How do you solve Triples with Bitwise AND Equal To Zero on LeetCode?

Triples with Bitwise AND Equal To Zero (LeetCode #982) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: The bitwise AND operation results in zero when at least one bit position is zero in all three numbers.

What is the brute force solution for Triples with Bitwise AND Equal To Zero?

The brute force approach for Triples with Bitwise AND Equal To Zero is Brute Force, with O(n³) time complexity and O(1) space complexity. The brute force approach involves checking every possible combination of three indices in the array. For each combination, we compute the bitwise AND and check if it equals zero. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Triples with Bitwise AND Equal To Zero?

Triples with Bitwise AND Equal To Zero uses the following concepts: Array, Hash Table, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Triples with Bitwise AND Equal To Zero?

Always start with a brute force solution to understand the problem better. Look for patterns in the data that can help optimize your solution. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Triples with Bitwise AND Equal To Zero?

Not considering the case where all three indices are the same. Forgetting to check combinations of different numbers correctly.

#982

Triples with Bitwise AND Equal To Zero

Hard

ArrayHash TableBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n³)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

The optimal solution leverages the properties of bitwise operations and counts how many numbers can form valid triples without checking every combination. We can use a hashmap to store the frequency of each number and calculate combinations based on their counts.

⚙️

Algorithm

4 steps

1Step 1: Create a hashmap to count the frequency of each number in the array.
2Step 2: For each unique pair of numbers, calculate their bitwise AND and check if it equals zero.
3Step 3: If it does, calculate the number of valid triples using the counts from the hashmap.
4Step 4: Return the total count of valid triples.

solution.py14 lines

1from collections import Counter
2
3def countTriplets(nums):
4    count = 0
5    freq = Counter(nums)
6    unique_nums = list(freq.keys())
7    for i in range(len(unique_nums)):
8        for j in range(i, len(unique_nums)):
9            if unique_nums[i] & unique_nums[j] == 0:
10                if i == j:
11                    count += freq[unique_nums[i]] ** 3
12                else:
13                    count += 3 * freq[unique_nums[i]] ** 2 * freq[unique_nums[j]]
14    return count

ℹ

Complexity note: The complexity is O(n²) due to the nested loops over unique numbers, but we only consider unique pairs, making it more efficient than the brute force approach.

1The bitwise AND operation results in zero when at least one bit position is zero in all three numbers.
2Using frequency counts can significantly reduce the number of checks needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.