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Nov 21, 2023
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feat: update solutions to lc problem: No.2304 #1991

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52 changes: 22 additions & 30 deletions solution/2300-2399/2304.Minimum Path Cost in a Grid/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -67,9 +67,9 @@ $$

其中 $\text{moveCost}[grid[i - 1][k]][j]$ 表示从第 $i - 1$ 行第 $k$ 列移动到第 $i$ 行第 $j$ 列的代价。

最终答案即为 $f[m - 1][j]$ 的最小值，其中 $j$ 的取值范围为 $[0, n - 1]$。
最终答案即为 $\min_{0 \leq j < n} \{f[m - 1][j]\}$。

由于每次转移只需要用到上一行的状态，因此可以将空间复杂度优化到 $O(n)$。
由于每次转移只需要用到上一行的状态，因此我们可以使用滚动数组的方式，将空间复杂度优化到 $O(n)$。

时间复杂度 $O(m \times n^2)$，空间复杂度 $O(n)$。其中 $m$ 和 $n$ 分别为网格的行数和列数。

Expand Down Expand Up @@ -172,25 +172,21 @@ func minPathCost(grid [][]int, moveCost [][]int) int {
```rust
impl Solution {
pub fn min_path_cost(grid: Vec<Vec<i32>>, move_cost: Vec<Vec<i32>>) -> i32 {
let (m, n) = (grid.len(), grid[0].len());
let mut dp = vec![0; n];
for i in 0..m - 1 {
let mut counter = vec![i32::MAX; n];
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 {
let val = grid[i][j];
for k in 0..n {
counter[k] = counter[k].min(val + move_cost[val as usize][k] + dp[j]);
g[j] = g[j].min(f[k] + move_cost[grid[i - 1][k] as usize][j] + grid[i][j]);
}
}
for j in 0..n {
dp[j] = counter[j];
}
f.copy_from_slice(&g);
}
let mut res = i32::MAX;
for i in 0..n {
res = res.min(dp[i] + grid[m - 1][i]);
}
res

f.iter().cloned().min().unwrap_or(0)
}
}
```
Expand All @@ -199,23 +195,19 @@ impl Solution {

```ts
function minPathCost(grid: number[][], moveCost: number[][]): number {
const m = grid.length,
n = grid[0].length;
let pre = grid[0].slice();
for (let i = 1; i < m; i++) {
let next = new Array(n);
for (let j = 0; j < n; j++) {
const key = grid[i - 1][j];
for (let k = 0; k < n; k++) {
let sum = pre[j] + moveCost[key][k] + grid[i][k];
if (j == 0 || next[k] > sum) {
next[k] = sum;
}
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]);
}
}
pre = next;
f.splice(0, n, ...g);
}
return Math.min(...pre);
return Math.min(...f);
}
```

Expand Down
64 changes: 36 additions & 28 deletions solution/2300-2399/2304.Minimum Path Cost in a Grid/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -49,6 +49,22 @@ So the total cost of this path is 5 + 1 = 6.

## 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.

<!-- tabs:start -->

### **Python3**
Expand Down Expand Up @@ -144,25 +160,21 @@ func minPathCost(grid [][]int, moveCost [][]int) int {
```rust
impl Solution {
pub fn min_path_cost(grid: Vec<Vec<i32>>, move_cost: Vec<Vec<i32>>) -> i32 {
let (m, n) = (grid.len(), grid[0].len());
let mut dp = vec![0; n];
for i in 0..m - 1 {
let mut counter = vec![i32::MAX; n];
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 {
let val = grid[i][j];
for k in 0..n {
counter[k] = counter[k].min(val + move_cost[val as usize][k] + dp[j]);
g[j] = g[j].min(f[k] + move_cost[grid[i - 1][k] as usize][j] + grid[i][j]);
}
}
for j in 0..n {
dp[j] = counter[j];
}
f.copy_from_slice(&g);
}
let mut res = i32::MAX;
for i in 0..n {
res = res.min(dp[i] + grid[m - 1][i]);
}
res

f.iter().cloned().min().unwrap_or(0)
}
}
```
Expand All @@ -171,23 +183,19 @@ impl Solution {

```ts
function minPathCost(grid: number[][], moveCost: number[][]): number {
const m = grid.length,
n = grid[0].length;
let pre = grid[0].slice();
for (let i = 1; i < m; i++) {
let next = new Array(n);
for (let j = 0; j < n; j++) {
const key = grid[i - 1][j];
for (let k = 0; k < n; k++) {
let sum = pre[j] + moveCost[key][k] + grid[i][k];
if (j == 0 || next[k] > sum) {
next[k] = sum;
}
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]);
}
}
pre = next;
f.splice(0, n, ...g);
}
return Math.min(...pre);
return Math.min(...f);
}
```

Expand Down
24 changes: 10 additions & 14 deletions solution/2300-2399/2304.Minimum Path Cost in a Grid/Solution.rs
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Original file line number Diff line number Diff line change
@@ -1,23 +1,19 @@
impl Solution {
pub fn min_path_cost(grid: Vec<Vec<i32>>, move_cost: Vec<Vec<i32>>) -> i32 {
let (m, n) = (grid.len(), grid[0].len());
let mut dp = vec![0; n];
for i in 0..m - 1 {
let mut counter = vec![i32::MAX; n];
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 {
let val = grid[i][j];
for k in 0..n {
counter[k] = counter[k].min(val + move_cost[val as usize][k] + dp[j]);
g[j] = g[j].min(f[k] + move_cost[grid[i - 1][k] as usize][j] + grid[i][j]);
}
}
for j in 0..n {
dp[j] = counter[j];
}
}
let mut res = i32::MAX;
for i in 0..n {
res = res.min(dp[i] + grid[m - 1][i]);
f.copy_from_slice(&g);
}
res

f.iter().cloned().min().unwrap_or(0)
}
}
24 changes: 10 additions & 14 deletions solution/2300-2399/2304.Minimum Path Cost in a Grid/Solution.ts
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,19 +1,15 @@
function minPathCost(grid: number[][], moveCost: number[][]): number {
const m = grid.length,
n = grid[0].length;
let pre = grid[0].slice();
for (let i = 1; i < m; i++) {
let next = new Array(n);
for (let j = 0; j < n; j++) {
const key = grid[i - 1][j];
for (let k = 0; k < n; k++) {
let sum = pre[j] + moveCost[key][k] + grid[i][k];
if (j == 0 || next[k] > sum) {
next[k] = sum;
}
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]);
}
}
pre = next;
f.splice(0, n, ...g);
}
return Math.min(...pre);
return Math.min(...f);
}