Count Triplets That Can Form Two Arrays of Equal XOR

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;
    }
};

The provided C++ solution efficiently counts the number of triplets within an array numbers, such that two subarrays formed by those triplets have the same XOR. This solution utilizes cumulative XOR calculations to identify equal XOR subarrays efficiently without having to recalculate XOR for every possible subarray.

• Initialize length as the size of the numbers vector.
• Use tripletCount to track the count of valid triplets.
• cumulativeXOR represents the XOR from the start of the vector up to the current processing index.
• Two hash maps, xorCountMap and xorIndexSumMap, are used:
  • xorCountMap stores how many times a particular XOR result has appeared.
  • xorIndexSumMap keeps track of the summation of indices where each XOR result has appeared.
• Loop through each element of the numbers vector:
  • Compute cumulative XOR up to the current index.
  • If the cumulative XOR has been seen before, calculate potential triplets by subtracting the pre-recorded index sum from index * count of previous occurrences.
  • Update the maps to reflect the current state.
• Finally, return tripletCount which now contains the number of valid triplet indices.

This approach avoids the naive method of checking each possible triplet explicitly, thus providing a more optimized solution using hashmap to exploit properties of XOR in constant time operations.

class Solution {

    public int findTriplets(int[] data) {
        int length = data.length;
        int tripletCount = 0;
        int xorSum = 0;

        // Hashmaps for tracking prefix XOR values and their indices
        Map<Integer, Integer> frequencyMap = new HashMap<>();
        frequencyMap.put(0, 1);
        Map<Integer, Integer> sumMap = new HashMap<>();

        // Processing each element in the array
        for (int index = 0; index < length; ++index) {
            // Current prefix XOR calculation
            xorSum ^= data[index];

            // Computing the current potential triplets count contribution
            tripletCount +=
            frequencyMap.getOrDefault(xorSum, 0) * index -
            sumMap.getOrDefault(xorSum, 0);

            // Updating maps for encountered XOR prefix
            sumMap.put(xorSum, sumMap.getOrDefault(xorSum, 0) + index + 1);
            frequencyMap.put(xorSum, frequencyMap.getOrDefault(xorSum, 0) + 1);
        }

        return tripletCount;
    }
}

The given problem titled "Count Triplets That Can Form Two Arrays of Equal XOR" is implemented in Java. The presented solution defines a method findTriplets that counts such triplets in an array data where the XOR of elements between two positions i and j is equal to the XOR of elements between positions j+1 and k.

• Initialize an integer variable tripletCount to count the number of valid triplets.
• Maintain a prefix XOR sum xorSum to keep track of XOR up to the current index in the array.
• Use two hashmaps:
  • frequencyMap to store the frequency of occurrence of each XOR sum.
  • sumMap to store the sum of indices where each XOR sum is encountered.

The core idea is to iterate through the array and for each element:

• Update the xorSum with the XOR of the current element.
• Calculate the potential contributions to tripletCount from the current xorSum by checking how many times xorSum has appeared before the current index and where it appeared.
• Store the frequency of the current xorSum in frequencyMap and the sum of indices in sumMap.

The method finally returns tripletCount indicating the number of triplets fulfilling the condition. The solution employs efficient use of hashmaps to store pre-computed values, reducing the complexity compared to a naive approach that checks all possible triplets explicitly.

class Solution:
    def countTriplets(self, array: List[int]) -> int:
        n = len(array)
        triplet_count = 0
        xor_sum = 0

        # Using dictionaries to store counts and totals based on XOR results
        count_tracker = defaultdict(int)
        count_tracker[0] = 1
        sum_tracker = defaultdict(int)

        # Looping over the array elements
        for index in range(n):
            # XOR accumulation
            xor_sum ^= array[index]

            # Count triplets contribution from this position
            triplet_count += count_tracker[xor_sum] * index - sum_tracker[xor_sum]

            # Update maps with the current XOR and indices
            sum_tracker[xor_sum] += index + 1
            count_tracker[xor_sum] += 1

        return triplet_count

This Python 3 solution tackles the problem of counting triplets in an array such that they can equate to two arrays of equal XOR. The approach leverages dictionary structures to efficiently compute the desired counts without explicitly considering all possible triplet combinations. Here’s a clear breakdown of how the solution operates:

• Initialize Variables: xor_sum is used to hold the cumulative XOR of array elements as the array is processed element by element. A triplet_count starts at zero and increments based on certain conditions discussed below.

• Use Two Dictionaries:

  • count_tracker keeps a running tally of how many times each XOR result has appeared.
  • sum_tracker captures the sum of indices where each XOR result appeared.

• Iterate Through the Array: For each element in the array, update the xor_sum.

• Count Contributions: For the current XOR result, calculate contributions to the triplet count. This is achieved by the formula where contributions are added based on previous appearances (index times the count from count_tracker minus the sum stored in sum_tracker).

• Update Mappings: Increase counters in both dictionaries for tracking future iterations.

• Return the Result: The accumulated triplet_count provides the number of valid triplet indices that meet the prompt’s condition.

Efficiency Considerations: This solution leverages the linear complexity of array traversal alongside dictionary operations, which theoretically are average-case constant time, leading to significant efficiency improvements over brute-force solutions.

This code effectively eliminates the need for nested loops through clever use of cumulative property storage and leveraging data structures like dictionaries, which are inherently more efficient for these types of look-up operations. Thus, you achieve both clarity and performance.