What is the time complexity of Path with Maximum Gold?

The optimal solution has a time complexity of O(m*n*4^(m*n)).

What is the space complexity of Path with Maximum Gold?

The space complexity is O(m*n).

What data structures are used to solve Path with Maximum Gold?

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

What are the key insights for solving Path with Maximum Gold?

Insight 1: The problem can be solved using Depth-First Search (DFS) because we need to explore all possible paths from each starting cell. Insight 2: We need to temporarily mark visited cells to avoid cycles in our paths. This can be done by setting the cell value to 0 during the DFS call and restoring it afterward (backtracking).

Which companies ask LeetCode 1219. Path with Maximum Gold in interviews?

This problem is frequently asked at Geico.

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Path with Maximum Gold - LeetCode 1219 Solution

Path with Maximum Gold - Complete Solution Guide

Path with Maximum Gold is LeetCode problem 1219, 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

In a gold mine grid of size m x n , each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty. Return the maximum amount of gold you can collect under the conditions: Every time you are located in a cell you will collect all the gold in that cell. From your position, you can walk one step to the left, right, up, or down. You can't visit the same cell more than once. Never visit a cell with 0 gold. You can start and stop collecting gold from any position

Detailed Explanation

The problem asks us to find the maximum amount of gold we can collect in a gold mine represented by a grid. Each cell in the grid contains a certain amount of gold (or 0 if empty). We can start at any cell with gold and move one step in the four cardinal directions (up, down, left, right). We collect the gold from each cell we visit, but we cannot visit the same cell twice, and we can't visit cells with 0 gold. The goal is to find the path that yields the maximum total gold collected.

Solution Approach

The provided code uses a Depth-First Search (DFS) algorithm to explore all possible paths from each starting cell containing gold. For each cell containing gold, the `getMaximumGold` function initiates a DFS traversal. The DFS function explores the adjacent cells recursively, only moving to valid cells (within bounds, contain gold, and haven't been visited in the current path). It temporarily marks a cell as visited by setting its value to 0. After exploring all possible paths from a cell, the original gold value is restored (backtracking), allowing other paths to explore the same cell. The function returns the maximum gold collected along any path starting from that cell. The `getMaximumGold` function then compares the maximum gold found from each starting cell to find the global maximum.

Step-by-Step Algorithm

Step 1: Iterate through each cell of the grid.
Step 2: If a cell contains gold (grid[r][c] > 0), initiate a DFS traversal from that cell.
Step 3: The DFS function checks if the current cell (r, c) is valid (within grid boundaries and contains gold). If not, it returns 0.
Step 4: If the cell is valid, store the original gold value and mark the cell as visited by setting its value to 0.
Step 5: Recursively call DFS on the four neighboring cells (up, down, left, right).
Step 6: Calculate the maximum gold collected from the neighboring cells.
Step 7: Restore the original gold value of the current cell (backtracking).
Step 8: Return the sum of the original gold value and the maximum gold collected from the neighbors.
Step 9: In the main function, update the `max_total_gold` with the maximum value returned by the DFS for each starting cell.
Step 10: Return the `max_total_gold`.

Key Insights

Insight 1: The problem can be solved using Depth-First Search (DFS) because we need to explore all possible paths from each starting cell.
Insight 2: We need to temporarily mark visited cells to avoid cycles in our paths. This can be done by setting the cell value to 0 during the DFS call and restoring it afterward (backtracking).
Insight 3: Since we can start at any cell with gold, we need to iterate through all such cells and initiate a DFS from each one to find the global maximum.

Complexity Analysis

Time Complexity: O(m*n*4^(m*n))

Space Complexity: O(m*n)

Topics

This problem involves: Array, Backtracking, Matrix.

Companies

Asked at: Geico.

