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feat: add solutions to lc problems: No.0807,2170 #3269

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feat: add solutions to lc problems: No.0807,2170
yanglbme Jul 13, 2024
7aed8e8
fix: update code
yanglbme Jul 13, 2024
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90 changes: 48 additions & 42 deletions solution/0800-0899/0807.Max Increase to Keep City Skyline/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -67,7 +67,13 @@ gridNew = [ [8, 4, 8, 7],

<!-- solution:start -->

### 方法一
### 方法一：贪心

根据题目描述，我们可以将每个单元格 $(i, j)$ 的值增加至第 $i$ 行的最大值和第 $j$ 列的最大值中的较小值，这样可以保证不影响天际线，即每个单元格增加的高度为 $\min(\text{rowMax}[i], \text{colMax}[j]) - \text{grid}[i][j]$。

因此，我们可以先遍历一次矩阵，分别计算出每行和每列的最大值，记录在数组 $\text{rowMax}$ 和 $\text{colMax}$ 中，然后再遍历一次矩阵，计算出答案即可。

时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 为矩阵 $\text{grid}$ 的边长。

<!-- tabs:start -->

Expand All @@ -76,12 +82,12 @@ gridNew = [ [8, 4, 8, 7],
```python
class Solution:
def maxIncreaseKeepingSkyline(self, grid: List[List[int]]) -> int:
rmx = [max(row) for row in grid]
cmx = [max(col) for col in zip(*grid)]
row_max = [max(row) for row in grid]
col_max = [max(col) for col in zip(*grid)]
return sum(
(min(rmx[i], cmx[j]) - grid[i][j])
for i in range(len(grid))
for j in range(len(grid[0]))
min(row_max[i], col_max[j]) - x
for i, row in enumerate(grid)
for j, x in enumerate(row)
)
```

Expand All @@ -91,18 +97,18 @@ class Solution:
class Solution {
public int maxIncreaseKeepingSkyline(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[] rmx = new int[m];
int[] cmx = new int[n];
int[] rowMax = new int[m];
int[] colMax = new int[n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
rmx[i] = Math.max(rmx[i], grid[i][j]);
cmx[j] = Math.max(cmx[j], grid[i][j]);
rowMax[i] = Math.max(rowMax[i], grid[i][j]);
colMax[j] = Math.max(colMax[j], grid[i][j]);
}
}
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans += Math.min(rmx[i], cmx[j]) - grid[i][j];
ans += Math.min(rowMax[i], colMax[j]) - grid[i][j];
}
}
return ans;
Expand All @@ -116,19 +122,22 @@ class Solution {
class Solution {
public:
int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
vector<int> rmx(m, 0);
vector<int> cmx(n, 0);
int m = grid.size();
int n = grid[0].size();
vector<int> rowMax(m);
vector<int> colMax(n);
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
rmx[i] = max(rmx[i], grid[i][j]);
cmx[j] = max(cmx[j], grid[i][j]);
rowMax[i] = max(rowMax[i], grid[i][j]);
colMax[j] = max(colMax[j], grid[i][j]);
}
}
int ans = 0;
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
ans += min(rmx[i], cmx[j]) - grid[i][j];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans += min(rowMax[i], colMax[j]) - grid[i][j];
}
}
return ans;
}
};
Expand All @@ -137,45 +146,42 @@ public:
#### Go

```go
func maxIncreaseKeepingSkyline(grid [][]int) int {
m, n := len(grid), len(grid[0])
rmx := make([]int, m)
cmx := make([]int, n)
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
rmx[i] = max(rmx[i], grid[i][j])
cmx[j] = max(cmx[j], grid[i][j])
func maxIncreaseKeepingSkyline(grid [][]int) (ans int) {
rowMax := make([]int, len(grid))
colMax := make([]int, len(grid[0]))
for i, row := range grid {
for j, x := range row {
rowMax[i] = max(rowMax[i], x)
colMax[j] = max(colMax[j], x)
}
}
ans := 0
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
ans += min(rmx[i], cmx[j]) - grid[i][j]
for i, row := range grid {
for j, x := range row {
ans += min(rowMax[i], colMax[j]) - x
}
}
return ans
return
}
```

