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Commit 8d6b198

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feat: add solutions to lc problem: No.1588 (doocs#3888)
No.1588.Sum of All Odd Length Subarrays
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‎solution/1500-1599/1585.Check If String Is Transformable With Substring Sort Operations/README_EN.md‎

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@@ -79,7 +79,17 @@ tags:
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Bubble Sort
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The problem is essentially equivalent to determining whether any substring of length 2 in string $s$ can be swapped using bubble sort to obtain $t$.
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Therefore, we use an array $pos$ of length 10 to record the indices of each digit in string $s,ドル where $pos[i]$ represents the list of indices where digit $i$ appears, sorted in ascending order.
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Next, we iterate through string $t$. For each character $t[i]$ in $t,ドル 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,ドル so we return `false`. Otherwise, to swap the character at the first index of $pos[x]$ to index $i,ドル 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 $t$.
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After the iteration, we return `true`.
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The time complexity is $O(n \times C),ドル and the space complexity is $O(n)$. Here, $n$ is the length of string $s,ドル and $C$ is the size of the digit set, which is 10 in this problem.
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<!-- tabs:start -->
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‎solution/1500-1599/1588.Sum of All Odd Length Subarrays/README.md‎

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<!-- solution:start -->
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### 方法一:枚举 + 前缀和
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### 方法一:动态规划
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85-
我们可以枚举子数组的起点 $i$ 和终点 $j$,其中 $i \leq j,ドル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。
85+
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],ドル而 $g[0] = 0$。答案即为 $\sum_{i=0}^{n-1} f[i]$。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(1)$。其中 $n$ 是数组的长度。
87+
当 $i > 0$ 时,考虑 $f[i]$ 和 $g[i]$ 如何进行状态转移:
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对于状态 $f[i],ドル元素 $\textit{arr}[i]$ 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 $f[i] = g[i-1] + \textit{arr}[i] \times ((i / 2) + 1)$。
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对于状态 $g[i],ドル当 $i = 0$ 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 $g[i] = f[i-1] + \textit{arr}[i] \times ((i + 1) / 2)$。
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最终答案即为 $\sum_{i=0}^{n-1} f[i]$。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{arr}$ 的长度。
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<!-- tabs:start -->
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@@ -93,13 +101,14 @@ tags:
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```python
94102
class Solution:
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def sumOddLengthSubarrays(self, arr: List[int]) -> int:
96-
ans, n = 0, len(arr)
97-
for i in range(n):
98-
s = 0
99-
for j in range(i, n):
100-
s += arr[j]
101-
if (j - i + 1) & 1:
102-
ans += s
104+
n = len(arr)
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f = [0] * n
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g = [0] * n
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ans = f[0] = arr[0]
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for i in range(1, n):
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f[i] = g[i - 1] + arr[i] * (i // 2 + 1)
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g[i] = f[i - 1] + arr[i] * ((i + 1) // 2)
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ans += f[i]
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return ans
104113
```
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@@ -109,15 +118,13 @@ class Solution:
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class Solution {
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public int sumOddLengthSubarrays(int[] arr) {
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int n = arr.length;
112-
int ans = 0;
113-
for (int i = 0; i < n; ++i) {
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int s = 0;
