Given an array of integers A
, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0
, where&
represents the bitwise-AND operator.
Example 1:
Input: [2,1,3] Output: 12 Explanation: We could choose the following i, j, k triples: (i=0, j=0, k=1) : 2 & 2 & 1 (i=0, j=1, k=0) : 2 & 1 & 2 (i=0, j=1, k=1) : 2 & 1 & 1 (i=0, j=1, k=2) : 2 & 1 & 3 (i=0, j=2, k=1) : 2 & 3 & 1 (i=1, j=0, k=0) : 1 & 2 & 2 (i=1, j=0, k=1) : 1 & 2 & 1 (i=1, j=0, k=2) : 1 & 2 & 3 (i=1, j=1, k=0) : 1 & 1 & 2 (i=1, j=2, k=0) : 1 & 3 & 2 (i=2, j=0, k=1) : 3 & 2 & 1 (i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16
Solution: Counting
Time complexity: O(n^2 + n * max(A))
Space complexity: O(max(A))
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
// Author: Huahua, running time: 60 ms class Solution { public: int countTriplets(vector<int>& A) { const int n = *max_element(begin(A), end(A)); vector<int> count(n + 1); for (int a: A) for (int b: A) ++count[a & b]; int ans = 0; for (int a : A) for (int i = 0; i <= n; ++i) if ((a & i) == 0) ans += count[i]; return ans; } }; |
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