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feat: add solutions to lc problem: No.3637 #4618

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122 changes: 118 additions & 4 deletions solution/3600-3699/3637.Trionic Array I/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -73,32 +73,146 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3637.Tr

<!-- solution:start -->

### 方法一
### 方法一：一次遍历

我们首先定义一个指针 $p$，初始时 $p = 0$，表示当前指向数组的第一个元素。我们将 $p$ 向右移动，直到找到第一个不满足严格递增的元素，即 $nums[p] \geq nums[p + 1]$。如果此时 $p = 0$，说明数组的前半部分没有严格递增的部分，因此直接返回 $\text{false}$。

接下来，我们定义另一个指针 $q$，初始时 $q = p$，表示当前指向数组的第二个部分的第一个元素。我们将 $q$ 向右移动，直到找到第一个不满足严格递减的元素，即 $nums[q] \leq nums[q + 1]$。如果此时 $q = p$ 或者 $q = n - 1$，说明数组的第二部分没有严格递减的部分或者没有第三部分，因此直接返回 $\text{false}$。

如果以上条件都满足，说明数组是三段式的，返回 $\text{true}$。

时间复杂度 $O(n)$，其中 $n$ 是数组的长度。空间复杂度 $O(1)$，只使用了常数级别的额外空间。

<!-- tabs:start -->

#### Python3

```python

class Solution:
def isTrionic(self, nums: List[int]) -> bool:
n = len(nums)
p = 0
while p < n - 2 and nums[p] < nums[p + 1]:
p += 1
if p == 0:
return False
q = p
while q < n - 1 and nums[q] > nums[q + 1]:
q += 1
if q == p or q == n - 1:
return False
while q < n - 1 and nums[q] < nums[q + 1]:
q += 1
return q == n - 1
```

#### Java

```java

class Solution {
public boolean isTrionic(int[] nums) {
int n = nums.length;
int p = 0;
while (p < n - 2 && nums[p] < nums[p + 1]) {
p++;
}
if (p == 0) {
return false;
}
int q = p;
while (q < n - 1 && nums[q] > nums[q + 1]) {
q++;
}
if (q == p || q == n - 1) {
return false;
}
while (q < n - 1 && nums[q] < nums[q + 1]) {
q++;
}
return q == n - 1;
}
}
```

#### C++

```cpp

class Solution {
public:
bool isTrionic(vector<int>& nums) {
int n = nums.size();
int p = 0;
while (p < n - 2 && nums[p] < nums[p + 1]) {
p++;
}
if (p == 0) {
return false;
}
int q = p;
while (q < n - 1 && nums[q] > nums[q + 1]) {
q++;
}
if (q == p || q == n - 1) {
return false;
}
while (q < n - 1 && nums[q] < nums[q + 1]) {
q++;
}
return q == n - 1;
}
};
```

#### Go

```go
func isTrionic(nums []int) bool {
n := len(nums)
p := 0
for p < n-2 && nums[p] < nums[p+1] {
p++
}
if p == 0 {
return false
}
q := p
for q < n-1 && nums[q] > nums[q+1] {
q++
}
if q == p || q == n-1 {
return false
}
for q < n-1 && nums[q] < nums[q+1] {
q++
}
return q == n-1
}
```

#### TypeScript

```ts
function isTrionic(nums: number[]): boolean {
const n = nums.length;
let p = 0;
while (p < n - 2 && nums[p] < nums[p + 1]) {
p++;
}
if (p === 0) {
return false;
}
let q = p;
while (q < n - 1 && nums[q] > nums[q + 1]) {
q++;
}
if (q === p || q === n - 1) {
return false;
}
while (q < n - 1 && nums[q] < nums[q + 1]) {
q++;
}
return q === n - 1;
}
```

<!-- tabs:end -->
Expand Down
122 changes: 118 additions & 4 deletions solution/3600-3699/3637.Trionic Array I/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -71,32 +71,146 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3637.Tr

<!-- solution:start -->

### Solution 1
### Solution 1: Single Pass

We first define a pointer $p$, initially $p = 0$, pointing to the first element of the array. We move $p$ to the right until we find the first element that doesn't satisfy strict increasing order, i.e., $nums[p] \geq nums[p + 1]$. If $p = 0$ at this point, it means the first part of the array doesn't have a strictly increasing section, so we return $\text{false}$ directly.

Next, we define another pointer $q$, initially $q = p$, pointing to the first element of the second part of the array. We move $q$ to the right until we find the first element that doesn't satisfy strict decreasing order, i.e., $nums[q] \leq nums[q + 1]$. If $q = p$ or $q = n - 1$ at this point, it means the second part of the array doesn't have a strictly decreasing section or there's no third part, so we return $\text{false}$ directly.

If all the above conditions are satisfied, it means the array is trionic, and we return $\text{true}$.

