Count Distinct XOR Values from Index Triples
Company: Google
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Other
### Problem
You are given an array `nums` of length `n` that is a permutation of the integers `1` through `n`.
For every index triple satisfying `0 <= i <= j <= k < n`, consider the value:
```text
nums[i] XOR nums[j] XOR nums[k]
```
Return the number of distinct values obtainable from all such triples. Equal indices are allowed, so the same array element may be used more than once in a triple. Count resulting XOR values, rather than the number of triples producing them.
### C++ Interface
```cpp
class Solution {
public:
int uniqueXorTriplets(vector<int>& nums) {
}
};
```
### Examples
**Example 1**
```text
Input: nums = [1]
Output: 1
```
The only index triple is `(0,0,0)`, whose value is `1 XOR 1 XOR 1 = 1`.
**Example 2**
```text
Input: nums = [4,2,1,3]
Output: 8
```
The distinct values are `0,1,2,3,4,5,6,7`. Each input value is obtained by using its own index three times. Additional witnesses include `1 XOR 2 XOR 3 = 0`, `4 XOR 2 XOR 3 = 5`, `4 XOR 1 XOR 3 = 6`, and `4 XOR 1 XOR 2 = 7`. Each witness can use indices in nondecreasing order. No larger value is possible for these inputs.
### Constraints
- `1 <= n == nums.length <= 100000`.
- `1 <= nums[i] <= n`.
- `nums` contains every integer from `1` through `n` exactly once.
### Clarifying Questions
- Must the three indices be different? No. The inequalities are non-strict.
- Does the array have to be sorted? No. Any permutation is valid.
- Is zero an eligible XOR result even though it is absent from the input? Yes, if a valid triple produces it.
- Should duplicate XOR results be counted again for different triples? No. Each resulting value counts once.
### What a Strong Answer Covers
- The permutation promise and the permitted repeated indices.
- The distinction between triples and their distinct XOR results.
- Completeness of the claimed result count for all valid triples.
- Small arrays and resource and numeric bounds up to `100000` elements.
### Follow-up Questions
- What would change in the definition if the three indices had to be strictly increasing?
- How would the requested result differ if the task counted index triples instead of distinct XOR values?
```hint Use the full value set
Every value from 1 through n is available, and XOR is unchanged when the three operands are reordered. Bound the possible result bits, then prove which values inside that bound can actually be formed; treat n = 1 and n = 2 separately.
```
Overview: Count distinct XOR results from nondecreasing index triples of a permutation. Allow repeated indices and preserve the distinction between counting triples and counting their resulting values.