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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

• Java
• C++
• Python
• Go
• TypeScript

• 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;
  }
  
  

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