The problem goes like this:
You are given an integer array nums.
A nice permutation is defined as a permutation of any non-empty subset of nums such that the absolute difference between any two consecutive elements is exactly 1. A permutation of size 1 is considered nice by definition. A subset is any non-empty selection of elements from nums (regardless of order). A permutation of a subset is any reordering of the elements of a subset.
Your task is to compute the sum of all elements across all such nice permutations.
Since the total may be very large, output the result modulo $$$10^9 + 7$$$.
Input
- An array
numsof integers. - $$$1 \leq \text{nums.length} \leq 10^5$$$ (maybe no solution exists though ... I can't guarantee anything lol)
- $$$0 \leq \text{nums}[i] \leq 10^5$$$ (maybe no solution exists though ... I can't guarantee anything lol)
Output
- A single integer: the sum of all elements in all nice permutations of all non-empty subsets of
nums, taken modulo $$$10^9 + 7$$$.
Examples
nums = [1, 2, 3]
[1],[2],[3],[1,2],[2,1],[2,3],[3,2],[1,2,3],[3,2,1]are all valid.- The sum is 34.
[1, 3],[3, 1],[2,3,1],[2,1,3],[3,1,2], and[1,3,2]are not valid.
nums = [2, 3, 2]
[2],[2],[3],[2,3],[3,2],[2,3],[3,2],[2,3,2],[2,3,2]are all valid.- The sum is 41.
[2,2],[2,2],[2,2,3],[2,2,3],[3,2,2], and[3,2,2]are not valid.
Apart from brute-forcing all $$$\sum_{k=1}^{n} \binom{n}{k} \cdot k!$$$ possibilities, I can't find any way to do it. I feel like we should start by sorting and making clusters (for example [1, 2, 9, 10, 11, 12] would have two clusters, [1, 2] and [9, 10, 11, 12]), but that kind of got me nowhere.








Very stupid question, but are all the elements distinct? Then it could be easier. With duplicates, it won't. Even counting the possible nice permutatiosn would be difficult then.
No they’re not necessarily distinct
Well, if they were distinct, then any nice permuation subsets must be in increasing order or decreasing order. Then, sorting the original permutation would be a very good idea, because it could help you find those permutations easily
Stuck on duplicates
I could thought of a solution of n*2^n...
if you choose an element as starting point and then do recursion to choose the -1 or +1 (if exists) then keep a sum tracker go on with it.
Then it may be solvable for n<=12
Also starting point are distinct so , if there are multpile same numbers we only have to choose the distinct ones.
The problem has a solution.