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

• 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;
}
}

• 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;
}
};

• 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)


• 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
}

• 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;
}