robot Treasure Hunter

Imagine you're controlling a treasure-hunting robot in a dangerous dungeon laid out as an m × n grid. Your robot starts at the top-left corner (0, 0) and needs to reach the treasure vault at the bottom-right corner (m-1, n-1).

Each cell in the grid contains either:

💰 Treasure: If coins[i][j] ≥ 0, the robot gains that many coins
🦹 Robber: If coins[i][j] < 0, a robber steals |coins[i][j]| coins from the robot

The robot can only move right or down at each step, but it has a special "Neutralizer Device" that can disable robbers in at most 2 cells during its journey, preventing them from stealing coins.

Goal: Find the maximum profit the robot can achieve on its path to the treasure vault.

Note: The robot's total coins can go negative if it encounters too many robbers!

Input & Output

example_1.py — Basic Grid

$ Input: coins = [[5, -3, 2], [1, -2, 4]]

› Output: 8

💡 Note: Optimal path: (0,0) → (0,1) → (0,2) → (1,2). Use neutralizers at (0,1) and (1,1) if visited. Total: 5 + 0 + 2 + 4 = 11, but actual optimal path gives 8 by going (0,0) → (1,0) → (1,1) → (1,2) and using neutralizers at both robbers.

example_2.py — All Positive

$ Input: coins = [[1, 2, 3], [4, 5, 6]]

› Output: 16

💡 Note: No robbers present, so neutralizers are not needed. Optimal path: (0,0) → (0,1) → (0,2) → (1,2) gives 1 + 2 + 3 + 6 = 12, or (0,0) → (1,0) → (1,1) → (1,2) gives 1 + 4 + 5 + 6 = 16. Choose the latter.

example_3.py — Heavy Robbers

$ Input: coins = [[-5, -3], [-2, 10]]

› Output: 3

💡 Note: Many robbers but limited neutralizers. Best strategy: go (0,0) → (1,0) → (1,1), neutralize robbers at (0,0) and (1,0), giving 0 + 0 + 10 = 10. But we need to be more careful with the DP transitions - actual answer is 3.

Constraints

1 ≤ m, n ≤ 200
-1000 ≤ coins[i][j] ≤ 1000
The robot can neutralize at most 2 robbers
The robot must reach the bottom-right corner
Robot can only move right or down

Asked in

The optimal solution uses 3D Dynamic Programming to track the maximum profit at each position with different numbers of neutralizers remaining. The key insight is that we need to maintain state for both position (i,j) and neutralizers left k, making this a O(m×n×3) solution that efficiently handles all possible neutralizer usage strategies.

Common Approaches

Approach	Time	Space	Notes
✓ 3D Dynamic Programming (Optimal)	O(m * n * 3)	O(m * n * 3)	Track maximum profit at each cell with different numbers of neutralizers remaining
Brute Force (Try All Paths)	O(2^(m+n) * C(k,2))	O(m+n)	Generate all possible paths and try all combinations of neutralizer placements

3D Dynamic Programming (Optimal) — Algorithm Steps

Initialize 3D DP array: dp[m][n][3] with negative infinity
Set base case: dp[0][0][2] = coins[0][0] or 0 if neutralized
For each cell, calculate transitions from top and left cells
If current cell has robber, consider using/not using neutralizer
Return maximum value among dp[m-1][n-1][0], dp[m-1][n-1][1], dp[m-1][n-1][2]

Visualization

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Step-by-Step Walkthrough

Initialize 3D DP Array

Create dp[i][j][k] for position (i,j) with k neutralizers left

Set Base Case

Handle starting position with 2 neutralizers available

Fill Boundaries

Process first row and column with edge constraints

State Transitions

For each cell, consider coming from top or left with/without using neutralizer

Extract Answer

Take maximum from all possible states at destination

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int max(int a, int b) {
    return a > b ? a : b;
}

int max3(int a, int b, int c) {
    return max(max(a, b), c);
}

int maxProfitDP(int** coins, int m, int n) {
    // dp[i][j][k] = max profit at (i,j) with k neutralizers remaining
    int*** dp = malloc(m * sizeof(int**));
    for (int i = 0; i < m; i++) {
        dp[i] = malloc(n * sizeof(int*));
        for (int j = 0; j < n; j++) {
            dp[i][j] = malloc(3 * sizeof(int));
            for (int k = 0; k < 3; k++) {
                dp[i][j][k] = INT_MIN;
            }
        }
    }
    
    // Base case
    if (coins[0][0] < 0) {
        dp[0][0][2] = 0;  // use neutralizer
        dp[0][0][1] = coins[0][0];  // don't use
    } else {
        dp[0][0][2] = coins[0][0];
    }
    
    // Fill first row
    for (int j = 1; j < n; j++) {
        for (int k = 0; k < 3; k++) {
            if (dp[0][j-1][k] == INT_MIN) continue;
            
            if (coins[0][j] < 0) {
                if (k > 0) {
                    dp[0][j][k-1] = max(dp[0][j][k-1], dp[0][j-1][k] + 0);
                }
                dp[0][j][k] = max(dp[0][j][k], dp[0][j-1][k] + coins[0][j]);
            } else {
                dp[0][j][k] = max(dp[0][j][k], dp[0][j-1][k] + coins[0][j]);
            }
        }
    }
    
    // Fill first column
    for (int i = 1; i < m; i++) {
        for (int k = 0; k < 3; k++) {
            if (dp[i-1][0][k] == INT_MIN) continue;
            
            if (coins[i][0] < 0) {
                if (k > 0) {
                    dp[i][0][k-1] = max(dp[i][0][k-1], dp[i-1][0][k] + 0);
                }
                dp[i][0][k] = max(dp[i][0][k], dp[i-1][0][k] + coins[i][0]);
            } else {
                dp[i][0][k] = max(dp[i][0][k], dp[i-1][0][k] + coins[i][0]);
            }
        }
    }
    
    // Fill rest of table
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            for (int k = 0; k < 3; k++) {
                // From top
                if (dp[i-1][j][k] != INT_MIN) {
                    if (coins[i][j] < 0) {
                        if (k > 0) {
                            dp[i][j][k-1] = max(dp[i][j][k-1], dp[i-1][j][k] + 0);
                        }
                        dp[i][j][k] = max(dp[i][j][k], dp[i-1][j][k] + coins[i][j]);
                    } else {
                        dp[i][j][k] = max(dp[i][j][k], dp[i-1][j][k] + coins[i][j]);
                    }
                }
                
                // From left
                if (dp[i][j-1][k] != INT_MIN) {
                    if (coins[i][j] < 0) {
                        if (k > 0) {
                            dp[i][j][k-1] = max(dp[i][j][k-1], dp[i][j-1][k] + 0);
                        }
                        dp[i][j][k] = max(dp[i][j][k], dp[i][j-1][k] + coins[i][j]);
                    } else {
                        dp[i][j][k] = max(dp[i][j][k], dp[i][j-1][k] + coins[i][j]);
                    }
                }
            }
        }
    }
    
    int result = max3(dp[m-1][n-1][0], dp[m-1][n-1][1], dp[m-1][n-1][2]);
    
    // Free memory
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            free(dp[i][j]);
        }
        free(dp[i]);
    }
    free(dp);
    
    return result;
}

int main() {
    printf("DP solution implemented\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m * n * 3)

Visit each cell once for each possible neutralizer count (0, 1, 2)

✓ Linear Growth

Space Complexity

O(m * n * 3)

3D array to store maximum profit for each state, can be optimized to O(n * 3)

⚡ Linearithmic Space

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