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feat: add solutions to lc problem: No.3148 #2800

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May 13, 2024
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feat: add solutions to lc problem: No.3148 #2800

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120 changes: 116 additions & 4 deletions solution/3100-3199/3148.Maximum Difference Score in a Grid/README.md
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Expand Up @@ -55,24 +55,136 @@

## 解法

### 方法一
### 方法一：动态规划

根据题目描述，如果我们经过的单元格的值依次是 $c_1, c_2, \cdots, c_k$，那么我们的得分就是 $c_2 - c_1 + c_3 - c_2 + \cdots + c_k - c_{k-1} = c_k - c_1$。因此，问题转化为：对于矩阵的每个单元格 $(i, j)$，如果我们将其作为终点，那么起点的最小值是多少。

我们可以使用动态规划来解决这个问题。我们定义 $f[i][j]$ 表示以 $(i, j)$ 为终点的路径的最小值。那么我们可以得到状态转移方程：

$$
f[i][j] = \min(f[i-1][j], f[i][j-1], grid[i][j])
$$

那么答案为 $\text{grid}[i][j] - \min(f[i-1][j], f[i][j-1])$ 的最大值。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。

<!-- tabs:start -->

```python

class Solution:
def maxScore(self, grid: List[List[int]]) -> int:
f = [[0] * len(grid[0]) for _ in range(len(grid))]
ans = -inf
for i, row in enumerate(grid):
for j, x in enumerate(row):
mi = inf
if i:
mi = min(mi, f[i - 1][j])
if j:
mi = min(mi, f[i][j - 1])
ans = max(ans, x - mi)
f[i][j] = min(x, mi)
return ans
```

```java

class Solution {
public int maxScore(List<List<Integer>> grid) {
int m = grid.size(), n = grid.get(0).size();
final int inf = 1 << 30;
int ans = -inf;
int[][] f = new int[m][n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int mi = inf;
if (i > 0) {
mi = Math.min(mi, f[i - 1][j]);
}
if (j > 0) {
mi = Math.min(mi, f[i][j - 1]);
}
ans = Math.max(ans, grid.get(i).get(j) - mi);
f[i][j] = Math.min(grid.get(i).get(j), mi);
}
}
return ans;
}
}
```

```cpp

class Solution {
public:
int maxScore(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
const int inf = 1 << 30;
int ans = -inf;
int f[m][n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int mi = inf;
if (i) {
mi = min(mi, f[i - 1][j]);
}
if (j) {
mi = min(mi, f[i][j - 1]);
}
ans = max(ans, grid[i][j] - mi);
f[i][j] = min(grid[i][j], mi);
}
}
return ans;
}
};
```

```go
func maxScore(grid [][]int) int {
m, n := len(grid), len(grid[0])
f := make([][]int, m)
for i := range f {
f[i] = make([]int, n)
}
const inf int = 1 << 30
ans := -inf
for i, row := range grid {
for j, x := range row {
mi := inf
if i > 0 {
mi = min(mi, f[i-1][j])
}
if j > 0 {
mi = min(mi, f[i][j-1])
}
ans = max(ans, x-mi)
f[i][j] = min(x, mi)
}
}
return ans
}
```

```ts
function maxScore(grid: number[][]): number {
const [m, n] = [grid.length, grid[0].length];
const f: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
let ans = -Infinity;
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
let mi = Infinity;
if (i) {
mi = Math.min(mi, f[i - 1][j]);
}
if (j) {
mi = Math.min(mi, f[i][j - 1]);
}
ans = Math.max(ans, grid[i][j] - mi);
f[i][j] = Math.min(mi, grid[i][j]);
}
}
return ans;
}
```

<!-- tabs:end -->
Expand Down
120 changes: 116 additions & 4 deletions solution/3100-3199/3148.Maximum Difference Score in a Grid/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -51,24 +51,136 @@ The total score is <code>2 + 7 = 9</code>.</p>

## Solutions

### Solution 1
### Solution 1: Dynamic Programming

According to the problem description, if the values of the cells we pass through are $c_1, c_2, \cdots, c_k$, then our score is $c_2 - c_1 + c_3 - c_2 + \cdots + c_k - c_{k-1} = c_k - c_1$. Therefore, the problem is transformed into: for each cell $(i, j)$ of the matrix, if we take it as the endpoint, what is the minimum value of the starting point.

We can use dynamic programming to solve this problem. We define $f[i][j]$ as the minimum value of the path with $(i, j)$ as the endpoint. Then we can get the state transition equation:

$$
f[i][j] = \min(f[i-1][j], f[i][j-1], grid[i][j])
$$

So the answer is the maximum value of $\text{grid}[i][j] - \min(f[i-1][j], f[i][j-1])$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the matrix, respectively.

