What is the time complexity of Maximum Amount of Money Robot Can Earn?

The optimal solution has a time complexity of O(m*n).

What data structures are used to solve Maximum Amount of Money Robot Can Earn?

This problem uses Array, Dynamic Programming, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Amount of Money Robot Can Earn?

Insight 1: Dynamic Programming is suitable because the optimal path to a cell depends on the optimal paths to the cells above and to the left. Insight 2: A 3D DP table is necessary to track the maximum coins at each cell for 0, 1, and 2 robber neutralizations used.

Which companies ask LeetCode 3418. Maximum Amount of Money Robot Can Earn in interviews?

This problem is frequently asked at PhonePe.

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Maximum Amount of Money Robot Can Earn - LeetCode 3418 Solution

Q: How to solve Maximum Amount of Money Robot Can Earn?

The solution uses dynamic programming with space optimization. It maintains two rows, `dp_prev` and `dp_curr`, to store the maximum coins for each cell, considering the number of neutralizations used so far (0, 1, or 2). For each cell, it checks if the cell contains a coin or a robber. If it's a coin, the robot gains the coin. If it's a robber, the robot can either neutralize the robber (if neutralizations are available) or have the coins stolen. The maximum coins for the current cell are calculated based on the maximum coins from the previous row and the previous column, considering both scenarios: neutralizing the robber or not.

Maximum Amount of Money Robot Can Earn - Complete Solution Guide

Maximum Amount of Money Robot Can Earn is LeetCode problem 3418, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 cel

Detailed Explanation

The problem presents a robot navigating a grid from the top-left to the bottom-right corner, moving only right or down. Each cell in the grid contains a value representing coins gained (positive) or coins stolen by a robber (negative). The robot can neutralize the robbers in at most two cells along its path. The goal is to find the maximum total coins the robot can have at the end of its journey.

Solution Approach

The solution uses dynamic programming with space optimization. It maintains two rows, `dp_prev` and `dp_curr`, to store the maximum coins for each cell, considering the number of neutralizations used so far (0, 1, or 2). For each cell, it checks if the cell contains a coin or a robber. If it's a coin, the robot gains the coin. If it's a robber, the robot can either neutralize the robber (if neutralizations are available) or have the coins stolen. The maximum coins for the current cell are calculated based on the maximum coins from the previous row and the previous column, considering both scenarios: neutralizing the robber or not.

Step-by-Step Algorithm

Step 1: Initialize `dp_prev` with negative infinity, representing an unreachable state, except for the starting cell (0,0).
Step 2: If the starting cell (0,0) has positive coins, initialize dp_prev[0][0] to that coin value. If it has negative coins, initialize dp_prev[0][0] to the coin value and dp_prev[0][1] to 0 (representing using one neutralization).
Step 3: Iterate through the grid row by row, starting from the second row.
Step 4: For each cell, calculate `dp_curr` based on `dp_prev` and the value of the current cell (coin or robber).
Step 5: If the current cell contains a coin (positive value), update `dp_curr` with the maximum value obtained from the top or left cell in the previous row, plus the current cell's value.
Step 6: If the current cell contains a robber (negative value), calculate two possible values: either neutralize the robber (if neutralization is available) or take the loss. Update `dp_curr` with the maximum value of these two scenarios.
Step 7: After each row is processed, update `dp_prev` with the calculated `dp_curr` to prepare for the next row.
Step 8: After processing the entire grid, the maximum coin value among dp_prev[n-1][0], dp_prev[n-1][1], and dp_prev[n-1][2] represents the solution. Return the solution.

Key Insights

Insight 1: Dynamic Programming is suitable because the optimal path to a cell depends on the optimal paths to the cells above and to the left.
Insight 2: A 3D DP table is necessary to track the maximum coins at each cell for 0, 1, and 2 robber neutralizations used.
Insight 3: Since the robot can only move right or down, the current state only depends on the previous row and the previous column, which enables space optimization using only two rows of the DP table.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: PhonePe.

