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

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

 

Constraints:

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

 

Solutions

Solution: Prefix Sum

• Time complexity: O(n^2)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {number[]} arr
 * @return {number}
 */
const countTriplets = function (arr) {
  let result = 0;

  for (let i = 0; i < arr.length - 1; i++) {
    let current = arr[i];

    for (let k = i + 1; k < arr.length; k++) {
      current ^= arr[k];
      if (current !== 0) continue;
      result += k - i;
    }
  }
  return result;
};