#### TypeScript

```ts
function maxIncreaseKeepingSkyline(grid: number[][]): number {
let rows = grid.map(arr => Math.max(...arr)),
cols = [];
let m = grid.length,
n = grid[0].length;
for (let j = 0; j < n; ++j) {
cols[j] = grid[0][j];
for (let i = 1; i < m; ++i) {
cols[j] = Math.max(cols[j], grid[i][j]);
const m = grid.length;
const n = grid[0].length;
const rowMax = Array(m).fill(0);
const colMax = Array(n).fill(0);
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
rowMax[i] = Math.max(rowMax[i], grid[i][j]);
colMax[j] = Math.max(colMax[j], grid[i][j]);
}
}

let ans = 0;
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
ans += Math.min(rows[i], cols[j]) - grid[i][j];
ans += Math.min(rowMax[i], colMax[j]) - grid[i][j];
}
}
return ans;
Expand Down
90 changes: 48 additions & 42 deletions solution/0800-0899/0807.Max Increase to Keep City Skyline/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -65,7 +65,13 @@ gridNew = [ [8, 4, 8, 7],

<!-- solution:start -->

### Solution 1
### Solution 1: Greedy

According to the problem description, we can increase the value of each cell $(i, j)$ to the smaller value between the maximum value of the $i$-th row and the $j$-th column, ensuring it does not affect the skyline. Thus, the height added to each cell is $\min(\text{rowMax}[i], \text{colMax}[j]) - \text{grid}[i][j]$.

Therefore, we can first traverse the matrix once to calculate the maximum value of each row and column, storing them in the arrays $\text{rowMax}$ and $\text{colMax}$, respectively. Then, we traverse the matrix again to compute the answer.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$, where $n$ is the side length of the matrix $\text{grid}$.

<!-- tabs:start -->

Expand All @@ -74,12 +80,12 @@ gridNew = [ [8, 4, 8, 7],
```python
class Solution:
def maxIncreaseKeepingSkyline(self, grid: List[List[int]]) -> int:
rmx = [max(row) for row in grid]
cmx = [max(col) for col in zip(*grid)]
row_max = [max(row) for row in grid]
col_max = [max(col) for col in zip(*grid)]
return sum(
(min(rmx[i], cmx[j]) - grid[i][j])
for i in range(len(grid))
for j in range(len(grid[0]))
min(row_max[i], col_max[j]) - x
for i, row in enumerate(grid)
for j, x in enumerate(row)
)
```

Expand All @@ -89,18 +95,18 @@ class Solution:
class Solution {
public int maxIncreaseKeepingSkyline(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[] rmx = new int[m];
int[] cmx = new int[n];
int[] rowMax = new int[m];
int[] colMax = new int[n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
rmx[i] = Math.max(rmx[i], grid[i][j]);
cmx[j] = Math.max(cmx[j], grid[i][j]);
rowMax[i] = Math.max(rowMax[i], grid[i][j]);
colMax[j] = Math.max(colMax[j], grid[i][j]);
}
}
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans += Math.min(rmx[i], cmx[j]) - grid[i][j];
ans += Math.min(rowMax[i], colMax[j]) - grid[i][j];
}
}
return ans;
Expand All @@ -114,19 +120,22 @@ class Solution {
class Solution {
public:
int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
vector<int> rmx(m, 0);
vector<int> cmx(n, 0);
int m = grid.size();
int n = grid[0].size();
vector<int> rowMax(m);
vector<int> colMax(n);
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
rmx[i] = max(rmx[i], grid[i][j]);
cmx[j] = max(cmx[j], grid[i][j]);
rowMax[i] = max(rowMax[i], grid[i][j]);
colMax[j] = max(colMax[j], grid[i][j]);
}
}
int ans = 0;
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
ans += min(rmx[i], cmx[j]) - grid[i][j];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans += min(rowMax[i], colMax[j]) - grid[i][j];
}
}
return ans;
}
};
Expand All @@ -135,45 +144,42 @@ public:
#### Go

