Formatted question description: https://leetcode.ca/all/1442.html

1442. Count Triplets That Can Form Two Arrays of Equal XOR

Level

Medium

Description

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

Solution

Create an array leftXOR to store the prefix XOR results. Use the property that the XOR result of arr[i..j] equals leftXOR[j] if i == 0 or leftXOR[i - 1] ^ leftXOR[j] if i > 0. Use three nested loops to find the triplets and return the number of such triplets.

class Solution {
    public int countTriplets(int[] arr) {
        int length = arr.length;
        int[] leftXOR = new int[length];
        leftXOR[0] = arr[0];
        for (int i = 1; i < length; i++)
            leftXOR[i] = leftXOR[i - 1] ^ arr[i];
        int count = 0;
        for (int i = 0; i < length; i++) {
            for (int j = i + 1; j < length; j++) {
                int xor1 = i == 0 ? leftXOR[j - 1] : leftXOR[j - 1] ^ leftXOR[i - 1];
                for (int k = j; k < length; k++) {
                    int xor2 = leftXOR[k] ^ leftXOR[j - 1];
                    if (xor1 == xor2)
                        count++;
                }
            }
        }
        return count;
    }
}

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