Count Triplets That Can Form Two Arrays of Equal XOR

Problem Statement

In this task, we are given an array of integers named arr. We need to determine the count of specific triplets within this array. These triplets consist of indices (i, j, k) that satisfy the condition 0 <= i < j <= k < arr.length. For each triplet, we define two values, a and b, based on the bitwise XOR operation (^). a is calculated as the XOR of all elements from index i to j-1, and b is derived from the XOR of elements from index j to k. The goal is to find the number of such triplets where a equals b.

Examples

Example 1

Input:

arr = [2,3,1,6,7]

Output:

4

Explanation:

The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)

Example 2

Input:

arr = [1,1,1,1,1]

Output:

10

Constraints

• 1 <= arr.length <= 300
• 1 <= arr[i] <= 108

Approach and Intuition

The intention is to identify the number of triplets (i, j, k) such that the XOR from i to j-1 equals the XOR from j to k. A brute-force approach could be to use three nested loops to check each combination, but that would be inefficient with larger arrays. Based on the constraints:

1. Understand that XOR is cumulative and associative, which can simplify the computation utilizing a prefix XOR array.
2. Construct a prefix XOR where the value at each index i is the XOR of all elements from the start of the array up to i.
3. A useful property of XOR is that for any segment of the array l to r, the XOR can be computed as prefixXOR[r] ^ prefixXOR[l-1].
4. Using the prefixXOR, reduce the complexity of the solution by using only a dual loop. The first loop would fix the end index k, and an inner loop would iterate for possible values of j.
5. Each possible (j, k) combination suggests a potential index i, such that the segments i to j-1 and j to k have the same XOR: effectively, requiring prefixXOR[j-1] == prefixXOR[k] given the starting index 0.
6. Use a dictionary or hashmap to efficiently count and quickly look up occurrences of each XOR value at indices, which helps in computing the required triplet count dynamically.

Following this approach helps us overcome the brute-force limitations and exploits properties of the XOR operation to gain efficiency.

Solutions

class Solution {
public:
    int countTriplets(vector<int>& numbers) {
        int length = numbers.size();
        int tripletCount = 0;
        int cumulativeXOR = 0;
    
        unordered_map<int, int> xorCountMap = {{0, 1}}, xorIndexSumMap;
    
        for (int index = 0; index < length; ++index) {
            cumulativeXOR ^= numbers[index];
            tripletCount += xorCountMap[cumulativeXOR]++ * index - xorIndexSumMap[cumulativeXOR];
            xorIndexSumMap[cumulativeXOR] += index + 1;
        }
    
        return tripletCount;
    }
};