LeetCode 0982. Triples with Bitwise AND Equal To Zero Solution in Java, C++, Python & More | Explanation + Code

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(n² + n × 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 ≤ 2¹⁶ in this problem.

PythonJavaC++GoTypeScript

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)(code-box)

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; } }(code-box)

class Solution { public: int countTriplets(vector<int>& nums) { int mx = ranges::max(nums); 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; } };(code-box)

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 }(code-box)

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; }(code-box)