Maximum Amount of Money Robot Can Earn | LeetCode 3418

Medium

Array Dynamic Programming Matrix

Time: O(m*n)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maximumAmount(int[][] coins) {
        int m = coins.length;
        int n = coins[0].length;

        double[][] dp_prev = new double[n][3];
        for (int i = 0; i < n; i++) {
            Arrays.fill(dp_prev[i], Double.NEGATIVE_INFINITY);
        }

        // Initialize first row (i=0)
        int c00 = coins[0][0];
        if (c00 >= 0) {
            dp_prev[0][0] = c00;
        } else {
            dp_prev[0][0] = c00;
            dp_prev[0][1] = 0;
        }

        for (int j = 1; j < n; j++) {
            int c = coins[0][j];
            for (int k = 0; k < 3; k++) {
                if (c >= 0) {
                    if (dp_prev[j - 1][k] > Double.NEGATIVE_INFINITY) {
                        dp_prev[j][k] = dp_prev[j - 1][k] + c;
                    }
                } else {
                    double val1 = Double.NEGATIVE_INFINITY;
                    if (dp_prev[j - 1][k] > Double.NEGATIVE_INFINITY) {
                        val1 = dp_prev[j - 1][k] + c;
                    }

                    double val2 = Double.NEGATIVE_INFINITY;
                    if (k > 0 && dp_prev[j - 1][k - 1] > Double.NEGATIVE_INFINITY) {
                        val2 = dp_prev[j - 1][k - 1];
                    }

                    dp_prev[j][k] = Math.max(val1, val2);
                }
            }
        }

        for (int i = 1; i < m; i++) {
            double[][] dp_curr = new double[n][3];
            for (int l = 0; l < n; l++) {
                Arrays.fill(dp_curr[l], Double.NEGATIVE_INFINITY);
            }

            // First column of current row (j=0)
            int c = coins[i][0];
            for (int k = 0; k < 3; k++) {
                if (c >= 0) {
                    if (dp_prev[0][k] > Double.NEGATIVE_INFINITY) {
                        dp_curr[0][k] = dp_prev[0][k] + c;
                    }
                } else {
                    double val1 = Double.NEGATIVE_INFINITY;
                    if (dp_prev[0][k] > Double.NEGATIVE_INFINITY) {
                        val1 = dp_prev[0][k] + c;
                    }
                    double val2 = Double.NEGATIVE_INFINITY;
                    if (k > 0 && dp_prev[0][k - 1] > Double.NEGATIVE_INFINITY) {
                        val2 = dp_prev[0][k - 1];
                    }
                    dp_curr[0][k] = Math.max(val1, val2);
                }
            }

            // Rest of the row (j=1 to n-1)
            for (int j = 1; j < n; j++) {
                c = coins[i][j];
                for (int k = 0; k < 3; k++) {
                    if (c >= 0) {
                        double prev_max = Math.max(dp_prev[j][k], dp_curr[j - 1][k]);
                        if (prev_max > Double.NEGATIVE_INFINITY) {
                            dp_curr[j][k] = prev_max + c;
                        }
                    } else {
                        double prev_max_k = Math.max(dp_prev[j][k], dp_curr[j - 1][k]);
                        double val1 = Double.NEGATIVE_INFINITY;
                        if (prev_max_k > Double.NEGATIVE_INFINITY) {
                            val1 = prev_max_k + c;
                        }

                        double val2 = Double.NEGATIVE_INFINITY;
                        if (k > 0) {
                            double prev_max_k_minus_1 = Math.max(dp_prev[j][k - 1], dp_curr[j - 1][k - 1]);
                            if (prev_max_k_minus_1 > Double.NEGATIVE_INFINITY) {
                                val2 = prev_max_k_minus_1;
                            }
                        }

