Count Triplets That Can Form Two Arrays of Equal XOR

Given an array of integers arr.

We want to select three indices i, j and k where (0 <= i < j <= k < arr.length).

Let's define a and b as follows:

• a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]

• b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]

Note that ^ denotes the bitwise-xor operation.

Return the number of triplets (i, j and k) Where a == b.

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

Example 3:

Input: arr = [2,3]
Output: 0

Example 4:

Input: arr = [1,3,5,7,9]
Output: 3

Example 5:

Input: arr = [7,11,12,9,5,2,7,17,22]
Output: 8

Constraints:

• 1 <= arr.length <= 300

• 1 <= arr[i] <= 10^8

/*
Now say currently we are at index i and let xor([0...i]) = x.

Now say x has occurred 3 times previously at indices (i1, i2, i3)

our answer for i will be = (i - i1 - 1) + (i - i2 - 1) + (i - i3 - 1)

if you simplify this further you get f * i - (i1 + i2 + i3) - f = (i - 1) * f - (i1 + i2 + i3)

f = no. of times x has occurred previously.

(i1 + i2 + i3) = sum of all the indices where x has occurred previously.

*/
class Solution {
    public int countTriplets(int[] A) {
        int n = A.length, res = 0, prefix = 0;
        // Maps of XORs
        Map<Integer, Integer> freqMap = new HashMap<>(), sumOfIndices = new HashMap<>();
        freqMap.put(0, 1);
        for (int i = 0; i < n; ++i) {
            prefix ^= A[i];
            int freq = freqMap.getOrDefault(prefix, 0);
            int sum = sumOfIndices.getOrDefault(prefix, 0);
            res += freq * i - sum;
            freqMap.put(prefix, freq + 1);
            sumOfIndices.put(prefix, sum + i + 1);
        }
        return res;
    }
}