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feat: add solutions to lc problem: No.1588 #3888

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12 changes: 11 additions & 1 deletion ...85.Check If String Is Transformable With Substring Sort Operations/README_EN.md
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Expand Up @@ -79,7 +79,17 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Bubble Sort

The problem is essentially equivalent to determining whether any substring of length 2 in string $s$ can be swapped using bubble sort to obtain $t$.

Therefore, we use an array $pos$ of length 10 to record the indices of each digit in string $s$, where $pos[i]$ represents the list of indices where digit $i$ appears, sorted in ascending order.

Next, we iterate through string $t$. For each character $t[i]$ in $t$, we convert it to the digit $x$. We check if $pos[x]$ is empty. If it is, it means that the digit in $t$ does not exist in $s$, so we return `false`. Otherwise, to swap the character at the first index of $pos[x]$ to index $i$, all indices of digits less than $x$ must be greater than or equal to the first index of $pos[x]. If this condition is not met, we return `false`. Otherwise, we pop the first index from $pos[x]$ and continue iterating through string $t$.

After the iteration, we return `true`.

The time complexity is $O(n \times C)$, and the space complexity is $O(n)$. Here, $n$ is the length of string $s$, and $C$ is the size of the digit set, which is 10 in this problem.

<!-- tabs:start -->

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259 changes: 197 additions & 62 deletions solution/1500-1599/1588.Sum of All Odd Length Subarrays/README.md
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Expand Up @@ -80,11 +80,19 @@ tags:

<!-- solution:start -->

### 方法一：枚举 + 前缀和
### 方法一：动态规划

我们可以枚举子数组的起点 $i$ 和终点 $j$，其中 $i \leq j$，维护每个子数组的和，然后判断子数组的长度是否为奇数，如果是，则将子数组的和加入答案。
我们定义两个长度为 $n$ 的数组 $f$ 和 $g$，其中 $f[i]$ 表示以 $\textit{arr}[i]$ 结尾的长度为奇数的子数组的和，而 $g[i]$ 表示以 $\textit{arr}[i]$ 结尾的长度为偶数的子数组的和。初始时 $f[0] = \textit{arr}[0]$，而 $g[0] = 0$。答案即为 $\sum_{i=0}^{n-1} f[i]$。

时间复杂度 $O(n^2)$，空间复杂度 $O(1)$。其中 $n$ 是数组的长度。
当 $i > 0$ 时，考虑 $f[i]$ 和 $g[i]$ 如何进行状态转移：

对于状态 $f[i]$，元素 $\textit{arr}[i]$ 可以与前面的 $g[i-1]$ 组成一个长度为奇数的子数组，一共可以组成的子数组个数为 $(i / 2) + 1$ 个，因此 $f[i] = g[i-1] + \textit{arr}[i] \times ((i / 2) + 1)$。

对于状态 $g[i]$，当 $i = 0$ 时，没有长度为偶数的子数组，因此 $g[0] = 0$；当 $i > 0$ 时，元素 $\textit{arr}[i]$ 可以与前面的 $f[i-1]$ 组成一个长度为偶数的子数组，一共可以组成的子数组个数为 $(i + 1) / 2$ 个，因此 $g[i] = f[i-1] + \textit{arr}[i] \times ((i + 1) / 2)$。

最终答案即为 $\sum_{i=0}^{n-1} f[i]$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{arr}$ 的长度。

<!-- tabs:start -->

Expand All @@ -93,13 +101,14 @@ tags:
```python
class Solution:
def sumOddLengthSubarrays(self, arr: List[int]) -> int:
ans, n = 0, len(arr)
for i in range(n):
s = 0
for j in range(i, n):
s += arr[j]
if (j - i + 1) & 1:
ans += s
n = len(arr)
f = [0] * n
g = [0] * n
ans = f[0] = arr[0]
for i in range(1, n):
f[i] = g[i - 1] + arr[i] * (i // 2 + 1)
g[i] = f[i - 1] + arr[i] * ((i + 1) // 2)
ans += f[i]
return ans
```

