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

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題目敘述

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

解題思路

Solution

class Solution {
    public int countTriplets(int[] arr) {
        int[] prefix = new int[arr.length + 1];

        for (int i = 0; i < arr.length; i++) {
            prefix[i + 1] = prefix[i] ^ arr[i];
        }

        int count = 0;

        for (int i = 0; i < arr.length; i++) {
            for (int k = i + 1; k < arr.length; k++) {
                if (prefix[i] == prefix[k + 1]) {
                    count += k - i;
                }
            }
        }

        return count;
    }
}

class Solution {
    public int countTriplets(int[] arr) {
        int count = 0;

        for (int i = 0; i < arr.length; i++) {
            int curr_XOR = 0;
            for (int k = i; k < arr.length; k++) {
                curr_XOR ^= arr[k];
                if(curr_XOR == 0) count += k - i;
            }
        }

        return count;
    }
}