2304. Minimum Path Cost in a Grid
Description
You are given a 0-indexed m x n
integer matrix grid
consisting of distinct integers from 0
to m * n - 1
. You can move in this matrix from a cell to any other cell in the next row. That is, if you are in cell (x, y)
such that x < m - 1
, you can move to any of the cells (x + 1, 0)
, (x + 1, 1)
, ..., (x + 1, n - 1)
. Note that it is not possible to move from cells in the last row.
Each possible move has a cost given by a 0-indexed 2D array moveCost
of size (m * n) x n
, where moveCost[i][j]
is the cost of moving from a cell with value i
to a cell in column j
of the next row. The cost of moving from cells in the last row of grid
can be ignored.
The cost of a path in grid
is the sum of all values of cells visited plus the sum of costs of all the moves made. Return the minimum cost of a path that starts from any cell in the first row and ends at any cell in the last row.
Example 1:
![](https://spcdn.pages.dev/leetcode/problems/2304.Minimum%20Path%20Cost%20in%20a%20Grid/images/griddrawio-2.png)
Input: grid = [[5,3],[4,0],[2,1]], moveCost = [[9,8],[1,5],[10,12],[18,6],[2,4],[14,3]] Output: 17 Explanation: The path with the minimum possible cost is the path 5 -> 0 -> 1. - The sum of the values of cells visited is 5 + 0 + 1 = 6. - The cost of moving from 5 to 0 is 3. - The cost of moving from 0 to 1 is 8. So the total cost of the path is 6 + 3 + 8 = 17.
Example 2:
Input: grid = [[5,1,2],[4,0,3]], moveCost = [[12,10,15],[20,23,8],[21,7,1],[8,1,13],[9,10,25],[5,3,2]] Output: 6 Explanation: The path with the minimum possible cost is the path 2 -> 3. - The sum of the values of cells visited is 2 + 3 = 5. - The cost of moving from 2 to 3 is 1. So the total cost of this path is 5 + 1 = 6.
Constraints:
m == grid.length
n == grid[i].length
2 <= m, n <= 50
grid
consists of distinct integers from0
tom * n - 1
.moveCost.length == m * n
moveCost[i].length == n
1 <= moveCost[i][j] <= 100
Solutions
Solution 1: Dynamic Programming
We define $f[i][j]$ to represent the minimum path cost from the first row to the $i$th row and $j$th column. Since we can only move from a column in the previous row to a column in the current row, the value of $f[i][j]$ can be transferred from $f[i - 1][k]$, where the range of $k$ is $[0, n - 1]$. Therefore, the state transition equation is:
$$ f[i][j] = \min_{0 \leq k < n} {f[i - 1][k] + \text{moveCost}[grid[i - 1][k]][j] + grid[i][j]} $$
where $\text{moveCost}[grid[i - 1][k]][j]$ represents the cost of moving from the $k$th column of the $i - 1$th row to the $j$th column of the $i$th row.
The final answer is $\min_{0 \leq j < n} {f[m - 1][j]}$.
Since each transition only needs the state of the previous row, we can use a rolling array to optimize the space complexity to $O(n)$.
The time complexity is $O(m \times n^2)$, and the space complexity is $O(n)$. Here, $m$ and $n$ are the number of rows and columns of the grid, respectively.
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