LeetCode #982 — HARD

Triples with Bitwise AND Equal To Zero

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Bit Manipulation

Example 1

[2,1,3]

Example 2

[0,0,0]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #982: Triples with Bitwise AND Equal To Zero
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;
    }
}

// Accepted solution for LeetCode #982: Triples with Bitwise AND Equal To Zero
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
}

# Accepted solution for LeetCode #982: Triples with Bitwise AND Equal To Zero
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)

// Accepted solution for LeetCode #982: Triples with Bitwise AND Equal To Zero
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn count_triplets(a: Vec<i32>) -> i32 {
        let n = a.len();
        let mut hm: HashMap<i32, usize> = HashMap::new();
        for i in 0..n {
            for j in 0..n {
                *hm.entry(a[i] & a[j]).or_default() += 1;
            }
        }
        let mut res = 0;
        for i in 0..n {
            for (&k, &v) in hm.iter() {
                if a[i] & k == 0 {
                    res += v;
                }
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let a = vec![2, 1, 3];
    let res = 12;
    assert_eq!(Solution::count_triplets(a), res);
}

// Accepted solution for LeetCode #982: Triples with Bitwise AND Equal To Zero
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2 + n × M)

Space

O(M)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.