```go
func maxIncreaseKeepingSkyline(grid [][]int) int {
m, n := len(grid), len(grid[0])
rmx := make([]int, m)
cmx := make([]int, n)
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
rmx[i] = max(rmx[i], grid[i][j])
cmx[j] = max(cmx[j], grid[i][j])
func maxIncreaseKeepingSkyline(grid [][]int) (ans int) {
rowMax := make([]int, len(grid))
colMax := make([]int, len(grid[0]))
for i, row := range grid {
for j, x := range row {
rowMax[i] = max(rowMax[i], x)
colMax[j] = max(colMax[j], x)
}
}
ans := 0
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
ans += min(rmx[i], cmx[j]) - grid[i][j]
for i, row := range grid {
for j, x := range row {
ans += min(rowMax[i], colMax[j]) - x
}
}
return ans
return
}
```

#### TypeScript

```ts
function maxIncreaseKeepingSkyline(grid: number[][]): number {
let rows = grid.map(arr => Math.max(...arr)),
cols = [];
let m = grid.length,
n = grid[0].length;
for (let j = 0; j < n; ++j) {
cols[j] = grid[0][j];
for (let i = 1; i < m; ++i) {
cols[j] = Math.max(cols[j], grid[i][j]);
const m = grid.length;
const n = grid[0].length;
const rowMax = Array(m).fill(0);
const colMax = Array(n).fill(0);
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
rowMax[i] = Math.max(rowMax[i], grid[i][j]);
colMax[j] = Math.max(colMax[j], grid[i][j]);
}
}

let ans = 0;
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
ans += Math.min(rows[i], cols[j]) - grid[i][j];
ans += Math.min(rowMax[i], colMax[j]) - grid[i][j];
}
}
return ans;
Expand Down
19 changes: 11 additions & 8 deletions solution/0800-0899/0807.Max Increase to Keep City Skyline/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,19 +1,22 @@
class Solution {
public:
int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
vector<int> rmx(m, 0);
vector<int> cmx(n, 0);
int m = grid.size();
int n = grid[0].size();
vector<int> rowMax(m);
vector<int> colMax(n);
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
rmx[i] = max(rmx[i], grid[i][j]);
cmx[j] = max(cmx[j], grid[i][j]);
rowMax[i] = max(rowMax[i], grid[i][j]);
colMax[j] = max(colMax[j], grid[i][j]);
}
}
int ans = 0;
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
ans += min(rmx[i], cmx[j]) - grid[i][j];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans += min(rowMax[i], colMax[j]) - grid[i][j];
}
}
return ans;
}
};
24 changes: 11 additions & 13 deletions solution/0800-0899/0807.Max Increase to Keep City Skyline/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,18 +1,16 @@
func maxIncreaseKeepingSkyline(grid [][]int) int {
m, n := len(grid), len(grid[0])
rmx := make([]int, m)
cmx := make([]int, n)
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
rmx[i] = max(rmx[i], grid[i][j])
cmx[j] = max(cmx[j], grid[i][j])
func maxIncreaseKeepingSkyline(grid [][]int) (ans int) {
rowMax := make([]int, len(grid))
colMax := make([]int, len(grid[0]))
for i, row := range grid {
for j, x := range row {
rowMax[i] = max(rowMax[i], x)
colMax[j] = max(colMax[j], x)
}
}
ans := 0
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
ans += min(rmx[i], cmx[j]) - grid[i][j]
for i, row := range grid {
for j, x := range row {
ans += min(rowMax[i], colMax[j]) - x
}
}
return ans
return
}
10 changes: 5 additions & 5 deletions solution/0800-0899/0807.Max Increase to Keep City Skyline/Solution.java
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,18 +1,18 @@
class Solution {
public int maxIncreaseKeepingSkyline(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[] rmx = new int[m];
int[] cmx = new int[n];
int[] rowMax = new int[m];
int[] colMax = new int[n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
rmx[i] = Math.max(rmx[i], grid[i][j]);
cmx[j] = Math.max(cmx[j], grid[i][j]);
rowMax[i] = Math.max(rowMax[i], grid[i][j]);
colMax[j] = Math.max(colMax[j], grid[i][j]);
}
}
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans += Math.min(rmx[i], cmx[j]) - grid[i][j];
ans += Math.min(rowMax[i], colMax[j]) - grid[i][j];
}
}
return ans;
Expand Down
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