982. Triples with Bitwise AND Equal To Zero

Problem

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

Solution

class Solution {
    public int countTriplets(int[] nums) {
        int[] arr = new int[1 << 17];
        for (int num : nums) {
            int mask = 0;
            for (int i = 0; i < 16; i++) {
                if ((num & (1 << i)) == 0) {
                    mask |= (1 << i);
                }
            }
            int s = mask;
            while (s > 0) {
                arr[s]++;
                s = (s - 1) & mask;
            }
        }
        int count = 0;
        for (int j : nums) {
            for (int num : nums) {
                int val = j & num;
                if (val == 0) {
                    count += nums.length;
                } else {
                    int mask = 0;
                    for (int k = 0; k < 16; k++) {
                        if ((val & (1 << k)) > 0) {
                            mask |= (1 << k);
                        }
                    }
                    count += arr[mask];
                }
            }
        }
        return count;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).