Problem
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.
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
Solution: Math
Sort the array, for A[i]:
- i numbers <= A[i]. A[i] is the upper bound of 2^i subsequences.
- n – i – 1 numbers >= A[i]. A[i] is the lower bound of 2^(n – i – 1) subsequences.
- A[i] contributes A[i] * 2^i – A[i] * 2^(n – i – 1) to the ans.
Time complexity: O(nlogn)
Space complexity: O(1)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 |
// Author: Huahua // Running time: 56 ms class Solution { public: int sumSubseqWidths(vector<int>& A) { constexpr long kMod = 1e9 + 7; const int n = A.size(); sort(begin(A), end(A)); long ans = 0; long p = 1; for (int i = 0; i < n; ++i) { ans = (ans + (A[i] - A[n - i - 1]) * p) % kMod; p = (p << 1) % kMod; } return (ans + kMod) % kMod; } }; |
Time complexity: O(n)
Space complexity: O(n)
Counting sort
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 |
// Author: Huahua // Running time: 16 / 44 ms (beats 100%) static auto x = [](){std::ios::sync_with_stdio(false);cin.tie(nullptr);return nullptr;}(); class Solution { public: int sumSubseqWidths(vector<int>& A) { constexpr long kMod = 1e9 + 7; const int n = A.size(); countingSort(A); long ans = 0; long p = 1; for (int i = 0; i < n; ++i) { ans = (ans + (A[i] - A[n - i - 1]) * p) % kMod; p = (p << 1) % kMod; } return (ans + kMod) % kMod; } private: void countingSort(vector<int>& A) { int max_v = INT_MIN; int min_v = INT_MAX; for (int a : A) { if (a > max_v) max_v = a; if (a < min_v) min_v = a; } vector<int> c(max_v - min_v + 1); for (int a : A) ++c[a - min_v]; int i = 0; int j = 0; while (i < c.size()) { if (--c[i] >= 0) A[j++] = i + min_v; else ++i; } } }; |
请尊重作者的劳动成果,转载请注明出处!花花保留对文章/视频的所有权利。
如果您喜欢这篇文章/视频,欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Be First to Comment