Welcome to Subscribe On Youtube
2304. Minimum Path Cost in a Grid
Description
You are given a 0indexed 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 0indexed 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:
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.

class Solution { public int minPathCost(int[][] grid, int[][] moveCost) { int m = grid.length, n = grid[0].length; int[] f = grid[0]; final int inf = 1 << 30; for (int i = 1; i < m; ++i) { int[] g = new int[n]; Arrays.fill(g, inf); for (int j = 0; j < n; ++j) { for (int k = 0; k < n; ++k) { g[j] = Math.min(g[j], f[k] + moveCost[grid[i  1][k]][j] + grid[i][j]); } } f = g; } // return Arrays.stream(f).min().getAsInt(); int ans = inf; for (int v : f) { ans = Math.min(ans, v); } return ans; } }

class Solution { public: int minPathCost(vector<vector<int>>& grid, vector<vector<int>>& moveCost) { int m = grid.size(), n = grid[0].size(); const int inf = 1 << 30; vector<int> f = grid[0]; for (int i = 1; i < m; ++i) { vector<int> g(n, inf); for (int j = 0; j < n; ++j) { for (int k = 0; k < n; ++k) { g[j] = min(g[j], f[k] + moveCost[grid[i  1][k]][j] + grid[i][j]); } } f = move(g); } return *min_element(f.begin(), f.end()); } };

class Solution: def minPathCost(self, grid: List[List[int]], moveCost: List[List[int]]) > int: m, n = len(grid), len(grid[0]) f = grid[0] for i in range(1, m): g = [inf] * n for j in range(n): for k in range(n): g[j] = min(g[j], f[k] + moveCost[grid[i  1][k]][j] + grid[i][j]) f = g return min(f)

func minPathCost(grid [][]int, moveCost [][]int) int { m, n := len(grid), len(grid[0]) f := grid[0] for i := 1; i < m; i++ { g := make([]int, n) for j := 0; j < n; j++ { g[j] = 1 << 30 for k := 0; k < n; k++ { g[j] = min(g[j], f[k]+moveCost[grid[i1][k]][j]+grid[i][j]) } } f = g } return slices.Min(f) }

function minPathCost(grid: number[][], moveCost: number[][]): number { const m = grid.length; const n = grid[0].length; const f = grid[0]; for (let i = 1; i < m; ++i) { const g: number[] = Array(n).fill(Infinity); for (let j = 0; j < n; ++j) { for (let k = 0; k < n; ++k) { g[j] = Math.min(g[j], f[k] + moveCost[grid[i  1][k]][j] + grid[i][j]); } } f.splice(0, n, ...g); } return Math.min(...f); }

impl Solution { pub fn min_path_cost(grid: Vec<Vec<i32>>, move_cost: Vec<Vec<i32>>) > i32 { let m = grid.len(); let n = grid[0].len(); let mut f = grid[0].clone(); for i in 1..m { let mut g: Vec<i32> = vec![i32::MAX; n]; for j in 0..n { for k in 0..n { g[j] = g[j].min(f[k] + move_cost[grid[i  1][k] as usize][j] + grid[i][j]); } } f.copy_from_slice(&g); } f.iter().cloned().min().unwrap_or(0) } }