README.md

982. Triples with Bitwise AND Equal To Zero

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 (Rust)

1. Solution

use std::collections::HashMap;

impl Solution {
    pub fn count_triplets(nums: Vec<i32>) -> i32 {
        let mut count = HashMap::new();
        let mut ret = 0;

        for i in 0..nums.len() {
            for j in 0..nums.len() {
                *count.entry(nums[i] & nums[j]).or_insert(0) += 1;
            }
        }

        for k in 0..nums.len() {
            for (x, c) in count.iter() {
                if x & nums[k] == 0 {
                    ret += c;
                }
            }
        }

        ret
    }
}