LeetCode in Kotlin

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

Medium

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

Solution

class Solution {
    fun countTriplets(arr: IntArray): Int {
        var count = 0
        for (i in arr.indices) {
            var xor = arr[i]
            for (k in i + 1 until arr.size) {
                xor = xor xor arr[k]
                if (xor == 0) {
                    count += k - i
                }
            }
        }
        return count
    }
}

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