feat: update solutions to lc problems: No.0903~0910

solution/0900-0999/0903.Valid Permutations for DI Sequence/README.md

@@ -73,8 +73,6 @@
 
 时间复杂度 $O(n^3)$，空间复杂度 $O(n^2)$。其中 $n$ 是字符串的长度。
 
-我们可以用前缀和优化时间复杂度，使得时间复杂度降低到 $O(n^2)$。另外，我们也可以用滚动数组优化空间复杂度，使得空间复杂度降低到 $O(n)$。
-
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 ```python
@@ -224,7 +222,7 @@ function numPermsDISequence(s: string): number {
 
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-### 方法二
+我们可以用前缀和优化时间复杂度，使得时间复杂度降低到 $O(n^2)$。
 
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@@ -372,7 +370,7 @@ function numPermsDISequence(s: string): number {
 
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-### 方法三
+另外，我们也可以用滚动数组优化空间复杂度，使得空间复杂度降低到 $O(n)$。
 
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solution/0900-0999/0903.Valid Permutations for DI Sequence/README_EN.md

@@ -68,8 +68,6 @@ The final answer is $\sum_{j=0}^n f[n][j]$.
 
 The time complexity is $O(n^3)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the string.
 
-We can optimize the time complexity to $O(n^2)$ using prefix sums. Additionally, we can optimize the space complexity to $O(n)$ using a rolling array.
-
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 ```python
@@ -219,7 +217,7 @@ function numPermsDISequence(s: string): number {
 
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-### Solution 2
+We can optimize the time complexity to $O(n^2)$ using prefix sums.
 
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@@ -367,7 +365,7 @@ function numPermsDISequence(s: string): number {
 
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-### Solution 3
+Additionally, we can optimize the space complexity to $O(n)$ using a rolling array.
 
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solution/0900-0999/0904.Fruit Into Baskets/README.md

