How do you solve Maximum Path Score in a Grid on LeetCode?

Maximum Path Score in a Grid (LeetCode #3742) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * k) time complexity and O(m * n * k) space complexity. Key insight: Dynamic programming reduces redundant calculations.

What is the time complexity of Maximum Path Score in a Grid?

The optimal solution for Maximum Path Score in a Grid runs in O(m * n * k) time and uses O(m * n * k) space. The complexity is driven by the need to fill a 3D DP table, iterating over all cells and possible costs.

What is the brute force solution for Maximum Path Score in a Grid?

The brute force approach for Maximum Path Score in a Grid is Brute Force, with O(2^(m+n)) time complexity and O(m+n) space complexity. Explore all possible paths from the top-left to the bottom-right corner, keeping track of scores and costs. This method checks every combination of moves, which can be inefficient. The optimal Optimal Solution approach improves this to O(m * n * k).

What algorithm or data structure is used to solve Maximum Path Score in a Grid?

Maximum Path Score in a Grid uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: Dynamic Programming, Grid Traversal.

What are the interview tips for Maximum Path Score in a Grid?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar DP problems.

What are common mistakes when solving Maximum Path Score in a Grid?

Not considering paths that exceed cost k. Forgetting to initialize the DP array correctly.

#3742

Maximum Path Score in a Grid

Medium

ArrayDynamic ProgrammingMatrixDynamic ProgrammingGrid Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^(m+n))	O(m * n * k)
Space	O(m+n)	O(m * n * k)

💡

Intuition

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

Use dynamic programming to store the maximum score at each cell for all possible costs. This avoids redundant calculations and efficiently finds the maximum score.

⚙️

Algorithm

3 steps

1Step 1: Initialize a 3D DP array where dp[i][j][c] represents the maximum score at cell (i, j) with cost c.
2Step 2: Fill the DP table by iterating through each cell and updating scores based on possible moves from the top or left cells.
3Step 3: Return the maximum score from the bottom-right cell for all costs up to k.

solution.py14 lines

1def maxPathScore(grid, k):
2    m, n = len(grid), len(grid[0])
3    dp = [[[-1] * (k + 1) for _ in range(n)] for _ in range(m)]
4    dp[0][0][0] = 0
5    for i in range(m):
6        for j in range(n):
7            for c in range(k + 1):
8                if dp[i][j][c] != -1:
9                    score = grid[i][j]
10                    cost = 1 if score > 0 else 0
11                    for ni, nj in [(i + 1, j), (i, j + 1)]:
12                        if ni < m and nj < n and c + cost <= k:
13                            dp[ni][nj][c + cost] = max(dp[ni][nj][c + cost], dp[i][j][c] + score)
14    return max(dp[m - 1][n - 1]) if max(dp[m - 1][n - 1]) != -1 else -1

ℹ

Complexity note: The complexity is driven by the need to fill a 3D DP table, iterating over all cells and possible costs.

1Dynamic programming reduces redundant calculations.
2Tracking costs alongside scores is essential.

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