982. Triples with Bitwise AND Equal To Zero

📋 Đề Bài

Given an integer array nums, return the number of AND triples.

An AND triple is a triple of indices (i, j, k) such that:

0 <= i < nums.length
0 <= j < nums.length
0 <= k < nums.length
nums[i] & nums[j] & nums[k] == 0, where & represents the bitwise-AND operator.

Example 1:

Input: nums = [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

Example 2:

Input: nums = [0,0,0]
Output: 27

Constraints:

1 <= nums.length <= 1000
0 <= nums[i] < 2¹⁶

🧠 Thuật Toán & Kỹ Thuật

Hash Table (Bảng băm)

⏱️ Thời gian O(n²)

💾 Không gian O(n)

💻 Lời Giải

Python 0982-triples-with-bitwise-and-equal-to-zero.py

class Solution:
    def countTriplets(self, nums: List[int]) -> int:
        n = len(nums)
        h = defaultdict(int)
        
        for i in range(n):
            for j in range(n):
                h[nums[i] & nums[j]] += 1
                
        ans = 0
        
        for k in range(n):
            for numIJ, cnt in h.items():
                if nums[k] & numIJ:
                    continue
                ans += cnt
                
        return ans