How do you solve Minimum Path Cost in a Grid on LeetCode?

Minimum Path Cost in a Grid (LeetCode #2304) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n²) time complexity and O(n) space complexity. Key insight: Dynamic programming helps avoid redundant calculations.

What is the brute force solution for Minimum Path Cost in a Grid?

The brute force approach for Minimum Path Cost in a Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves exploring all possible paths from the first row to the last row. We calculate the total cost for each path and return the minimum cost found. The optimal Optimal Solution approach improves this to O(m * n²).

What algorithm or data structure is used to solve Minimum Path Cost in a Grid?

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

What are the interview tips for Minimum Path Cost in a Grid?

Explain your thought process clearly. Consider edge cases, like a single row or column. Discuss time and space complexity during your solution.

What are common mistakes when solving Minimum Path Cost in a Grid?

Not considering all possible moves from a cell. Failing to properly initialize the DP array.

#2304

Minimum Path Cost in a Grid

Medium

ArrayDynamic ProgrammingMatrixDynamic ProgrammingMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n²)
Space	O(1)	O(n)

💡

Intuition

Time O(m * n²)Space O(n)

Using dynamic programming, we can build up the minimum cost for each cell in the grid row by row, avoiding redundant calculations and significantly reducing the number of operations.

⚙️

Algorithm

4 steps

1Step 1: Create a DP array to store the minimum cost to reach each cell in the current row.
2Step 2: Initialize the first row of the DP array with the values from the grid.
3Step 3: For each subsequent row, calculate the minimum cost for each cell based on the previous row's DP values and the move costs.
4Step 4: Return the minimum value from the last row of the DP array.

solution.py14 lines

1# Full working Python code
2from typing import List
3
4
5def minPathCost(grid: List[List[int]], moveCost: List[List[int]]) -> int:
6    m, n = len(grid), len(grid[0])
7    dp = grid[0][:]
8    for row in range(1, m):
9        new_dp = [float('inf')] * n
10        for col in range(n):
11            for next_col in range(n):
12                new_dp[next_col] = min(new_dp[next_col], dp[col] + moveCost[grid[row - 1][col]][next_col] + grid[row][next_col])
13        dp = new_dp
14    return min(dp)

ℹ

Complexity note: The time complexity is O(m * n²) because for each row (m), we iterate through each cell (n) and check all possible next cells (n) for the minimum cost.

1Dynamic programming helps avoid redundant calculations.
2Understanding the cost structure is crucial for efficient pathfinding.

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