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Apr 4, 2024
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feat: add solutions to lc problem: No.2209 #2537

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2 changes: 1 addition & 1 deletion solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README.md
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Expand Up @@ -57,7 +57,7 @@

### 方法一：记忆化搜索

设计函数 $dfs(i, j)$ 表示从下标 $i$ 开始，使用 $j$ 条地毯，最少有多少个白色砖块没有被覆盖。答案即为 $dfs(0, numCarpets)$。
我们设计一个函数 $dfs(i, j)$ 表示从下标 $i$ 开始，使用 $j$ 条地毯，最少有多少个白色砖块没有被覆盖。答案即为 $dfs(0, numCarpets)$。

对于下标 $i$，我们分情况讨论：

Expand Down
108 changes: 53 additions & 55 deletions solution/2200-2299/2210.Count Hills and Valleys in an Array/README.md
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Expand Up @@ -56,21 +56,33 @@

## 解法

### 方法一
### 方法一：遍历

我们初始化一个指针 $j$ 指向下标 $0$ 的位置，然后在 $[1, n-1]$ 的范围内遍历数组。对于每一个位置 $i$：

- 如果 $nums[i] = nums[i+1]$，则跳过。
- 否则，如果 $nums[i]$ 大于 $nums[j]$ 且 $nums[i]$ 大于 $nums[i+1]$，则 $i$ 是一个峰；如果 $nums[i]$ 小于 $nums[j]$ 且 $nums[i]$ 小于 $nums[i+1]$，则 $i$ 是一个谷。
- 然后，我们将 $j$ 更新为 $i$，继续遍历。

遍历结束后，我们就可以得到峰和谷的数量。

时间复杂度 $O(n)$，其中 $n$ 是数组的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

```python
class Solution:
def countHillValley(self, nums: List[int]) -> int:
arr = [nums[0]]
for v in nums[1:]:
if v != arr[-1]:
arr.append(v)
return sum(
(arr[i] < arr[i - 1]) == (arr[i] < arr[i + 1])
for i in range(1, len(arr) - 1)
)
ans = j = 0
for i in range(1, len(nums) - 1):
if nums[i] == nums[i + 1]:
continue
if nums[i] > nums[j] and nums[i] > nums[i + 1]:
ans += 1
if nums[i] < nums[j] and nums[i] < nums[i + 1]:
ans += 1
j = i
return ans
```

```java
Expand Down Expand Up @@ -100,9 +112,15 @@ public:
int countHillValley(vector<int>& nums) {
int ans = 0;
for (int i = 1, j = 0; i < nums.size() - 1; ++i) {
if (nums[i] == nums[i + 1]) continue;
if (nums[i] > nums[j] && nums[i] > nums[i + 1]) ++ans;
if (nums[i] < nums[j] && nums[i] < nums[i + 1]) ++ans;
if (nums[i] == nums[i + 1]) {
continue;
}
if (nums[i] > nums[j] && nums[i] > nums[i + 1]) {
++ans;
}
if (nums[i] < nums[j] && nums[i] < nums[i + 1]) {
++ans;
}
j = i;
}
return ans;
Expand Down Expand Up @@ -131,67 +149,47 @@ func countHillValley(nums []int) int {

```ts
function countHillValley(nums: number[]): number {
const n = nums.length;
let res = 0;
let prev = nums[0];
for (let i = 1; i < n - 1; i++) {
const num = nums[i];
const next = nums[i + 1];
if (num == next) {
let ans = 0;
for (let i = 1, j = 0; i < nums.length - 1; ++i) {
if (nums[i] === nums[i + 1]) {
continue;
}
if ((num > prev && num > next) || (num < prev && num < next)) {
res += 1;
if (nums[i] > nums[j] && nums[i] > nums[i + 1]) {
ans++;
}
if (nums[i] < nums[j] && nums[i] < nums[i + 1]) {
ans++;
}
prev = num;
j = i;
}
return res;
return ans;
}
```

