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feat: add solutions to lc problems: No.2908,2909 #1864

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110 changes: 107 additions & 3 deletions solution/2900-2999/2908.Minimum Sum of Mountain Triplets I/README.md
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Expand Up @@ -62,34 +62,138 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一：预处理 + 枚举**

我们可以预处理出每个位置右侧的最小值，记录在数组 $right[i]$ 中，即 $right[i]$ 表示 $nums[i+1..n-1]$ 中的最小值。

接下来，我们从左到右枚举山形三元组的中间元素 $nums[i]$，用一个变量 $left$ 表示 $nums[0..i-1]$ 中的最小值，用一个变量 $ans$ 表示当前找到的最小元素和。对于每个 $i$，我们需要找到满足 $left < nums[i]$ 且 $right[i+1] < nums[i]$ 的元素 $nums[i]$，并更新 $ans$。

最后，如果 $ans$ 仍然为初始值，则说明不存在山形三元组，返回 $-1$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组长度。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def minimumSum(self, nums: List[int]) -> int:
n = len(nums)
right = [inf] * (n + 1)
for i in range(n - 1, -1, -1):
right[i] = min(right[i + 1], nums[i])
ans = left = inf
for i, x in enumerate(nums):
if left < x and right[i + 1] < x:
ans = min(ans, left + x + right[i + 1])
left = min(left, x)
return -1 if ans == inf else ans
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java

class Solution {
public int minimumSum(int[] nums) {
int n = nums.length;
int[] right = new int[n + 1];
final int inf = 1 << 30;
right[n] = inf;
for (int i = n - 1; i >= 0; --i) {
right[i] = Math.min(right[i + 1], nums[i]);
}
int ans = inf, left = inf;
for (int i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = Math.min(ans, left + nums[i] + right[i + 1]);
}
left = Math.min(left, nums[i]);
}
return ans == inf ? -1 : ans;
}
}
```

### **C++**

```cpp

class Solution {
public:
int minimumSum(vector<int>& nums) {
int n = nums.size();
const int inf = 1 << 30;
int right[n + 1];
right[n] = inf;
for (int i = n - 1; ~i; --i) {
right[i] = min(right[i + 1], nums[i]);
}
int ans = inf, left = inf;
for (int i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = min(ans, left + nums[i] + right[i + 1]);
}
left = min(left, nums[i]);
}
return ans == inf ? -1 : ans;
}
};
```

### **Go**

```go
func minimumSum(nums []int) int {
n := len(nums)
const inf = 1 << 30
right := make([]int, n+1)
right[n] = inf
for i := n - 1; i >= 0; i-- {
right[i] = min(right[i+1], nums[i])
}
ans, left := inf, inf
for i, x := range nums {
if left < x && right[i+1] < x {
ans = min(ans, left+x+right[i+1])
}
left = min(left, x)
}
if ans == inf {
return -1
}
return ans
}

func min(a, b int) int {
if a < b {
return a
}
return b
}
```

### **TypeScript**

```ts
function minimumSum(nums: number[]): number {
const n = nums.length;
const right: number[] = Array(n + 1).fill(Infinity);
for (let i = n - 1; ~i; --i) {
right[i] = Math.min(right[i + 1], nums[i]);
}
let [ans, left] = [Infinity, Infinity];
for (let i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = Math.min(ans, left + nums[i] + right[i + 1]);
}
left = Math.min(left, nums[i]);
}
return ans === Infinity ? -1 : ans;
}
```

### **...**
Expand Down
110 changes: 107 additions & 3 deletions solution/2900-2999/2908.Minimum Sum of Mountain Triplets I/README_EN.md
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Expand Up @@ -56,30 +56,134 @@ And the sum of this triplet is nums[1] + nums[3] + nums[5] = 13. It can be shown

## Solutions

**Solution 1: Preprocessing + Enumeration**

We can preprocess the minimum value on the right side of each position and record it in the array $right[i]$, where $right[i]$ represents the minimum value in $nums[i+1..n-1]$.

Next, we enumerate the middle element $nums[i]$ of the mountain triplet from left to right, and use a variable $left$ to represent the minimum value in $ums[0..i-1]$, and a variable $ans$ to represent the current minimum element sum found. For each $i$, we need to find the element $nums[i]$ that satisfies $left < nums[i]$ and $right[i+1] < nums[i]$, and update $ans$.

Finally, if $ans$ is still the initial value, it means that there is no mountain triplet, and we return $-1$.

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

<!-- tabs:start -->

### **Python3**

```python

class Solution:
def minimumSum(self, nums: List[int]) -> int:
n = len(nums)
right = [inf] * (n + 1)
for i in range(n - 1, -1, -1):
right[i] = min(right[i + 1], nums[i])
ans = left = inf
for i, x in enumerate(nums):
if left < x and right[i + 1] < x:
ans = min(ans, left + x + right[i + 1])
left = min(left, x)
return -1 if ans == inf else ans
```

### **Java**

```java

class Solution {
public int minimumSum(int[] nums) {
int n = nums.length;
int[] right = new int[n + 1];
final int inf = 1 << 30;
right[n] = inf;
for (int i = n - 1; i >= 0; --i) {
right[i] = Math.min(right[i + 1], nums[i]);
}
int ans = inf, left = inf;
for (int i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = Math.min(ans, left + nums[i] + right[i + 1]);
}
left = Math.min(left, nums[i]);
}
return ans == inf ? -1 : ans;
}
}
```

### **C++**

