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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
}

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