Number of Unique XOR Triplets I

Number of Unique XOR Triplets I

What is this problem about?

The Number of Unique XOR Triplets I problem asks you to count the number of distinct XOR values achievable by any triplet (i, j, k) with i ≤ j ≤ k where the XOR of three elements gives a unique value. For a specially structured input (array is [1, 2, 3, ..., n]), there's a mathematical insight that gives the answer directly. This coding problem tests bit manipulation reasoning on XOR across ranges.

Why is this asked in interviews?

Meesho asks this to test the mathematical insight for XOR properties of the sequence 1..n. All XOR values achievable by triplets of elements from 1..n form a specific range. The array, math, and bit manipulation interview pattern is applied with formula derivation.

Algorithmic pattern used

Mathematical observation for 1..n sequences. For the array [1..n]: any value from 0 to the next power of 2 above n minus 1 can be achieved as XOR of three elements. The count of distinct achievable XOR values is the smallest power of 2 ≥ n. More precisely: if n=1, answer=1. If n is at least 2, answer = 2^(floor(log2(n))+1) because with 3 elements from [1..n], every value achievable from XOR of all such values covers up to the full bit range.

Example explanation

n=4. Array=[1,2,3,4]. Maximum XOR with three elements: 1^2^4=7, 1^3^4=6, 2^3^4=5, etc. Values achievable: 0 through 7. Distinct count = 8 = 2^3 (next power of 2 after 4 is 8, bit length = 3).

n=2. Array=[1,2]. XOR triplets (with repetition since i≤j≤k): 1^1^1=1, 1^1^2=2, 1^2^2=1, 2^2^2=2, 1^2^1=2... Distinct: {1,2} = 2 = 2^1?

Actually for n=1: answer=1. For n≥2: answer = 2^(bit_length of n).

Common mistakes candidates make

• Brute-force O(n³) enumeration of all XOR triplets.
• Not recognizing the mathematical pattern specific to consecutive integer arrays.
• Confusing bit_length(n) with floor(log2(n)).
• Forgetting that i ≤ j ≤ k allows repeated indices (XOR with itself = 0).

Interview preparation tip

XOR problems on structured arrays (1..n) often have elegant closed-form answers. The key observation: using repetition (i ≤ j ≤ k), you can achieve 0 (a^a^a) and any value up to the bit range of n. Practice analyzing XOR properties: XOR is self-inverse (a^a=0), commutative, and associative. These properties make XOR range analysis mathematical rather than computational.