```cpp

class Solution {
public:
int minimumSum(vector<int>& nums) {
int n = nums.size();
const int inf = 1 << 30;
int right[n + 1];
right[n] = inf;
for (int i = n - 1; ~i; --i) {
right[i] = min(right[i + 1], nums[i]);
}
int ans = inf, left = inf;
for (int i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = min(ans, left + nums[i] + right[i + 1]);
}
left = min(left, nums[i]);
}
return ans == inf ? -1 : ans;
}
};
```

### **Go**

```go
func minimumSum(nums []int) int {
n := len(nums)
const inf = 1 << 30
right := make([]int, n+1)
right[n] = inf
for i := n - 1; i >= 0; i-- {
right[i] = min(right[i+1], nums[i])
}
ans, left := inf, inf
for i, x := range nums {
if left < x && right[i+1] < x {
ans = min(ans, left+x+right[i+1])
}
left = min(left, x)
}
if ans == inf {
return -1
}
return ans
}

func min(a, b int) int {
if a < b {
return a
}
return b
}
```

### **TypeScript**

```ts
function minimumSum(nums: number[]): number {
const n = nums.length;
const right: number[] = Array(n + 1).fill(Infinity);
for (let i = n - 1; ~i; --i) {
right[i] = Math.min(right[i + 1], nums[i]);
}
let [ans, left] = [Infinity, Infinity];
for (let i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = Math.min(ans, left + nums[i] + right[i + 1]);
}
left = Math.min(left, nums[i]);
}
return ans === Infinity ? -1 : ans;
}
```

### **...**
Expand Down
20 changes: 20 additions & 0 deletions solution/2900-2999/2908.Minimum Sum of Mountain Triplets I/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -0,0 +1,20 @@
class Solution {
public:
int minimumSum(vector<int>& nums) {
int n = nums.size();
const int inf = 1 << 30;
int right[n + 1];
right[n] = inf;
for (int i = n - 1; ~i; --i) {
right[i] = min(right[i + 1], nums[i]);
}
int ans = inf, left = inf;
for (int i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = min(ans, left + nums[i] + right[i + 1]);
}
left = min(left, nums[i]);
}
return ans == inf ? -1 : ans;
}
};
27 changes: 27 additions & 0 deletions solution/2900-2999/2908.Minimum Sum of Mountain Triplets I/Solution.go
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Original file line number Diff line number Diff line change
@@ -0,0 +1,27 @@
func minimumSum(nums []int) int {
n := len(nums)
const inf = 1 << 30
right := make([]int, n+1)
right[n] = inf
for i := n - 1; i >= 0; i-- {
right[i] = min(right[i+1], nums[i])
}
ans, left := inf, inf
for i, x := range nums {
if left < x && right[i+1] < x {
ans = min(ans, left+x+right[i+1])
}
left = min(left, x)
}
if ans == inf {
return -1
}
return ans
}

func min(a, b int) int {
if a < b {
return a
}
return b
}
19 changes: 19 additions & 0 deletions solution/2900-2999/2908.Minimum Sum of Mountain Triplets I/Solution.java
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Original file line number Diff line number Diff line change
@@ -0,0 +1,19 @@
class Solution {
public int minimumSum(int[] nums) {
int n = nums.length;
int[] right = new int[n + 1];
final int inf = 1 << 30;
right[n] = inf;
for (int i = n - 1; i >= 0; --i) {
right[i] = Math.min(right[i + 1], nums[i]);
}
int ans = inf, left = inf;
for (int i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = Math.min(ans, left + nums[i] + right[i + 1]);
}
left = Math.min(left, nums[i]);
}
return ans == inf ? -1 : ans;
}
}
12 changes: 12 additions & 0 deletions solution/2900-2999/2908.Minimum Sum of Mountain Triplets I/Solution.py
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Original file line number Diff line number Diff line change
@@ -0,0 +1,12 @@
class Solution:
def minimumSum(self, nums: List[int]) -> int:
n = len(nums)
right = [inf] * (n + 1)
for i in range(n - 1, -1, -1):
right[i] = min(right[i + 1], nums[i])
ans = left = inf
for i, x in enumerate(nums):
if left < x and right[i + 1] < x:
ans = min(ans, left + x + right[i + 1])
left = min(left, x)
return -1 if ans == inf else ans
15 changes: 15 additions & 0 deletions solution/2900-2999/2908.Minimum Sum of Mountain Triplets I/Solution.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
function minimumSum(nums: number[]): number {
const n = nums.length;
const right: number[] = Array(n + 1).fill(Infinity);
for (let i = n - 1; ~i; --i) {
right[i] = Math.min(right[i + 1], nums[i]);
}
let [ans, left] = [Infinity, Infinity];
for (let i = 0; i < n; ++i) {
if (left < nums[i] && right[i + 1] < nums[i]) {
ans = Math.min(ans, left + nums[i] + right[i + 1]);
}
left = Math.min(left, nums[i]);
}
return ans === Infinity ? -1 : ans;
}
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