Question

982. Triples with Bitwise AND Equal To Zero

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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

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)

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;
  }
};

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;
}