Sum of Subsequence Widths

Problem Statement

In this problem, the concept of a sequence "width" is defined as the difference between the maximum and minimum elements within that sequence. We need to consider the array of integers named nums and calculate the sum of the "widths" for all possible non-empty subsequences it can generate. Due to the potentially large result size, the final answer should be returned modulo 10^9 + 7.

A subsequence in this context can be derived by either omitting some elements or no elements at all from the array without changing the order of the remaining elements. Each subsequence will have a calculated width based on its elements, and the objective is to sum these widths across all possible subsequences derived from the initial array.

Examples

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

Approach and Intuition

1. Understanding Width Calculation: For any subsequence, the width is directly calculated as the difference between the largest and smallest element present in it.

2. Subsequences Overview:

   • Due to properties of subsets, an array of size N can generate 2^N possible subsets (including the empty set). However, the task only concerns non-empty subsequences.
   • For example, an array of three elements would have 2^3 - 1 = 7 non-empty subsequences.

3. Calculating Sum of Widths:

   • Sorting for Easier Calculations: By sorting the array, determining min and max for any subsequence becomes more structured, particularly because the smallest value will be towards the beginning and the largest towards the end of any subsequence.
   • Using Powers of Two:
     • Every element in a sorted array contributes to many subsequences. The number of these subsequences varies exponentially as more elements from the array are considered for deletion. This is a key insight that can be utilized to optimize the calculation.
     • Specifically, for any element at index I, it can appear in 2^(I) subsequences as the maximum if we view the elements to the right (inclusive) and in 2^(N-I-1) subsequences as the minimum if we look at the elements to the left (inclusive).
   • More specifically, this allows us to determine how many times an element contributes to the width sum depending on its position in the sorted array.

By sorting the array and using the efficient counting of subsequences for max and min contributions per element, the solution can effectively and efficiently compute the required sum of widths. Additionally, utilizing modular arithmetic helps manage large numbers resulting from the calculations, especially given the constraints.

Solutions

class Solution {
    public int computeWidths(int[] nums) {
        int MODULO = 1_000_000_007;
        int length = nums.length;
        Arrays.sort(nums);
    
        long[] powerOfTwo = new long[length];
        powerOfTwo[0] = 1;
        for (int index = 1; index < length; ++index)
            powerOfTwo[index] = powerOfTwo[index - 1] * 2 % MODULO;
    
        long result = 0;
        for (int i = 0; i < length; ++i)
            result = (result + (powerOfTwo[i] - powerOfTwo[length - 1 - i]) * nums[i]) % MODULO;
    
        return (int) result;
    }
}