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

Number of Unique XOR Triplets I (LeetCode #3513) 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 of identical numbers is 0.

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

The optimal solution for Number of Unique XOR Triplets I runs in O(n) time and uses O(n) space. We only iterate through the range [1, n] and leverage the properties of XOR, reducing complexity.

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

The brute force approach for Number of Unique XOR Triplets I is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible triplets and calculate their XOR values. Store unique results in a set. The optimal Optimal Solution approach improves this to O(n).

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

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

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

Understand XOR properties thoroughly. Practice generating combinations efficiently. Focus on optimizing nested loops.

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

Not considering all triplet combinations. Overlooking duplicates in XOR results.

#3513

Number of Unique XOR Triplets I

Medium

ArrayMathBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Utilize the properties of XOR and the range of numbers to derive all possible XOR values without generating triplets.

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Algorithm

3 steps

1Step 1: Initialize a set to store unique XOR values.
2Step 2: Iterate through all possible combinations of three numbers in the range [1, n].
3Step 3: Calculate the XOR for each combination and add it to the set.

solution.py8 lines

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

ℹ

Complexity note: We only iterate through the range [1, n] and leverage the properties of XOR, reducing complexity.

1XOR of identical numbers is 0.
2XOR is commutative and associative.

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