982. Triples with Bitwise AND Equal To Zero

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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. 1 <= A.length <= 1000
2. 0 <= A[i] < 2^16

思路：直接暴力遍历一遍两两求&的count，然后再和原始数组遍历一下

讨论去有用DP，但是复杂度跟上面暴力差不多https://leetcode.com/problems/triples-with-bitwise-and-equal-to-zero/discuss/226721/Java-DP-O(3-*-216-*-n)-time-O(n)-space

也有O(n log n)的解法：https://leetcode.com/problems/triples-with-bitwise-and-equal-to-zero/discuss/226832/Fastest-Solution-Using-Fast-Walsh-Hadamard-Transform-32ms

class Solution(object):
    def countTriplets(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        cnt = {}
        for i in A:
            for j in A:
                a=i&j
                cnt[a] = cnt.get(a, 0)+1
        
        res=0
        for i in A:
            for j in cnt:
                if i&j==0: res+=cnt[j]
        
        return res