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

Posted 2020-12-15 CNoodle

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

形成两个异或相等数组的三元组数目。题意是请你返回一个三元组（i, j, k），这三个数字都是数组里面的index，请你返回三元组使得a == b并且a, b满足

• a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
• b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]

思路是位运算。既然a和b都是位运算的结果，而且a == b所以得出a ^ b = 0的结论，因为两数相同异或为0，这个结论是可以被反推的。所以这个题是在找是否能满足a ^ b = 0的三元组。同时因为a和b是由很多数字互相异或XOR组成的，XOR操作又是有结合律的，比如(a ^ b) ^ c = a ^ (b ^ c)。所以一旦发现有一段下标从a到c的数字XOR的结果为0，中间b的位置其实可以随便放，那么就有c - a种可能了。

时间O(n^2)

空间O(1)

Java实现

 1 class Solution {
 2     public int countTriplets(int[] arr) {
 3         int len = arr.length;
 4         if (len < 2) {
 5             return 0;
 6         }
 7         int res = 0;
 8         for (int i = 0; i < len; i++) {
 9             int temp = arr[i];
10             for (int j = i + 1; j < len; j++) {
11                 temp = temp ^ arr[j];
12                 if (temp == 0) {
13                     res += j - i;
14                 }
15             }
16         }
17         return res;
18     }
19 }

 

LeetCode 题目总结