LeetCode 2304. Minimum Path Cost in a Grid Solution in Java, C++, Python & More | Explanation + Code

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:

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 from 0 to m * 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 ith row and jth 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 ≤ k < n} {f[i - 1][k] + moveCost[grid[i - 1][k]][j] + grid[i][j]}

where moveCost[grid[i - 1][k]][j] represents the cost of moving from the kth column of the i - 1th row to the jth column of the ith row.

The final answer is min_{0 ≤ 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 × n²), and the space complexity is O(n). Here, m and n are the number of rows and columns of the grid, respectively.

PythonJavaC++GoTypeScriptRust

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)(code-box)

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; } }(code-box)

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()); } };(code-box)

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[i-1][k]][j]+grid[i][j]) } } f = g } return slices.Min(f) }(code-box)

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); }(code-box)

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) } }(code-box)