How do you solve Path with Maximum Gold on LeetCode?

Path with Maximum Gold (LeetCode #1219) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m*n) time complexity and O(m*n) space complexity. Key insight: Backtracking is essential for exploring all paths.

What is the brute force solution for Path with Maximum Gold?

The brute force approach for Path with Maximum Gold is Brute Force, with O(4^(m*n)) time complexity and O(m*n) space complexity. The brute force approach explores all possible paths in the grid starting from each cell that contains gold. It recursively visits adjacent cells, collecting gold until it can no longer move, then backtracks to explore other paths. The optimal Optimal Solution approach improves this to O(m*n).

What algorithm or data structure is used to solve Path with Maximum Gold?

Path with Maximum Gold uses the following concepts: Array, Backtracking, Matrix. The recommended patterns to study are: Backtracking, Recursion.

What are the interview tips for Path with Maximum Gold?

Explain your thought process clearly while coding. Discuss edge cases, such as grids with no gold cells. Practice writing clean and efficient recursive functions.

What are common mistakes when solving Path with Maximum Gold?

Not marking cells as visited properly, leading to incorrect results. Forgetting to backtrack and unmark cells after exploration.

#1219

Path with Maximum Gold

Medium

ArrayBacktrackingMatrixBacktrackingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(4^(m*n))	O(m*n)
Space	O(m*n)	O(m*n)

💡

Intuition

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

The optimal solution also uses backtracking but leverages memoization to avoid recalculating results for already visited paths. This significantly reduces the number of recursive calls and improves performance.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable to track the maximum gold collected.
2Step 2: For each cell in the grid, if it contains gold, start a backtracking search from that cell.
3Step 3: In the backtracking function, collect gold and mark the cell as visited.
4Step 4: Explore all four possible directions recursively, using memoization to store results of previously computed paths.
5Step 5: After exploring all paths from the current starting cell, backtrack by unmarking the cell as visited.

solution.py21 lines

1def getMaximumGold(grid):
2    def backtrack(x, y, visited):
3        if x < 0 or x >= len(grid) or y < 0 or y >= len(grid[0]) or (x, y) in visited or grid[x][y] == 0:
4            return 0
5        visited.add((x, y))
6        gold = grid[x][y]
7        max_gold = gold + max(
8            backtrack(x + 1, y, visited),
9            backtrack(x - 1, y, visited),
10            backtrack(x, y + 1, visited),
11            backtrack(x, y - 1, visited)
12        )
13        visited.remove((x, y))
14        return max_gold
15
16    max_gold = 0
17    for i in range(len(grid)):
18        for j in range(len(grid[0])):
19            if grid[i][j] > 0:
20                max_gold = max(max_gold, backtrack(i, j, set()))
21    return max_gold

ℹ

Complexity note: The time complexity is linear with respect to the number of cells because we visit each cell at most once during the backtracking process.

1Backtracking is essential for exploring all paths.
2Memoization can significantly improve performance.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.