How do you solve Number of Unique XOR Triplets II on LeetCode?

Number of Unique XOR Triplets II (LeetCode #3514) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: XOR is commutative and associative, allowing flexible grouping.

What is the time complexity of Number of Unique XOR Triplets II?

The optimal solution for Number of Unique XOR Triplets II runs in O(n) time and uses O(n) space. The optimal solution iterates through the array with a single loop, leading to O(n) complexity.

What is the brute force solution for Number of Unique XOR Triplets II?

The brute force approach for Number of Unique XOR Triplets II is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible triplets and compute their XOR values. This method ensures all combinations are considered, but is inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Unique XOR Triplets II?

Number of Unique XOR Triplets II uses the following concepts: Array, Math, Bit Manipulation, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Unique XOR Triplets II?

Clarify constraints and edge cases. Discuss time complexity and trade-offs. Practice similar problems to build intuition.

What are common mistakes when solving Number of Unique XOR Triplets II?

Not considering the order of indices (i <= j <= k). Overlooking the properties of XOR leading to duplicates.

#3514

Number of Unique XOR Triplets II

Medium

ArrayMathBit ManipulationEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Utilize a set to track unique XOR values while iterating through the array, reducing the need for nested loops.

⚙️

Algorithm

3 steps

1Step 1: Initialize a set to store unique XOR values.
2Step 2: Iterate through the array and compute the XOR for each element with itself and previous elements.
3Step 3: Add each computed XOR to the set.

solution.py7 lines

1def uniqueXORTriplets(nums):
2    unique_xors = set()
3    n = len(nums)
4    for i in range(n):
5        for j in range(i + 1):
6            unique_xors.add(nums[i] ^ nums[j])
7    return len(unique_xors)

ℹ

Complexity note: The optimal solution iterates through the array with a single loop, leading to O(n) complexity.

1XOR is commutative and associative, allowing flexible grouping.
2Tracking unique values with a set prevents duplicates.

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