# 891. Sum of Subsequence Widths

## Description

The width of a sequence is the difference between the maximum and minimum elements in the sequence.

Given an array of integers nums, return the sum of the widths of all the non-empty subsequences of nums. Since the answer may be very large, return it modulo 109 + 7.

A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].

Example 1:

Input: nums = [2,1,3]
Output: 6
Explanation: The subsequences are [1], [2], [3], [2,1], [2,3], [1,3], [2,1,3].
The corresponding widths are 0, 0, 0, 1, 1, 2, 2.
The sum of these widths is 6.


Example 2:

Input: nums = [2]
Output: 0


Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105

## Solutions

• class Solution {
private static final int MOD = (int) 1e9 + 7;

public int sumSubseqWidths(int[] nums) {
Arrays.sort(nums);
long ans = 0, p = 1;
int n = nums.length;
for (int i = 0; i < n; ++i) {
ans = (ans + (nums[i] - nums[n - i - 1]) * p + MOD) % MOD;
p = (p << 1) % MOD;
}
return (int) ans;
}
}

• class Solution {
public:
const int mod = 1e9 + 7;

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

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


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