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

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(javascript)

/**
 * @param {number[]} arr
 * @return {number}
 */
const countTriplets = function (arr) {
  let count = 0
  for (let i = 0; i < arr.length; i++) {
    let a = arr[i]
    for (let j = i + 1; j < arr.length; j++) {
      if (i <= j - 1) {
        a ^= arr[j]
      }
      let b = arr[j]
      for (let k = j; k < arr.length; k++) {
        b ^= arr[k]
        if (a === b) {
          count += 1
        }
      }
    }
  }
  return count
}