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for (int j = i; j < n; ++j) {
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s += arr[j];
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if ((j - i + 1) % 2 == 1) {
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ans += s;
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}
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}
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int[] f = new int[n];
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int[] g = new int[n];
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int ans = f[0] = arr[0];
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for (int i = 1; i < n; ++i) {
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f[i] = g[i - 1] + arr[i] * (i / 2 + 1);
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g[i] = f[i - 1] + arr[i] * ((i + 1) / 2);
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ans += f[i];
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}
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return ans;
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}
@@ -131,15 +138,13 @@ class Solution {
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public:
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int sumOddLengthSubarrays(vector<int>& arr) {
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int n = arr.size();
134-
int ans = 0;
135-
for (int i = 0; i < n; ++i) {
136-
int s = 0;
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for (int j = i; j < n; ++j) {
138-
s += arr[j];
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if ((j - i + 1) & 1) {
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ans += s;
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}
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}
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vector<int> f(n, arr[0]);
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vector<int> g(n);
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int ans = f[0];
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for (int i = 1; i < n; ++i) {
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f[i] = g[i - 1] + arr[i] * (i / 2 + 1);
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g[i] = f[i - 1] + arr[i] * ((i + 1) / 2);
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ans += f[i];
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}
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return ans;
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}
@@ -151,14 +156,14 @@ public:
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```go
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func sumOddLengthSubarrays(arr []int) (ans int) {
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n := len(arr)
154-
for i := range arr {
155-
s := 0
156-
for j := i; j < n; j++ {
157-
s += arr[j]
158-
if (j-i+1)%2 == 1 {
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ans += s
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}
161-
}
159+
f := make([]int, n)
160+
g := make([]int, n)
161+
f[0] = arr[0]
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ans = f[0]
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for i := 1; i < n; i++ {
164+
f[i] = g[i-1] + arr[i]*(i/2+1)
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g[i] = f[i-1] + arr[i]*((i+1)/2)
166+
ans += f[i]
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}
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return
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}
@@ -169,15 +174,13 @@ func sumOddLengthSubarrays(arr []int) (ans int) {
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```ts
170175
function sumOddLengthSubarrays(arr: number[]): number {
171176
const n = arr.length;
172-
let ans = 0;
173-
for (let i = 0; i < n; ++i) {
174-
let s = 0;
175-
for (let j = i; j < n; ++j) {
176-
s += arr[j];
177-
if ((j - i + 1) % 2 === 1) {
178-
ans += s;
179-
}
180-
}
177+
const f: number[] = Array(n).fill(arr[0]);
178+
const g: number[] = Array(n).fill(0);
179+
let ans = f[0];
180+
for (let i = 1; i < n; ++i) {
181+
f[i] = g[i - 1] + arr[i] * ((i >> 1) + 1);
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g[i] = f[i - 1] + arr[i] * ((i + 1) >> 1);
183+
ans += f[i];
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}
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return ans;
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}
@@ -189,15 +192,14 @@ function sumOddLengthSubarrays(arr: number[]): number {
