Watch 4 video solutions for Number of Unique XOR Triplets I, a medium level problem involving Array, Math, Bit Manipulation. This walkthrough by CodeWithMeGuys has 978 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given an integer array nums of length n, where nums is a permutation of the numbers in the range [1, n].
A XOR triplet is defined as the XOR of three elements nums[i] XOR nums[j] XOR nums[k] where i <= j <= k.
Return the number of unique XOR triplet values from all possible triplets (i, j, k).
Example 1:
Input: nums = [1,2]
Output: 2
Explanation:
The possible XOR triplet values are:
(0, 0, 0) → 1 XOR 1 XOR 1 = 1(0, 0, 1) → 1 XOR 1 XOR 2 = 2(0, 1, 1) → 1 XOR 2 XOR 2 = 1(1, 1, 1) → 2 XOR 2 XOR 2 = 2The unique XOR values are {1, 2}, so the output is 2.
Example 2:
Input: nums = [3,1,2]
Output: 4
Explanation:
The possible XOR triplet values include:
(0, 0, 0) → 3 XOR 3 XOR 3 = 3(0, 0, 1) → 3 XOR 3 XOR 1 = 1(0, 0, 2) → 3 XOR 3 XOR 2 = 2(0, 1, 2) → 3 XOR 1 XOR 2 = 0The unique XOR values are {0, 1, 2, 3}, so the output is 4.
Constraints:
1 <= n == nums.length <= 1051 <= nums[i] <= nnums is a permutation of integers from 1 to n.Problem Overview: You receive an integer array and must compute how many distinct values can be produced by XORing three different elements nums[i] ^ nums[j] ^ nums[k] with i < j < k. The challenge is not generating the triplets, but ensuring the XOR results are counted uniquely.
Approach 1: Brute Force Triplet Enumeration (O(n^3) time, O(U) space)
Iterate through every possible triplet using three nested loops. For each combination (i, j, k), compute nums[i] ^ nums[j] ^ nums[k] and store the result in a hash set. The set automatically removes duplicates, leaving only unique XOR values. This approach is easy to reason about and works for very small arrays, but the cubic iteration quickly becomes impractical as n grows.
Approach 2: Pair XOR + Third Element (O(n^2) time, O(U) space)
XOR has a useful associative property: a ^ b ^ c = (a ^ b) ^ c. Instead of building triplets directly, first compute XOR values for pairs. While scanning the array, maintain a collection of pair XORs from earlier indices. For the current element nums[k], combine it with each stored pair XOR and insert the result into a result set. Because each pair represents (i, j) with i < j < k, the index constraint holds automatically. This reduces the search space from three nested loops to two.
If the integer range is limited (common in XOR problems), the result space is also bounded. You can replace hash sets with a boolean array indexed by XOR value, reducing constant factors. The algorithm still iterates over roughly n^2 pair combinations but avoids redundant triplet enumeration.
The technique relies on properties from bit manipulation, especially XOR associativity. The iteration logic is straightforward using standard array traversal, while the uniqueness constraint is handled through set membership or a fixed boolean lookup. Some implementations also reason about the limited XOR value range using simple math observations.
Recommended for interviews: The pair-XOR approach. Start by mentioning the brute force to demonstrate baseline understanding. Then reduce the complexity by grouping two numbers first and combining them with the third using XOR associativity. Interviewers expect you to recognize that (a ^ b) ^ c allows collapsing a triple loop into pair preprocessing, bringing the runtime down to O(n^2).
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Triplets | O(n^3) | O(U) | Useful for understanding the problem or when the array size is very small |
| Pair XOR + Third Element | O(n^2) | O(U) | General optimal solution using XOR associativity and pair preprocessing |
| Pair XOR with Boolean Lookup | O(n^2) | O(K) | When XOR values are bounded (e.g., ≤1024), allowing faster constant-time checks |