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feat: add solutions to lc problem: No.0891
No.0891.Sum of Subsequence Widths
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‎solution/0700-0799/0795.Number of Subarrays with Bounded Maximum/README.md

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**方法一：单调栈**
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我们可以枚举数组中每个元素作为子数组的最大值，然后统计以该元素为最大值的子数组的个数。问题转化为求出每个元素 $nums[i]$ 左侧第一个大于该元素的下标 $l[i]$，右侧第一个大于等于该元素的下标 $r[i]$，则以该元素为最大值的子数组的个数为 $(i - left[i]) \times (right[i] - i)$。
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我们可以枚举数组中每个元素作为子数组的最大值，然后统计以该元素为最大值的子数组的个数。问题转化为求出每个元素 $nums[i]$ 左侧第一个大于该元素的下标 $l[i]$，右侧第一个大于等于该元素的下标 $r[i]$，则以该元素为最大值的子数组的个数为 $(i - l[i]) \times (r[i] - i)$。
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我们可以使用单调栈求出 $l[i]$ 和 $r[i]$。
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‎solution/0800-0899/0891.Sum of Subsequence Widths/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：排序 + 枚举贡献**
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题目求解的是数组 `nums` 中所有子序列中最大值与最小值差值之和，注意到“子序列”，并且涉及到“最大值”与“最小值”，我们考虑先对数组 `nums` 进行排序。
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然后我们枚举数组 `nums` 中的每个元素 $nums[i]$，该元素左侧的元素个数为 $i$，右侧的元素个数为 $n-i-1$。
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如果我们将元素 $nums[i]$ 作为子序列的最大值，总共有多少个满足条件的子序列呢？显然，子序列的其他元素应该从左侧的 $i$ 个元素中选取，每个元素有两种选择，即选或不选，因此总共有 $2^i$ 个子序列。同理，如果我们将元素 $nums[i]$ 作为子序列的最小值，那么总共有 $2^{n-i-1}$ 个满足条件的子序列。因此 $nums[i]$ 对答案的贡献为：
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$$
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\begin{aligned}
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nums[i] \times (2^i - 2^{n-i-1})
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\end{aligned}
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$$
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我们将数组 `nums` 中所有元素的贡献累加，即为答案：
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$$
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\begin{aligned}
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\sum_{i=0}^{n-1} nums[i] \times (2^i - 2^{n-i-1})
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\end{aligned}
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$$
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我们将上述式子展开，可以得到：
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$$
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\begin{aligned}
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nums[0] \times (2^0 - 2^{n-1}) + nums[1] \times (2^1 - 2^{n-2}) + ... + nums[n-1] \times (2^{n-1} - 2^0)
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\end{aligned}
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$$
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再将式子中相同的幂次项合并，可以得到：
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$$
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\begin{aligned}
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(nums[0] - nums[n-1]) \times 2^0 + (nums[1] - nums[n-2]) \times 2^1 + ... + (nums[n-1] - nums[0]) \times 2^{n-1}
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\end{aligned}
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$$
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因此我们只需要对数组 `nums` 进行排序，然后计算上述的贡献即可。注意答案的取模操作。
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时间复杂度 $O(n\times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组 `nums` 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def sumSubseqWidths(self, nums: List[int]) -> int:
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mod = 10**9 + 7
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nums.sort()
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ans, p = 0, 1
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for i, v in enumerate(nums):
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ans = (ans + (v - nums[-i - 1]) * p) % mod
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p = (p << 1) % mod
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private static final int MOD = (int) 1e9 + 7;
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public int sumSubseqWidths(int[] nums) {
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Arrays.sort(nums);
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long ans = 0, p = 1;
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int n = nums.length;
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for (int i = 0; i < n; ++i) {
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ans = (ans + (nums[i] - nums[n - i - 1]) * p + MOD) % MOD;
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p = (p << 1) % MOD;
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}
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return (int) ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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const int mod = 1e9 + 7;
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int sumSubseqWidths(vector<int>& nums) {
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sort(nums.begin(), nums.end());
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long ans = 0, p = 1;
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int n = nums.size();
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for (int i = 0; i < n; ++i) {
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ans = (ans + (nums[i] - nums[n - i - 1]) * p + mod) % mod;
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p = (p << 1) % mod;
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func sumSubseqWidths(nums []int) (ans int) {
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const mod int = 1e9 + 7
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sort.Ints(nums)
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p, n := 1, len(nums)
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for i, v := range nums {
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ans = (ans + (v-nums[n-i-1])*p + mod) % mod
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p = (p << 1) % mod
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}
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return
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}
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```
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### **...**