```rust
impl Solution {
pub fn count_hill_valley(nums: Vec<i32>) -> i32 {
let n = nums.len();
let mut res = 0;
let mut prev = nums[0];
for i in 1..n - 1 {
let num = nums[i];
let next = nums[i + 1];
if num == next {
let mut ans = 0;
let mut j = 0;

for i in 1..nums.len() - 1 {
if nums[i] == nums[i + 1] {
continue;
}
if (num > prev && num > next) || (num < prev && num < next) {
res += 1;
if nums[i] > nums[j] && nums[i] > nums[i + 1] {
ans += 1;
}
if nums[i] < nums[j] && nums[i] < nums[i + 1] {
ans += 1;
}
prev = num;
j = i;
}
res

ans
}
}
```

<!-- tabs:end -->

### 方法二

<!-- tabs:start -->

```python
class Solution:
def countHillValley(self, nums: List[int]) -> int:
ans = j = 0
for i in range(1, len(nums) - 1):
if nums[i] == nums[i + 1]:
continue
if nums[i] > nums[j] and nums[i] > nums[i + 1]:
ans += 1
if nums[i] < nums[j] and nums[i] < nums[i + 1]:
ans += 1
j = i
return ans
```

<!-- tabs:end -->

<!-- end -->
108 changes: 53 additions & 55 deletions solution/2200-2299/2210.Count Hills and Valleys in an Array/README_EN.md
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Expand Up @@ -53,21 +53,33 @@ There are 0 hills and valleys so we return 0.

## Solutions

### Solution 1
### Solution 1: Traversal

We initialize a pointer $j$ to point to the position with index $0$, and then traverse the array in the range $[1, n-1]$. For each position $i$:

- If $nums[i] = nums[i+1]$, then skip.
- Otherwise, if $nums[i]$ is greater than $nums[j]$ and $nums[i]$ is greater than $nums[i+1]$, then $i$ is a peak; if $nums[i]$ is less than $nums[j]$ and $nums[i]$ is less than $nums[i+1]$, then $i$ is a valley.
- Then, we update $j$ to $i$ and continue to traverse.

After the traversal, we can get the number of peaks and valleys.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

<!-- tabs:start -->

```python
class Solution:
def countHillValley(self, nums: List[int]) -> int:
arr = [nums[0]]
for v in nums[1:]:
if v != arr[-1]:
arr.append(v)
return sum(
(arr[i] < arr[i - 1]) == (arr[i] < arr[i + 1])
for i in range(1, len(arr) - 1)
)
ans = j = 0
for i in range(1, len(nums) - 1):
if nums[i] == nums[i + 1]:
continue
if nums[i] > nums[j] and nums[i] > nums[i + 1]:
ans += 1
if nums[i] < nums[j] and nums[i] < nums[i + 1]:
ans += 1
j = i
return ans
```

```java
Expand Down Expand Up @@ -97,9 +109,15 @@ public:
int countHillValley(vector<int>& nums) {
int ans = 0;
for (int i = 1, j = 0; i < nums.size() - 1; ++i) {
if (nums[i] == nums[i + 1]) continue;
if (nums[i] > nums[j] && nums[i] > nums[i + 1]) ++ans;
if (nums[i] < nums[j] && nums[i] < nums[i + 1]) ++ans;
if (nums[i] == nums[i + 1]) {
continue;
}
if (nums[i] > nums[j] && nums[i] > nums[i + 1]) {
++ans;
}
if (nums[i] < nums[j] && nums[i] < nums[i + 1]) {
++ans;
}
j = i;
}
return ans;
Expand Down Expand Up @@ -128,67 +146,47 @@ func countHillValley(nums []int) int {

```ts
function countHillValley(nums: number[]): number {
const n = nums.length;
let res = 0;
let prev = nums[0];
for (let i = 1; i < n - 1; i++) {
const num = nums[i];
const next = nums[i + 1];
if (num == next) {
let ans = 0;
for (let i = 1, j = 0; i < nums.length - 1; ++i) {
if (nums[i] === nums[i + 1]) {
continue;
}
if ((num > prev && num > next) || (num < prev && num < next)) {
res += 1;
if (nums[i] > nums[j] && nums[i] > nums[i + 1]) {
ans++;
}
if (nums[i] < nums[j] && nums[i] < nums[i + 1]) {
ans++;
}
prev = num;
j = i;
}
return res;
return ans;
}
```

```rust
impl Solution {
pub fn count_hill_valley(nums: Vec<i32>) -> i32 {
let n = nums.len();
let mut res = 0;
let mut prev = nums[0];
for i in 1..n - 1 {
let num = nums[i];
let next = nums[i + 1];
if num == next {
let mut ans = 0;
let mut j = 0;

for i in 1..nums.len() - 1 {
if nums[i] == nums[i + 1] {
continue;
}
if (num > prev && num > next) || (num < prev && num < next) {
res += 1;
if nums[i] > nums[j] && nums[i] > nums[i + 1] {
ans += 1;
}
if nums[i] < nums[j] && nums[i] < nums[i + 1] {
ans += 1;
}
prev = num;
j = i;
}
res

ans
}
}
```