<!-- tabs:start -->

```python

class Solution:
def maxScore(self, grid: List[List[int]]) -> int:
f = [[0] * len(grid[0]) for _ in range(len(grid))]
ans = -inf
for i, row in enumerate(grid):
for j, x in enumerate(row):
mi = inf
if i:
mi = min(mi, f[i - 1][j])
if j:
mi = min(mi, f[i][j - 1])
ans = max(ans, x - mi)
f[i][j] = min(x, mi)
return ans
```

```java

class Solution {
public int maxScore(List<List<Integer>> grid) {
int m = grid.size(), n = grid.get(0).size();
final int inf = 1 << 30;
int ans = -inf;
int[][] f = new int[m][n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int mi = inf;
if (i > 0) {
mi = Math.min(mi, f[i - 1][j]);
}
if (j > 0) {
mi = Math.min(mi, f[i][j - 1]);
}
ans = Math.max(ans, grid.get(i).get(j) - mi);
f[i][j] = Math.min(grid.get(i).get(j), mi);
}
}
return ans;
}
}
```

```cpp

class Solution {
public:
int maxScore(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
const int inf = 1 << 30;
int ans = -inf;
int f[m][n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int mi = inf;
if (i) {
mi = min(mi, f[i - 1][j]);
}
if (j) {
mi = min(mi, f[i][j - 1]);
}
ans = max(ans, grid[i][j] - mi);
f[i][j] = min(grid[i][j], mi);
}
}
return ans;
}
};
```

```go
func maxScore(grid [][]int) int {
m, n := len(grid), len(grid[0])
f := make([][]int, m)
for i := range f {
f[i] = make([]int, n)
}
const inf int = 1 << 30
ans := -inf
for i, row := range grid {
for j, x := range row {
mi := inf
if i > 0 {
mi = min(mi, f[i-1][j])
}
if j > 0 {
mi = min(mi, f[i][j-1])
}
ans = max(ans, x-mi)
f[i][j] = min(x, mi)
}
}
return ans
}
```

```ts
function maxScore(grid: number[][]): number {
const [m, n] = [grid.length, grid[0].length];
const f: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
let ans = -Infinity;
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
let mi = Infinity;
if (i) {
mi = Math.min(mi, f[i - 1][j]);
}
if (j) {
mi = Math.min(mi, f[i][j - 1]);
}
ans = Math.max(ans, grid[i][j] - mi);
f[i][j] = Math.min(mi, grid[i][j]);
}
}
return ans;
}
```

<!-- tabs:end -->
Expand Down
23 changes: 23 additions & 0 deletions solution/3100-3199/3148.Maximum Difference Score in a Grid/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,23 @@
class Solution {
public:
int maxScore(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
const int inf = 1 << 30;
int ans = -inf;
int f[m][n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int mi = inf;
if (i) {
mi = min(mi, f[i - 1][j]);
}
if (j) {
mi = min(mi, f[i][j - 1]);
}
ans = max(ans, grid[i][j] - mi);
f[i][j] = min(grid[i][j], mi);
}
}
return ans;
}
};
23 changes: 23 additions & 0 deletions solution/3100-3199/3148.Maximum Difference Score in a Grid/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,23 @@
func maxScore(grid [][]int) int {
m, n := len(grid), len(grid[0])
f := make([][]int, m)
for i := range f {
f[i] = make([]int, n)
}
const inf int = 1 << 30
ans := -inf
for i, row := range grid {
for j, x := range row {
mi := inf
if i > 0 {
mi = min(mi, f[i-1][j])
}
if j > 0 {
mi = min(mi, f[i][j-1])
}
ans = max(ans, x-mi)
f[i][j] = min(x, mi)
}
}
return ans
}
22 changes: 22 additions & 0 deletions solution/3100-3199/3148.Maximum Difference Score in a Grid/Solution.java
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,22 @@
class Solution {
public int maxScore(List<List<Integer>> grid) {
int m = grid.size(), n = grid.get(0).size();
final int inf = 1 << 30;
int ans = -inf;
int[][] f = new int[m][n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int mi = inf;
if (i > 0) {
mi = Math.min(mi, f[i - 1][j]);
}
if (j > 0) {
mi = Math.min(mi, f[i][j - 1]);
}
ans = Math.max(ans, grid.get(i).get(j) - mi);
f[i][j] = Math.min(grid.get(i).get(j), mi);
}
}
return ans;
}
}
14 changes: 14 additions & 0 deletions solution/3100-3199/3148.Maximum Difference Score in a Grid/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,14 @@
class Solution:
def maxScore(self, grid: List[List[int]]) -> int:
f = [[0] * len(grid[0]) for _ in range(len(grid))]
ans = -inf
for i, row in enumerate(grid):
for j, x in enumerate(row):
mi = inf
if i:
mi = min(mi, f[i - 1][j])
if j:
mi = min(mi, f[i][j - 1])
ans = max(ans, x - mi)
f[i][j] = min(x, mi)
return ans
19 changes: 19 additions & 0 deletions solution/3100-3199/3148.Maximum Difference Score in a Grid/Solution.ts
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,19 @@
function maxScore(grid: number[][]): number {
const [m, n] = [grid.length, grid[0].length];
const f: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
let ans = -Infinity;
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
let mi = Infinity;
if (i) {
mi = Math.min(mi, f[i - 1][j]);
}
if (j) {
mi = Math.min(mi, f[i][j - 1]);
}
ans = Math.max(ans, grid[i][j] - mi);
f[i][j] = Math.min(mi, grid[i][j]);
}
}
return ans;
}
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