‎solution/0800-0899/0891.Sum of Subsequence Widths/README_EN.md

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### **Python3**
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```python
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class Solution:
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def sumSubseqWidths(self, nums: List[int]) -> int:
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mod = 10**9 + 7
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nums.sort()
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ans, p = 0, 1
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for i, v in enumerate(nums):
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ans = (ans + (v - nums[-i - 1]) * p) % mod
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p = (p << 1) % mod
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return ans
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```
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### **Java**
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```java
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class Solution {
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private static final int MOD = (int) 1e9 + 7;
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public int sumSubseqWidths(int[] nums) {
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Arrays.sort(nums);
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long ans = 0, p = 1;
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int n = nums.length;
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for (int i = 0; i < n; ++i) {
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ans = (ans + (nums[i] - nums[n - i - 1]) * p + MOD) % MOD;
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p = (p << 1) % MOD;
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}
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return (int) ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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const int mod = 1e9 + 7;
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int sumSubseqWidths(vector<int>& nums) {
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sort(nums.begin(), nums.end());
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long ans = 0, p = 1;
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int n = nums.size();
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for (int i = 0; i < n; ++i) {
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ans = (ans + (nums[i] - nums[n - i - 1]) * p + mod) % mod;
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p = (p << 1) % mod;
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func sumSubseqWidths(nums []int) (ans int) {
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const mod int = 1e9 + 7
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sort.Ints(nums)
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p, n := 1, len(nums)
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for i, v := range nums {
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ans = (ans + (v-nums[n-i-1])*p + mod) % mod
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p = (p << 1) % mod
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}
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return
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}
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```
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### **...**

‎solution/0800-0899/0891.Sum of Subsequence Widths/Solution.cpp

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class Solution {
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public:
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const int mod = 1e9 + 7;
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int sumSubseqWidths(vector<int>& nums) {
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sort(nums.begin(), nums.end());
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long ans = 0, p = 1;
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int n = nums.size();
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for (int i = 0; i < n; ++i) {
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ans = (ans + (nums[i] - nums[n - i - 1]) * p + mod) % mod;
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p = (p << 1) % mod;
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}
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return ans;
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}
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};

‎solution/0800-0899/0891.Sum of Subsequence Widths/Solution.go

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func sumSubseqWidths(nums []int) (ans int) {
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const mod int = 1e9 + 7
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sort.Ints(nums)
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p, n := 1, len(nums)
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for i, v := range nums {
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ans = (ans + (v-nums[n-i-1])*p + mod) % mod
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p = (p << 1) % mod
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}
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return
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}

‎solution/0800-0899/0891.Sum of Subsequence Widths/Solution.java

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class Solution {
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public int sumSubseqWidths(int[] A) {
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final int MOD = (int) (1e9 + 7);
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Arrays.sort(A);
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int n = A.length;
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long res = 0;
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long p = 1;
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private static final int MOD = (int) 1e9 + 7;
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public int sumSubseqWidths(int[] nums) {
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Arrays.sort(nums);
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long ans = 0, p = 1;
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int n = nums.length;
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for (int i = 0; i < n; ++i) {
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res = (res + (A[i] - A[n - 1 - i]) * p) % MOD;
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ans = (ans + (nums[i] - nums[n - i - 1]) * p + MOD) % MOD;
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p = (p << 1) % MOD;
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}
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return (int) ((res + MOD) % MOD);
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return (int) ans;
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}
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}
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}

‎solution/0800-0899/0891.Sum of Subsequence Widths/Solution.py

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class Solution:
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def sumSubseqWidths(self, nums: List[int]) -> int:
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mod = 10**9 + 7
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nums.sort()
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ans, p = 0, 1
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for i, v in enumerate(nums):
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ans = (ans + (v - nums[-i - 1]) * p) % mod
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p = (p << 1) % mod
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return ans

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