Expand All @@ -109,15 +118,13 @@ class Solution:
class Solution {
public int sumOddLengthSubarrays(int[] arr) {
int n = arr.length;
int ans = 0;
for (int i = 0; i < n; ++i) {
int s = 0;
for (int j = i; j < n; ++j) {
s += arr[j];
if ((j - i + 1) % 2 == 1) {
ans += s;
}
}
int[] f = new int[n];
int[] g = new int[n];
int ans = f[0] = arr[0];
for (int i = 1; i < n; ++i) {
f[i] = g[i - 1] + arr[i] * (i / 2 + 1);
g[i] = f[i - 1] + arr[i] * ((i + 1) / 2);
ans += f[i];
}
return ans;
}
Expand All @@ -131,15 +138,13 @@ class Solution {
public:
int sumOddLengthSubarrays(vector<int>& arr) {
int n = arr.size();
int ans = 0;
for (int i = 0; i < n; ++i) {
int s = 0;
for (int j = i; j < n; ++j) {
s += arr[j];
if ((j - i + 1) & 1) {
ans += s;
}
}
vector<int> f(n, arr[0]);
vector<int> g(n);
int ans = f[0];
for (int i = 1; i < n; ++i) {
f[i] = g[i - 1] + arr[i] * (i / 2 + 1);
g[i] = f[i - 1] + arr[i] * ((i + 1) / 2);
ans += f[i];
}
return ans;
}
Expand All @@ -151,14 +156,14 @@ public:
```go
func sumOddLengthSubarrays(arr []int) (ans int) {
n := len(arr)
for i := range arr {
s := 0
for j := i; j < n; j++ {
s += arr[j]
if (j-i+1)%2 == 1 {
ans += s
}
}
f := make([]int, n)
g := make([]int, n)
f[0] = arr[0]
ans = f[0]
for i := 1; i < n; i++ {
f[i] = g[i-1] + arr[i]*(i/2+1)
g[i] = f[i-1] + arr[i]*((i+1)/2)
ans += f[i]
}
return
}
Expand All @@ -169,15 +174,13 @@ func sumOddLengthSubarrays(arr []int) (ans int) {
```ts
function sumOddLengthSubarrays(arr: number[]): number {
const n = arr.length;
let ans = 0;
for (let i = 0; i < n; ++i) {
let s = 0;
for (let j = i; j < n; ++j) {
s += arr[j];
if ((j - i + 1) % 2 === 1) {
ans += s;
}
}
const f: number[] = Array(n).fill(arr[0]);
const g: number[] = Array(n).fill(0);
let ans = f[0];
for (let i = 1; i < n; ++i) {
f[i] = g[i - 1] + arr[i] * ((i >> 1) + 1);
g[i] = f[i - 1] + arr[i] * ((i + 1) >> 1);
ans += f[i];
}
return ans;
}
Expand All @@ -189,15 +192,14 @@ function sumOddLengthSubarrays(arr: number[]): number {
impl Solution {
pub fn sum_odd_length_subarrays(arr: Vec<i32>) -> i32 {
let n = arr.len();
let mut ans = 0;
for i in 0..n {
let mut s = 0;
for j in i..n {
s += arr[j];
if (j - i + 1) % 2 == 1 {
ans += s;
}
}
let mut f = vec![0; n];
let mut g = vec![0; n];
let mut ans = arr[0];
f[0] = arr[0];
for i in 1..n {
f[i] = g[i - 1] + arr[i] * ((i as i32) / 2 + 1);
g[i] = f[i - 1] + arr[i] * (((i + 1) as i32) / 2);
ans += f[i];
}
ans
}
Expand All @@ -208,15 +210,148 @@ impl Solution {

```c
int sumOddLengthSubarrays(int* arr, int arrSize) {
int ans = 0;
for (int i = 0; i < arrSize; ++i) {
int s = 0;
for (int j = i; j < arrSize; ++j) {
s += arr[j];
if ((j - i + 1) % 2 == 1) {
ans += s;
}
int n = arrSize;
int f[n];
int g[n];
int ans = f[0] = arr[0];
g[0] = 0;
for (int i = 1; i < n; ++i) {
f[i] = g[i - 1] + arr[i] * (i / 2 + 1);
g[i] = f[i - 1] + arr[i] * ((i + 1) / 2);
ans += f[i];
}
return ans;
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### 方法二：动态规划（空间优化）

我们注意到，状态 $f[i]$ 和 $g[i]$ 的值只与 $f[i - 1]$ 和 $g[i - 1]$ 有关，因此我们可以使用两个变量 $f$ 和 $g$ 分别记录 $f[i - 1]$ 和 $g[i - 1]$ 的值，从而优化空间复杂度。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。

<!-- tabs:start -->

#### Python3

```python
class Solution:
def sumOddLengthSubarrays(self, arr: List[int]) -> int:
ans, f, g = arr[0], arr[0], 0
for i in range(1, len(arr)):
ff = g + arr[i] * (i // 2 + 1)
gg = f + arr[i] * ((i + 1) // 2)
f, g = ff, gg
ans += f
return ans
```

#### Java

```java
class Solution {
public int sumOddLengthSubarrays(int[] arr) {
int ans = arr[0], f = arr[0], g = 0;
for (int i = 1; i < arr.length; ++i) {
int ff = g + arr[i] * (i / 2 + 1);
int gg = f + arr[i] * ((i + 1) / 2);
f = ff;
g = gg;
ans += f;
}
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
int sumOddLengthSubarrays(vector<int>& arr) {
int ans = arr[0], f = arr[0], g = 0;
for (int i = 1; i < arr.size(); ++i) {
int ff = g + arr[i] * (i / 2 + 1);
int gg = f + arr[i] * ((i + 1) / 2);
f = ff;
g = gg;
ans += f;
}
return ans;
}
};
```

#### Go

```go
func sumOddLengthSubarrays(arr []int) (ans int) {
f, g := arr[0], 0
ans = f
for i := 1; i < len(arr); i++ {
ff := g + arr[i]*(i/2+1)
gg := f + arr[i]*((i+1)/2)
f, g = ff, gg
ans += f
}
return
}
```

#### TypeScript

```ts
function sumOddLengthSubarrays(arr: number[]): number {
const n = arr.length;
let [ans, f, g] = [arr[0], arr[0], 0];
for (let i = 1; i < n; ++i) {
const ff = g + arr[i] * (Math.floor(i / 2) + 1);
const gg = f + arr[i] * Math.floor((i + 1) / 2);
[f, g] = [ff, gg];
ans += f;
}
return ans;
}
```

#### Rust

```rust
impl Solution {
pub fn sum_odd_length_subarrays(arr: Vec<i32>) -> i32 {
let mut ans = arr[0];
let mut f = arr[0];
let mut g = 0;
for i in 1..arr.len() {
let ff = g + arr[i] * ((i as i32) / 2 + 1);
let gg = f + arr[i] * (((i + 1) as i32) / 2);
f = ff;
g = gg;
ans += f;
}
ans
}
}
```

#### C

```c
int sumOddLengthSubarrays(int* arr, int arrSize) {
int ans = arr[0], f = arr[0], g = 0;
for (int i = 1; i < arrSize; ++i) {
int ff = g + arr[i] * (i / 2 + 1);
int gg = f + arr[i] * ((i + 1) / 2);
f = ff;
g = gg;
ans += f;
}
return ans;
}
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