<!-- tabs:end -->

### Solution 2

<!-- tabs:start -->

```python
class Solution:
def countHillValley(self, nums: List[int]) -> int:
ans = j = 0
for i in range(1, len(nums) - 1):
if nums[i] == nums[i + 1]:
continue
if nums[i] > nums[j] and nums[i] > nums[i + 1]:
ans += 1
if nums[i] < nums[j] and nums[i] < nums[i + 1]:
ans += 1
j = i
return ans
```

<!-- tabs:end -->

<!-- end -->
12 changes: 9 additions & 3 deletions solution/2200-2299/2210.Count Hills and Valleys in an Array/Solution.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -3,9 +3,15 @@ class Solution {
int countHillValley(vector<int>& nums) {
int ans = 0;
for (int i = 1, j = 0; i < nums.size() - 1; ++i) {
if (nums[i] == nums[i + 1]) continue;
if (nums[i] > nums[j] && nums[i] > nums[i + 1]) ++ans;
if (nums[i] < nums[j] && nums[i] < nums[i + 1]) ++ans;
if (nums[i] == nums[i + 1]) {
continue;
}
if (nums[i] > nums[j] && nums[i] > nums[i + 1]) {
++ans;
}
if (nums[i] < nums[j] && nums[i] < nums[i + 1]) {
++ans;
}
j = i;
}
return ans;
Expand Down
4 changes: 2 additions & 2 deletions solution/2200-2299/2210.Count Hills and Valleys in an Array/Solution.java
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Original file line number Diff line number Diff line change
@@ -1,7 +1,7 @@
class Solution {
public int countHillValley(int[] nums) {
int ans = 0;
for (int i = 1, j = 0; i < nums.length - 1; ++i) {
int ans = 0, j = 0;
for (int i = 1; i < nums.length - 1; ++i) {
if (nums[i] == nums[i + 1]) {
continue;
}
Expand Down
18 changes: 10 additions & 8 deletions solution/2200-2299/2210.Count Hills and Valleys in an Array/Solution.py
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Original file line number Diff line number Diff line change
@@ -1,10 +1,12 @@
class Solution:
def countHillValley(self, nums: List[int]) -> int:
arr = [nums[0]]
for v in nums[1:]:
if v != arr[-1]:
arr.append(v)
return sum(
(arr[i] < arr[i - 1]) == (arr[i] < arr[i + 1])
for i in range(1, len(arr) - 1)
)
ans = j = 0
for i in range(1, len(nums) - 1):
if nums[i] == nums[i + 1]:
continue
if nums[i] > nums[j] and nums[i] > nums[i + 1]:
ans += 1
if nums[i] < nums[j] and nums[i] < nums[i + 1]:
ans += 1
j = i
return ans
24 changes: 13 additions & 11 deletions solution/2200-2299/2210.Count Hills and Valleys in an Array/Solution.rs
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Original file line number Diff line number Diff line change
@@ -1,19 +1,21 @@
impl Solution {
pub fn count_hill_valley(nums: Vec<i32>) -> i32 {
let n = nums.len();
let mut res = 0;
let mut prev = nums[0];
for i in 1..n - 1 {
let num = nums[i];
let next = nums[i + 1];
if num == next {
let mut ans = 0;
let mut j = 0;

for i in 1..nums.len() - 1 {
if nums[i] == nums[i + 1] {
continue;
}
if (num > prev && num > next) || (num < prev && num < next) {
res += 1;
if nums[i] > nums[j] && nums[i] > nums[i + 1] {
ans += 1;
}
prev = num;
if nums[i] < nums[j] && nums[i] < nums[i + 1] {
ans += 1;
}
j = i;
}
res

ans
}
}
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