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982. Triples with Bitwise AND Equal To Zero

Description

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

Solutions

Solution 1: Enumeration + Counting

First, we enumerate any two numbers $x$ and $y$, and use a hash table or array $cnt$ to count the occurrences of their bitwise AND result $x \& y$.

Then, we enumerate the bitwise AND result $xy$, and enumerate $z$. If $xy \& z = 0$, then we add the value of $cnt[xy]$ to the answer.

Finally, we return the answer.

The time complexity is $O(n^2 + n \times M)$, and the space complexity is $O(M)$, where $n$ is the length of the array $nums$; and $M$ is the maximum value in the array $nums$, with $M \leq 2^{16}$ in this problem.

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class Solution:
    def countTriplets(self, nums: List[int]) -> int:
        cnt = Counter(x & y for x in nums for y in nums)
        return sum(v for xy, v in cnt.items() for z in nums if xy & z == 0)

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class Solution {
    public int countTriplets(int[] nums) {
        int mx = 0;
        for (int x : nums) {
            mx = Math.max(mx, x);
        }
        int[] cnt = new int[mx + 1];
        for (int x : nums) {
            for (int y : nums) {
                cnt[x & y]++;
            }
        }
        int ans = 0;
        for (int xy = 0; xy <= mx; ++xy) {
            for (int z : nums) {
                if ((xy & z) == 0) {
                    ans += cnt[xy];
                }
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int countTriplets(vector<int>& nums) {
        int mx = *max_element(nums.begin(), nums.end());
        int cnt[mx + 1];
        memset(cnt, 0, sizeof cnt);
        for (int& x : nums) {
            for (int& y : nums) {
                cnt[x & y]++;
            }
        }
        int ans = 0;
        for (int xy = 0; xy <= mx; ++xy) {
            for (int& z : nums) {
                if ((xy & z) == 0) {
                    ans += cnt[xy];
                }
            }
        }
        return ans;
    }
};

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func countTriplets(nums []int) (ans int) {
    mx := slices.Max(nums)
    cnt := make([]int, mx+1)
    for _, x := range nums {
        for _, y := range nums {
            cnt[x&y]++
        }
    }
    for xy := 0; xy <= mx; xy++ {
        for _, z := range nums {
            if xy&z == 0 {
                ans += cnt[xy]
            }
        }
    }
    return
}

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function countTriplets(nums: number[]): number {
    const mx = Math.max(...nums);
    const cnt: number[] = Array(mx + 1).fill(0);
    for (const x of nums) {
        for (const y of nums) {
            cnt[x & y]++;
        }
    }
    let ans = 0;
    for (let xy = 0; xy <= mx; ++xy) {
        for (const z of nums) {
            if ((xy & z) === 0) {
                ans += cnt[xy];
            }
        }
    }
    return ans;
}

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