Solution Explanation: Counting AND Triples

This problem asks us to find the number of triplets (i, j, k) in an array nums such that the bitwise AND of nums[i], nums[j], and nums[k] is 0. A brute-force approach would have a time complexity of O(n³), which is inefficient for larger arrays. The provided solution optimizes this using a clever counting strategy.

Approach:

The solution leverages the commutative and associative properties of the bitwise AND operation. It breaks down the problem into two steps:

1. Count Pairwise ANDs: It first iterates through all possible pairs (i, j) in nums and calculates the bitwise AND of nums[i] and nums[j]. It uses a Counter (Python) or a frequency array (other languages) to store the frequency of each unique AND result. This step effectively pre-computes the results of pairwise AND operations.

2. Check for Zero AND: For each unique pairwise AND result (xy), it iterates through the array again. If the bitwise AND of xy and nums[k] is 0 for any k, it means we've found a triplet (i, j, k) where the overall AND is 0. The count of such triplets is added to the final answer, multiplying the frequency of the pair's AND result (cnt[xy]) by the number of times a suitable z was found.

Time and Space Complexity Analysis:

• Time Complexity: O(n² + nM), where n is the length of nums and M is the maximum value in nums. The first step (pairwise ANDs) takes O(n²) time. The second step iterates through nums for each unique pairwise AND result. In the worst case, the number of unique pairwise AND results is O(M), so the second step can take up to O(nM) time.

• Space Complexity: O(M). The space complexity is dominated by the cnt array (or Counter in Python) which stores the frequencies of pairwise AND results. The size of this array is proportional to the maximum value in nums (M).

Code Explanation (Python):

from collections import Counter
 
class Solution:
    def countTriplets(self, nums: List[int]) -> int:
        cnt = Counter(x & y for x in nums for y in nums) # Count pairwise ANDs
        return sum(v for xy, v in cnt.items() for z in nums if xy & z == 0) # Check for zero AND

The Python code is concise due to the use of list comprehensions and the Counter object. It elegantly handles the counting and summation steps. Other language versions achieve the same functionality using arrays and loops. The core logic remains consistent across all implementations.

Example Walkthrough (Python):

Let's consider nums = [2, 1, 3].

1. Pairwise ANDs:

   • 2 & 2 = 2 (count 1)
   • 2 & 1 = 0 (count 2) (appears twice: 2&1 and 1&2)
   • 2 & 3 = 2 (count 1)
   • 1 & 1 = 1 (count 1)
   • 1 & 3 = 1 (count 1)
   • 3 & 3 = 3 (count 1)

2. Check for Zero AND:

   • For xy = 0, the count is 2. 0 & z == 0 for all z in nums, so we add 2 * 3 = 6 to the answer.
   • For xy = 1, the count is 2. 1 & 1 == 1, 1 & 2 == 0, 1 & 3 == 1, so we add 2 * 1 = 2 to the answer.
   • For xy = 2, the count is 2. 2 & 1 == 0, 2 & 2 == 0, 2 & 3 == 2, so we add 2 * 1 = 2 to the answer.
   • For xy = 3, the count is 1. No z satisfies 3 & z == 0, so we add 0 to the answer.

The total answer becomes 6 + 2 + 2 = 12.

This detailed explanation provides a comprehensive understanding of the solution's approach, code implementation, and complexity analysis. The example walkthrough clarifies the process step-by-step.