Sum of Subsequence Widths

Given an array of integers A, consider all non-empty subsequences of A.

For any sequence S, let the width of S be the difference between the maximum and minimum element of S.

Return the sum of the widths of all subsequences of A.

As the answer may be very large, return the answer modulo 10^9 + 7.

Example 1:

```Input: [2,1,3]
Output: 6
Explanation:
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.
```

Note:

• 1 <= A.length <= 20000
• 1 <= A[i] <= 20000

Solution:

```class Solution {
long mod = (long) 1e9 + 7;

/**
Explanation
The order in initial arrays doesn't matter,
my first intuition is to sort the array.

For each number A[i]:

There are i smaller numbers,
so there are 2 ^ i sequences in which A[i] is maximum.
we should do res += A[i] * (2 ^ i)

There are n - i - 1 bigger numbers,
so there are 2 ^ (n - i - 1) sequences in which A[i] is minimum.
we should do res -= A[i] * 2 ^ (n - i - 1)

*/
public int sumSubseqWidths(int[] A) {
Arrays.sort(A);
int n = A.length;
long c = 1, res = 0;
for (int i = 0; i < n; i ++) {
res = (res + (A[i] * c) % mod) % mod;
res = (res - (A[n - i - 1] * c) % mod) % mod;
c = (c * 2) % mod;
}
return (int) ((res + mod) % mod);
}
}```