LeetCode #2304 — MEDIUM

Minimum Path Cost in a Grid

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[[5,3],[4,0],[2,1]]
[[9,8],[1,5],[10,12],[18,6],[2,4],[14,3]]

Example 2

[[5,1,2],[4,0,3]]
[[12,10,15],[20,23,8],[21,7,1],[8,1,13],[9,10,25],[5,3,2]]

Core Insight

What unlocks the optimal approach

What is the optimal cost to get to each of the cells in the second row? What about the third row?
Use dynamic programming to compute the optimal cost to get to each cell.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2304: Minimum Path Cost in a Grid
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;
    }
}

// Accepted solution for LeetCode #2304: Minimum Path Cost in a Grid
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)
}

# Accepted solution for LeetCode #2304: Minimum Path Cost in a Grid
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)

// Accepted solution for LeetCode #2304: Minimum Path Cost in a Grid
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)
    }
}

// Accepted solution for LeetCode #2304: Minimum Path Cost in a Grid
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);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n^2)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.