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

<!-- tabs:start -->

#### Python3

```python

class Solution:
def isTrionic(self, nums: List[int]) -> bool:
n = len(nums)
p = 0
while p < n - 2 and nums[p] < nums[p + 1]:
p += 1
if p == 0:
return False
q = p
while q < n - 1 and nums[q] > nums[q + 1]:
q += 1
if q == p or q == n - 1:
return False
while q < n - 1 and nums[q] < nums[q + 1]:
q += 1
return q == n - 1
```

#### Java

```java

class Solution {
public boolean isTrionic(int[] nums) {
int n = nums.length;
int p = 0;
while (p < n - 2 && nums[p] < nums[p + 1]) {
p++;
}
if (p == 0) {
return false;
}
int q = p;
while (q < n - 1 && nums[q] > nums[q + 1]) {
q++;
}
if (q == p || q == n - 1) {
return false;
}
while (q < n - 1 && nums[q] < nums[q + 1]) {
q++;
}
return q == n - 1;
}
}
```

#### C++

```cpp

class Solution {
public:
bool isTrionic(vector<int>& nums) {
int n = nums.size();
int p = 0;
while (p < n - 2 && nums[p] < nums[p + 1]) {
p++;
}
if (p == 0) {
return false;
}
int q = p;
while (q < n - 1 && nums[q] > nums[q + 1]) {
q++;
}
if (q == p || q == n - 1) {
return false;
}
while (q < n - 1 && nums[q] < nums[q + 1]) {
q++;
}
return q == n - 1;
}
};
```

#### Go

```go
func isTrionic(nums []int) bool {
n := len(nums)
p := 0
for p < n-2 && nums[p] < nums[p+1] {
p++
}
if p == 0 {
return false
}
q := p
for q < n-1 && nums[q] > nums[q+1] {
q++
}
if q == p || q == n-1 {
return false
}
for q < n-1 && nums[q] < nums[q+1] {
q++
}
return q == n-1
}
```

#### TypeScript

```ts
function isTrionic(nums: number[]): boolean {
const n = nums.length;
let p = 0;
while (p < n - 2 && nums[p] < nums[p + 1]) {
p++;
}
if (p === 0) {
return false;
}
let q = p;
while (q < n - 1 && nums[q] > nums[q + 1]) {
q++;
}
if (q === p || q === n - 1) {
return false;
}
while (q < n - 1 && nums[q] < nums[q + 1]) {
q++;
}
return q === n - 1;
}
```

<!-- tabs:end -->
Expand Down
24 changes: 24 additions & 0 deletions solution/3600-3699/3637.Trionic Array I/Solution.cpp
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Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,24 @@
class Solution {
public:
bool isTrionic(vector<int>& nums) {
int n = nums.size();
int p = 0;
while (p < n - 2 && nums[p] < nums[p + 1]) {
p++;
}
if (p == 0) {
return false;
}
int q = p;
while (q < n - 1 && nums[q] > nums[q + 1]) {
q++;
}
if (q == p || q == n - 1) {
return false;
}
while (q < n - 1 && nums[q] < nums[q + 1]) {
q++;
}
return q == n - 1;
}
};
21 changes: 21 additions & 0 deletions solution/3600-3699/3637.Trionic Array I/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,21 @@
func isTrionic(nums []int) bool {
n := len(nums)
p := 0
for p < n-2 && nums[p] < nums[p+1] {
p++
}
if p == 0 {
return false
}
q := p
for q < n-1 && nums[q] > nums[q+1] {
q++
}
if q == p || q == n-1 {
return false
}
for q < n-1 && nums[q] < nums[q+1] {
q++
}
return q == n-1
}
23 changes: 23 additions & 0 deletions solution/3600-3699/3637.Trionic Array I/Solution.java
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Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,23 @@
class Solution {
public boolean isTrionic(int[] nums) {
int n = nums.length;
int p = 0;
while (p < n - 2 && nums[p] < nums[p + 1]) {
p++;
}
if (p == 0) {
return false;
}
int q = p;
while (q < n - 1 && nums[q] > nums[q + 1]) {
q++;
}
if (q == p || q == n - 1) {
return false;
}
while (q < n - 1 && nums[q] < nums[q + 1]) {
q++;
}
return q == n - 1;
}
}
16 changes: 16 additions & 0 deletions solution/3600-3699/3637.Trionic Array I/Solution.py
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Original file line number Diff line number Diff line change
@@ -0,0 +1,16 @@
class Solution:
def isTrionic(self, nums: List[int]) -> bool:
n = len(nums)
p = 0
while p < n - 2 and nums[p] < nums[p + 1]:
p += 1
if p == 0:
return False
q = p
while q < n - 1 and nums[q] > nums[q + 1]:
q += 1
if q == p or q == n - 1:
return False
while q < n - 1 and nums[q] < nums[q + 1]:
q += 1
return q == n - 1
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