Leetcode: 982. Triples with Bitwise AND Equal To Zero

Description

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

 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

分析

之前想过用 bit 位分类，但是总有遗漏的情况无法计算～

class Solution(object):
    def _do(self, N):         
        m = len('{0:b}'.format(max(N)))
        ssum, total = 0, len(N)**3
        dp = []
        
        def rr(n, h, i):
            if i > 16 or len(n) == 0 or len(h) == 0:
                return 0
            v = 1 << (i-1)
            l1 = sum([1 for j in n if j & v != 0])
            l2 = sum([1 for j in h if j & v != 0])
            if l1*l2 == 0:
                return rr(n, h, i+1)
            return (l1+l2)**3 -l1**3 -l2**3 + rr(n, [j for j in h if j & v == 0] , i+1) + rr([j for j in n if j &v ==0], h , i+1)
                
        for i in range(1, m+1):
            v  = 1 << (i-1)
            has, noth = [], []
            for j in N:
                if j & v != 0: 
                    has.append(j)
                else:
                    noth.append(j)
            t = rr(has, [j for j in dp], i)
            ssum += len(has)**3 + t
            print(dp, has, t)
            N = noth
            dp += has
为了节省时间，结果写的贼复杂。还出错了～

code

class Solution(object):
    def countTriplets(self, N):
        """
        :type A: List[int]
        :rtype: int
        """
        L = len(N)
        m = max(N)
        dp = [0 for _ in range(m+1)]
        for i in range(L):
            for j in range(L):
                dp[N[i]&N[j]] += 1
        ssum = 0
        for j in range(len(dp)):
            if dp[j] == 0:
                continue
            for i in range(L):
                if N[i] & j == 0:
                    ssum += dp[j]
        return ssum

总结

Runtime: 3976 ms, faster than 61.54% of Python online submissions for Triples with Bitwise AND Equal To Zero.
Memory Usage: 15.5 MB, less than 100.00% of Python online submissions for Triples with Bitwise AND Equal To Zero.

• 之前一直在寻找有效的优化方法，但是自己思考的方法总有遗漏～
• 本题可以推广到 n 个数的 &， + 等等问题，只要先计算 2 个数的问题，依次循环。这样就等于是 log2 的降维，和快速幂有点类似

原文地址：https://www.cnblogs.com/tmortred/p/13275455.html