                        dp_curr[j][k] = Math.max(val1, val2);
                    }
                }
            }

            dp_prev = dp_curr;
        }

        double result = Double.NEGATIVE_INFINITY;
        for (int k = 0; k < 3; k++){
            result = Math.max(result, dp_prev[n - 1][k]);
        }
        return (int)result;
    }
}

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Problem Info

Description

Problem 3418: Maximum Amount of Money Robot Can Earn - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

PhonePe

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3418

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize `dp_prev` with negative infinity, representing an unreachable state, except for the starting cell (0,0).
2
Step 2: If the starting cell (0,0) has positive coins, initialize dp_prev[0][0] to that coin value. If it has negative coins, initialize dp_prev[0][0] to the coin value and dp_prev[0][1] to 0 (representing using one neutralization).
3
Step 3: Iterate through the grid row by row, starting from the second row.
4
Step 4: For each cell, calculate `dp_curr` based on `dp_prev` and the value of the current cell (coin or robber).
5
Step 5: If the current cell contains a coin (positive value), update `dp_curr` with the maximum value obtained from the top or left cell in the previous row, plus the current cell's value.
6
Step 6: If the current cell contains a robber (negative value), calculate two possible values: either neutralize the robber (if neutralization is available) or take the loss. Update `dp_curr` with the maximum value of these two scenarios.
7
Step 7: After each row is processed, update `dp_prev` with the calculated `dp_curr` to prepare for the next row.
8
Step 8: After processing the entire grid, the maximum coin value among dp_prev[n-1][0], dp_prev[n-1][1], and dp_prev[n-1][2] represents the solution. Return the solution.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loops that iterate over the grid's rows and columns. The space complexity is determined by the storage required for the dynamic programming table. Because only the previous row is needed for the current row's computation, the solution employs a space optimization technique using two arrays of size O(n), where n is the number of columns, resulting in O(n) overall space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach with Memoization

A recursive approach with memoization can also solve the problem. It explores all possible paths with and without neutralizing robbers and stores the results in a memoization table to avoid redundant calculations.

Time: O(m*n*3) - Similar to DP.Space: O(m*n*3) - Due to the memoization table and recursion stack.

Trade-offs: The recursive approach might be more intuitive to understand initially, but the DP approach is often more efficient due to the iterative nature and reduced overhead of function calls. The recursive approach uses more space, since it requires maintaining a call stack.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable because the optimal path to a cell depends on the optimal paths to the cells above and to the left.
2Insight 2: A 3D DP table is necessary to track the maximum coins at each cell for 0, 1, and 2 robber neutralizations used.
3Insight 3: Since the robot can only move right or down, the current state only depends on the previous row and the previous column, which enables space optimization using only two rows of the DP table.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Grid with only one cell. The solution correctly handles this case by initializing the dp table and returning the value of that single cell, potentially neutralized.
Edge case 2: Grid with all positive coins. The solution correctly calculates the sum of all coins along the path.
Edge case 3: Grid with all negative coins. The solution maximizes the coins by neutralizing up to two robbers.
Edge case 4: Empty grid (m=0 or n=0). Although not explicitly handled in the given code, a check can be added at the beginning of the function to return 0 if either m or n is zero. Otherwise, the code would likely result in an error.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the number of neutralizations available, leading to incorrect calculations when encountering robbers.

✗Common mistake 2: Incorrectly initializing the DP table, causing incorrect results for the base cases.

✗Common mistake 3: Not handling negative coin values correctly, treating them as zero or positive.

✗Common mistake 4: Not memoizing the intermediate results in recursive solutions, leading to Time Limit Exceeded (TLE) errors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Path planning with obstacles. This algorithm can be adapted to find the optimal path for a robot or vehicle, where some locations have costs associated with them, and the robot has a limited number of actions to avoid these costs.

🏦Real-world scenario 2: Resource allocation with constraints. This algorithm can be used to find the maximum profit or value in a scenario where resources have varying values, and there are constraints on resource usage (like the number of neutralizations).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 26, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Amount of Money Robot Can Earn"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?