@@ -91,7 +91,7 @@ j   i
   j     i
 ```
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `fruits` 的长度。
+时间复杂度 $O(n)$，其中 $n$ 为数组 `fruits` 的长度。空间复杂度 $O(1)$。
 
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@@ -231,7 +231,7 @@ impl Solution {
 
 但本题实际上求的是水果的最大数目，也就是“最大”的窗口，我们没有必要缩小窗口，只需要让窗口单调增大。于是代码就少了每次更新答案的操作，只需要在遍历结束后将此时的窗口大小作为答案返回即可。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `fruits` 的长度。
+时间复杂度 $O(n)$，其中 $n$ 为数组 `fruits` 的长度。空间复杂度 $O(1)$。
 
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solution/0900-0999/0907.Sum of Subarray Minimums/README.md

@@ -300,61 +300,4 @@ impl Solution {
 
 <!-- tabs:end -->
 
-### 方法二
-
-<!-- tabs:start -->
-
-```rust
-const MOD: i64 = (1e9 as i64) + 7;
-
-impl Solution {
-    pub fn sum_subarray_mins(arr: Vec<i32>) -> i32 {
-        let n: usize = arr.len();
-        let mut ret: i64 = 0;
-        let mut left: Vec<i32> = vec![-1; n];
-        let mut right: Vec<i32> = vec![n as i32; n];
-        // Index stack, store the index of the value in the given array
-        let mut stack: Vec<i32> = Vec::new();
-
-        // Find the first element that's less than the current value for the left side
-        // The default value of which is -1
-        for i in 0..n {
-            while !stack.is_empty() && arr[*stack.last().unwrap() as usize] >= arr[i] {
-                stack.pop();
-            }
-            if !stack.is_empty() {
-                left[i] = *stack.last().unwrap();
-            }
-            stack.push(i as i32);
-        }
-
-        stack.clear();
-
-        // Find the first element that's less or equal than the current value for the right side
-        // The default value of which is n
-        for i in (0..n).rev() {
-            while !stack.is_empty() && arr[*stack.last().unwrap() as usize] > arr[i] {
-                stack.pop();
-            }
-            if !stack.is_empty() {
-                right[i] = *stack.last().unwrap();
-            }
-            stack.push(i as i32);
-        }
-
-        // Traverse the array, to find the sum
-        for i in 0..n {
-            ret +=
-                ((((right[i] - (i as i32)) * ((i as i32) - left[i])) as i64) * (arr[i] as i64)) %
-                MOD;
-            ret %= MOD;
-        }
-
-        (ret % (MOD as i64)) as i32
-    }
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

solution/0900-0999/0907.Sum of Subarray Minimums/README_EN.md

@@ -37,7 +37,37 @@ Sum is 17.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Monotonic Stack
+
+The problem asks for the sum of the minimum values of each subarray, which is equivalent to finding the number of subarrays for which each element $arr[i]$ is the minimum, then multiplying by $arr[i]$, and finally summing these up.
+
+Therefore, the focus of the problem is to find the number of subarrays for which $arr[i]$ is the minimum. For $arr[i]$, we find the first position $left[i]$ to its left that is less than $arr[i]$, and the first position $right[i]$ to its right that is less than or equal to $arr[i]$. The number of subarrays for which $arr[i]$ is the minimum is $(i - left[i]) \times (right[i] - i)$.
+
+Note, why do we find the first position $right[i]$ to the right that is less than or equal to $arr[i]$, rather than less than $arr[i]$? This is because if we find the first position $right[i]$ to the right that is less than $arr[i]$, it will lead to duplicate calculations.
+
+Let's take an example to illustrate. For the following array:
+
+The element at index $3$ is $2$, the first element to its left that is less than $2$ is at index $0$. If we find the first element to its right that is less than $2$, we get index $7$. That is, the subarray interval is $(0, 7)$. Note that this is an open interval.
+
+```
+0 4 3 2 5 3 2 1
+*     ^       *
+```
+
+In the same way, we can find the subarray interval for the element at index $6$, and find that its subarray interval is also $(0, 7)$. That is, the subarray intervals for the elements at index $3$ and index $6$ are the same. This leads to duplicate calculations.
+
+```
+0 4 3 2 5 3 2 1
+*           ^ *
+```
+
+If we find the first element to its right that is less than or equal to its value, there will be no duplication, because the subarray interval for the element at index $3$ becomes $(0, 6)$, and the subarray interval for the element at index $6$ is $(0, 7)$, which are not the same.
+
+Back to this problem, we just need to traverse the array, for each element $arr[i]$, use a monotonic stack to find the first position $left[i]$ to its left that is less than $arr[i]$, and the first position $right[i]$ to its right that is less than or equal to $arr[i]$. The number of subarrays for which $arr[i]$ is the minimum is $(i - left[i]) \times (right[i] - i)$, then multiply by $arr[i]$, and finally sum these up.
+
+Be aware of data overflow and modulo operations.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $arr$.
 
 <!-- tabs:start -->
 
@@ -264,61 +294,4 @@ impl Solution {
 
 <!-- tabs:end -->
 
-### Solution 2
-
-<!-- tabs:start -->
-
-```rust
-const MOD: i64 = (1e9 as i64) + 7;
-
-impl Solution {
-    pub fn sum_subarray_mins(arr: Vec<i32>) -> i32 {
-        let n: usize = arr.len();
-        let mut ret: i64 = 0;
-        let mut left: Vec<i32> = vec![-1; n];
-        let mut right: Vec<i32> = vec![n as i32; n];
-        // Index stack, store the index of the value in the given array
-        let mut stack: Vec<i32> = Vec::new();
-
-        // Find the first element that's less than the current value for the left side
-        // The default value of which is -1
-        for i in 0..n {
-            while !stack.is_empty() && arr[*stack.last().unwrap() as usize] >= arr[i] {
-                stack.pop();
-            }
-            if !stack.is_empty() {
-                left[i] = *stack.last().unwrap();
-            }
-            stack.push(i as i32);
-        }
-
-        stack.clear();
-
-        // Find the first element that's less or equal than the current value for the right side
-        // The default value of which is n
-        for i in (0..n).rev() {
-            while !stack.is_empty() && arr[*stack.last().unwrap() as usize] > arr[i] {
-                stack.pop();
-            }
-            if !stack.is_empty() {
-                right[i] = *stack.last().unwrap();
-            }
-            stack.push(i as i32);
-        }
-
-        // Traverse the array, to find the sum
-        for i in 0..n {
-            ret +=
-                ((((right[i] - (i as i32)) * ((i as i32) - left[i])) as i64) * (arr[i] as i64)) %
-                MOD;
-            ret %= MOD;
-        }
-
-        (ret % (MOD as i64)) as i32
-    }
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

solution/0900-0999/0908.Smallest Range I/README.md

@@ -54,7 +54,13 @@
 
 ## 解法
 
-### 方法一
+### 方法一：数学
+
+根据题目描述，我们可以将数组中的最大值加上 $k$，最小值减去 $k$，这样可以使得数组中的最大值和最小值之差变小。
+
+因此，最终的答案就是 $\max(nums) - \min(nums) - 2 \times k$。
+
+时间复杂度 $O(n)$，其中 $n$ 为数组 `nums` 的长度。空间复杂度 $O(1)$。
 
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solution/0900-0999/0908.Smallest Range I/README_EN.md

@@ -50,7 +50,13 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Mathematics
+
+According to the problem description, we can add $k$ to the maximum value in the array and subtract $k$ from the minimum value. This can reduce the difference between the maximum and minimum values in the array.
+
+Therefore, the final answer is $\max(nums) - \min(nums) - 2 \times k$.
+
+The time complexity is $O(n)$, where $n$ is the length of the array `nums`. The space complexity is $O(1)$.
 
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solution/0900-0999/0910.Smallest Range II/README_EN.md

@@ -50,7 +50,13 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Greedy + Enumeration
+
+According to the problem requirements, we need to find the minimum difference between the maximum and minimum values in the array. Each element can be increased or decreased by $k$, so we can divide the elements in the array into two parts, one part increased by $k$ and the other part decreased by $k$. Therefore, we should decrease the larger values in the array by $k$ and increase the smaller values by $k$ to ensure the minimum difference between the maximum and minimum values.
+
+Therefore, we can first sort the array, then enumerate each element in the array, divide it into two parts, one part increased by $k$ and the other part decreased by $k$, and calculate the difference between the maximum and minimum values. Finally, take the minimum value among all differences.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$, where $n$ is the length of the array.
 
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