LeetCode #1219 — MEDIUM

Path with Maximum Gold

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

The Problem

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 in the grid that has some gold.

Example 1:

Input: grid = [[0,6,0],[5,8,7],[0,9,0]]
Output: 24
Explanation:
[[0,6,0],
 [5,8,7],
 [0,9,0]]
Path to get the maximum gold, 9 -> 8 -> 7.

Example 2:

Input: grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
Output: 28
Explanation:
[[1,0,7],
 [2,0,6],
 [3,4,5],
 [0,3,0],
 [9,0,20]]
Path to get the maximum gold, 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 15
0 <= grid[i][j] <= 100
There are at most 25 cells containing gold.

Step 01

Brute Force Baseline

Problem summary: 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 in the grid that has some gold.

Baseline thinking

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

Pattern signal: Array · Backtracking

Example 1

[[0,6,0],[5,8,7],[0,9,0]]

Example 2

[[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]

Step 02

Core Insight

What unlocks the optimal approach

Use recursion to try all such paths and find the one with the maximum value.

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 #1219: Path with Maximum Gold
class Solution {
    private final int[] dirs = {-1, 0, 1, 0, -1};
    private int[][] grid;
    private int m;
    private int n;

    public int getMaximumGold(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans = Math.max(ans, dfs(i, j));
            }
        }
        return ans;
    }

    private int dfs(int i, int j) {
        if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) {
            return 0;
        }
        int v = grid[i][j];
        grid[i][j] = 0;
        int ans = 0;
        for (int k = 0; k < 4; ++k) {
            ans = Math.max(ans, v + dfs(i + dirs[k], j + dirs[k + 1]));
        }
        grid[i][j] = v;
        return ans;
    }
}

// Accepted solution for LeetCode #1219: Path with Maximum Gold
func getMaximumGold(grid [][]int) (ans int) {
	m, n := len(grid), len(grid[0])
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0 {
			return 0
		}
		v := grid[i][j]
		grid[i][j] = 0
		ans := 0
		dirs := []int{-1, 0, 1, 0, -1}
		for k := 0; k < 4; k++ {
			ans = max(ans, v+dfs(i+dirs[k], j+dirs[k+1]))
		}
		grid[i][j] = v
		return ans
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			ans = max(ans, dfs(i, j))
		}
	}
	return
}

# Accepted solution for LeetCode #1219: Path with Maximum Gold
class Solution:
    def getMaximumGold(self, grid: List[List[int]]) -> int:
        def dfs(i: int, j: int) -> int:
            if not (0 <= i < m and 0 <= j < n and grid[i][j]):
                return 0
            v = grid[i][j]
            grid[i][j] = 0
            ans = max(dfs(i + a, j + b) for a, b in pairwise(dirs)) + v
            grid[i][j] = v
            return ans

        m, n = len(grid), len(grid[0])
        dirs = (-1, 0, 1, 0, -1)
        return max(dfs(i, j) for i in range(m) for j in range(n))

// Accepted solution for LeetCode #1219: Path with Maximum Gold
struct Solution;

impl Solution {
    fn get_maximum_gold(mut grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let m = grid[0].len();
        let mut sum = 0;
        let mut res = 0;
        for i in 0..n {
            for j in 0..m {
                Self::dfs(i, j, &mut sum, &mut res, &mut grid, n, m);
            }
        }
        res
    }

    fn dfs(
        i: usize,
        j: usize,
        sum: &mut i32,
        max: &mut i32,
        grid: &mut Vec<Vec<i32>>,
        n: usize,
        m: usize,
    ) {
        if grid[i][j] == 0 {
            return;
        }
        let val = grid[i][j];
        *sum += val;
        *max = (*max).max(*sum);
        grid[i][j] = 0;
        if i > 0 {
            Self::dfs(i - 1, j, sum, max, grid, n, m);
        }
        if j > 0 {
            Self::dfs(i, j - 1, sum, max, grid, n, m);
        }
        if i + 1 < n {
            Self::dfs(i + 1, j, sum, max, grid, n, m);
        }
        if j + 1 < m {
            Self::dfs(i, j + 1, sum, max, grid, n, m);
        }
        grid[i][j] = val;
        *sum -= grid[i][j];
    }
}

#[test]
fn test() {
    let grid = vec_vec_i32![[0, 6, 0], [5, 8, 7], [0, 9, 0]];
    let res = 24;
    assert_eq!(Solution::get_maximum_gold(grid), res);
    let grid = vec_vec_i32![[1, 0, 7], [2, 0, 6], [3, 4, 5], [0, 3, 0], [9, 0, 20]];
    let res = 28;
    assert_eq!(Solution::get_maximum_gold(grid), res);
}

// Accepted solution for LeetCode #1219: Path with Maximum Gold
function getMaximumGold(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    const dfs = (i: number, j: number): number => {
        if (i < 0 || i >= m || j < 0 || j >= n || !grid[i][j]) {
            return 0;
        }
        const v = grid[i][j];
        grid[i][j] = 0;
        let ans = v + Math.max(dfs(i - 1, j), dfs(i + 1, j), dfs(i, j - 1), dfs(i, j + 1));
        grid[i][j] = v;
        return ans;
    };
    let ans = 0;
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            ans = Math.max(ans, dfs(i, j));
        }
    }
    return ans;
}

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(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

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.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.