891. Sum of Subsequence Widths

Explanation

To solve this problem, we need to find the width of all possible subsequences of the given array and sum up these widths. The width of a subsequence is the difference between the maximum and minimum elements in that subsequence. We can achieve this by sorting the array and then calculating the contribution of each element to the sum.

Sort the array nums.
For each element num[i] in nums, calculate its contribution to the sum.
The contribution of num[i] is (2^i - 2^(n-1-i)) * num[i], where i is the index of the element in the sorted array and n is the length of the array.
Sum up the contributions for all elements and return the result modulo 10^9 + 7.

Time complexity: O(nlogn) where n is the length of the input array. Space complexity: O(1)

View Leetcode Problem

class Solution {
    public int sumSubseqWidths(int[] nums) {
        int mod = 1000000007;
        Arrays.sort(nums);
        int n = nums.length;
        long sum = 0;
        long pow2 = 1;
        
        for (int i = 0; i < n; i++) {
            sum = (sum + (pow2 - pow(2, n - 1 - i)) * nums[i]) % mod;
            pow2 = (pow2 * 2) % mod;
        }
        
        return (int)sum;
    }
    
    private long pow(int x, int n) {
        if (n == 0) return 1;
        long mod = 1000000007;
        long res = 1;
        long base = x;
        while (n > 0) {
            if (n % 2 == 1) {
                res = (res * base) % mod;
            }
            base = (base * base) % mod;
            n /= 2;
        }
        return res;
    }
}

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Explanation

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