Path with Maximum Gold | LeetCode 1219

Medium

Array Backtracking Matrix

Time: O(m*n*4^(m*n))Space: O(m*n)

Solution Code

Available in:

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

        int max_total_gold = 0;
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (grid[r][c] > 0) {
                    max_total_gold = Math.max(max_total_gold, dfs(grid, r, c));
                }
            }
        }

        return max_total_gold;
    }

    private int dfs(int[][] grid, int r, int c) {
        int m = grid.length;
        int n = grid[0].length;

        if (!(0 <= r && r < m && 0 <= c && c < n && grid[r][c] > 0)) {
            return 0;
        }

        int original_gold = grid[r][c];
        grid[r][c] = 0;

        int max_from_neighbors = 0;
        int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        for (int[] direction : directions) {
            int dr = direction[0];
            int dc = direction[1];
            max_from_neighbors = Math.max(max_from_neighbors, dfs(grid, r + dr, c + dc));
        }

        grid[r][c] = original_gold;

        return original_gold + max_from_neighbors;
    }
}

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Problem Info

Description

Problem 1219: Path with Maximum Gold - Detailed explanation and solution approaches.

Topics

Array Backtracking Matrix

Companies

Geico

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1219

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

10 steps

1
Step 1: Iterate through each cell of the grid.
2
Step 2: If a cell contains gold (grid[r][c] > 0), initiate a DFS traversal from that cell.
3
Step 3: The DFS function checks if the current cell (r, c) is valid (within grid boundaries and contains gold). If not, it returns 0.
4
Step 4: If the cell is valid, store the original gold value and mark the cell as visited by setting its value to 0.
5
Step 5: Recursively call DFS on the four neighboring cells (up, down, left, right).
6
Step 6: Calculate the maximum gold collected from the neighboring cells.
7
Step 7: Restore the original gold value of the current cell (backtracking).
8
Step 8: Return the sum of the original gold value and the maximum gold collected from the neighbors.
9
Step 9: In the main function, update the `max_total_gold` with the maximum value returned by the DFS for each starting cell.
10
Step 10: Return the `max_total_gold`.

💡

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*4^(m*n))

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity is exponential because the DFS explores every possible path. The space complexity is determined by the depth of the DFS recursion, which is bounded by the total number of cells in the grid.

📖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

Dynamic Programming with Memoization

Instead of recomputing the maximum gold obtainable from a cell for different paths, we can memoize the results. This involves using a table to store the maximum gold obtainable from each cell, given a set of visited cells.

Time: O(m*n*2^(m*n)), where m and n are the dimensions of the grid. The 2^(m*n) factor comes from representing visited states as a bitmask.Space: O(m*n*2^(m*n)) to store the memoization table.

Trade-offs: While memoization can reduce redundant computations, the high space complexity can be a bottleneck, especially for larger grids. The original backtracking approach is simpler to implement and might be preferable when memory is a concern.

✨

Key Insights

1Insight 1: The problem can be solved using Depth-First Search (DFS) because we need to explore all possible paths from each starting cell.
2Insight 2: We need to temporarily mark visited cells to avoid cycles in our paths. This can be done by setting the cell value to 0 during the DFS call and restoring it afterward (backtracking).
3Insight 3: Since we can start at any cell with gold, we need to iterate through all such cells and initiate a DFS from each one to find the global maximum.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: A grid with only one cell containing gold. The solution correctly returns the value of that cell.
Edge case 2: A grid with no cells containing gold. The solution correctly returns 0.
Edge case 3: A grid where gold cells are isolated and cannot form a path. The solution will correctly find the maximum individual cell value.
Edge case 4: A grid where all cells contain gold, forming a long path. The solution must correctly explore all possible paths without revisiting cells.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to restore the original gold value after the DFS call (not backtracking). This leads to incorrect results because paths might be blocked prematurely.

✗Common mistake 2: Not checking the bounds of the grid during the DFS call, leading to out-of-bounds errors.

✗Common mistake 3: Not considering all possible starting cells with gold, leading to potentially missing the global maximum.

✗Common mistake 4: Incorrectly implementing the DFS logic, resulting in infinite recursion or incorrect path traversal.

✅

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: Resource gathering in games. Finding the optimal path to collect resources in a game world while avoiding obstacles and restricted areas.

🏦Real-world scenario 2: Path planning for robots in warehouses. Optimizing the route for a robot to pick up items in a warehouse, considering the location of items, obstacles, and the robot's movement constraints.

💡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: October 18, 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

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

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid duplicate solutions in backtracking?

Sort the input first, then skip consecutive duplicates at the same recursion level: if (i > start && nums[i] == nums[i-1]) continue. For permutations, use a visited array or swap-based approach. Always consider if order matters (permutations vs combinations).

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Path with Maximum Gold"?

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 backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?