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impl Solution {
190193
pub fn sum_odd_length_subarrays(arr: Vec<i32>) -> i32 {
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let n = arr.len();
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let mut ans = 0;
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for i in 0..n {
194-
let mut s = 0;
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for j in i..n {
196-
s += arr[j];
197-
if (j - i + 1) % 2 == 1 {
198-
ans += s;
199-
}
200-
}
195+
let mut f = vec![0; n];
196+
let mut g = vec![0; n];
197+
let mut ans = arr[0];
198+
f[0] = arr[0];
199+
for i in 1..n {
200+
f[i] = g[i - 1] + arr[i] * ((i as i32) / 2 + 1);
201+
g[i] = f[i - 1] + arr[i] * (((i + 1) as i32) / 2);
202+
ans += f[i];
201203
}
202204
ans
203205
}
@@ -208,15 +210,148 @@ impl Solution {
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209211
```c
210212
int sumOddLengthSubarrays(int* arr, int arrSize) {
211-
int ans = 0;
212-
for (int i = 0; i < arrSize; ++i) {
213-
int s = 0;
214-
for (int j = i; j < arrSize; ++j) {
215-
s += arr[j];
216-
if ((j - i + 1) % 2 == 1) {
217-
ans += s;
218-
}
213+
int n = arrSize;
214+
int f[n];
215+
int g[n];
216+
int ans = f[0] = arr[0];
217+
g[0] = 0;
218+
for (int i = 1; i < n; ++i) {
219+
f[i] = g[i - 1] + arr[i] * (i / 2 + 1);
220+
g[i] = f[i - 1] + arr[i] * ((i + 1) / 2);
221+
ans += f[i];
222+
}
223+
return ans;
224+
}
225+
```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- solution:start -->
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### 方法二:动态规划(空间优化)
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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。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。
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<!-- tabs:start -->
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#### Python3
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```python
244+
class Solution:
245+
def sumOddLengthSubarrays(self, arr: List[int]) -> int:
246+
ans, f, g = arr[0], arr[0], 0
247+
for i in range(1, len(arr)):
248+
ff = g + arr[i] * (i // 2 + 1)
249+
gg = f + arr[i] * ((i + 1) // 2)
250+
f, g = ff, gg
251+
ans += f
252+
return ans
253+
```
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#### Java
256+
257+
```java
258+
class Solution {
259+
public int sumOddLengthSubarrays(int[] arr) {
260+
int ans = arr[0], f = arr[0], g = 0;
261+
for (int i = 1; i < arr.length; ++i) {
262+
int ff = g + arr[i] * (i / 2 + 1);
263+
int gg = f + arr[i] * ((i + 1) / 2);
264+
f = ff;
265+
g = gg;
266+
ans += f;
219267
}
268+
return ans;
269+
}
270+
}
271+
```
272+
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#### C++
274+
275+
```cpp
276+
class Solution {
277+
public:
278+
int sumOddLengthSubarrays(vector<int>& arr) {
279+
int ans = arr[0], f = arr[0], g = 0;
280+
for (int i = 1; i < arr.size(); ++i) {
281+
int ff = g + arr[i] * (i / 2 + 1);
282+
int gg = f + arr[i] * ((i + 1) / 2);
283+
f = ff;
284+
g = gg;
285+
ans += f;
286+
}
287+
return ans;
288+
}
289+
};
290+
```
291+
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#### Go
293+
294+
```go
295+
func sumOddLengthSubarrays(arr []int) (ans int) {
296+
f, g := arr[0], 0
297+
ans = f
298+
for i := 1; i < len(arr); i++ {
299+
ff := g + arr[i]*(i/2+1)
300+
gg := f + arr[i]*((i+1)/2)
301+
f, g = ff, gg
302+
ans += f
303+
}
304+
return
305+
}
306+
```
307+
308+
#### TypeScript
309+
310+
```ts
311+
function sumOddLengthSubarrays(arr: number[]): number {
312+
const n = arr.length;
313+
let [ans, f, g] = [arr[0], arr[0], 0];
314+
for (let i = 1; i < n; ++i) {
315+
const ff = g + arr[i] * (Math.floor(i / 2) + 1);
316+
const gg = f + arr[i] * Math.floor((i + 1) / 2);
317+
[f, g] = [ff, gg];
318+
ans += f;
319+
}
320+
return ans;
321+
}
322+
```
323+
324+
#### Rust
325+
326+
```rust
327+
impl Solution {
328+
pub fn sum_odd_length_subarrays(arr: Vec<i32>) -> i32 {
329+
let mut ans = arr[0];
330+
let mut f = arr[0];
331+
let mut g = 0;
332+
for i in 1..arr.len() {
333+
let ff = g + arr[i] * ((i as i32) / 2 + 1);
334+
let gg = f + arr[i] * (((i + 1) as i32) / 2);
335+
f = ff;
336+
g = gg;
337+
ans += f;
338+
}
339+
ans
340+
}
341+
}
342+
```
343+
344+
#### C
345+
346+
```c
347+
int sumOddLengthSubarrays(int* arr, int arrSize) {
348+
int ans = arr[0], f = arr[0], g = 0;
349+
for (int i = 1; i < arrSize; ++i) {
350+
int ff = g + arr[i] * (i / 2 + 1);
351+
int gg = f + arr[i] * ((i + 1) / 2);
352+
f = ff;
353+
g = gg;
354+
ans += f;
220355
}
221356
return ans;
222357
}

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