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.

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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

Idea 1.  刚开始想subset穷举, sort the array and get all the pair (0<= i < j <= n-1, A[i] < A[j]) such that (A[j] - A[i]) * 2^(j-i-1), saw an amazing online soloution, consider the contribution for each element, assume sequence is like A[0]...A[i-1]A[i]A[i+1]...A[n-1], on the left, there are i numbers < A[i], 2^(i) subsequence where A[i] is the maximu, on the right, there are n-1-i numbers > A[i], 2^(n-1-i) subsequence A[i] as minimum, hence we have

res = A[i]*2^(i) - A[i]*2^(n-1-i)

another trick to save compute 2^(n-1-i) and 2^(i) separately, sum(A[n-1-i]*2^(n-1-i)) = sum(A[n-1-i]*2^(i))

1 << i

(c=1 << 1) incrementely

Time complexity: O(nlogn)

Space complexity: O(1)

 1 class Solution {
 2     public int sumSubseqWidths(int[] A) {
 3         long res = 0;
 4         long mod = (long)1e9+7;
 5         long c = 1;
 6         int n = A.length;
 7         
 8         Arrays.sort(A);
 9         
10         for(int i = 0; i < A.length; ++i, c = (c << 1)%mod) {
11             res =  (res + (A[i] - A[n - 1 - i]) * c + mod)%mod;
12         }
13         
14         return (int)(res);
15     }
16 }
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