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97 lines
2.5 KiB
Markdown
97 lines
2.5 KiB
Markdown
# [1442. Count Triplets That Can Form Two Arrays of Equal XOR](https://leetcode.com/problems/count-triplets-that-can-form-two-arrays-of-equal-xor/)
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## 题目
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Given an array of integers `arr`.
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We want to select three indices `i`, `j` and `k` where `(0 <= i < j <= k < arr.length)`.
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Let's define `a` and `b` as follows:
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- `a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]`
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- `b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]`
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Note that **^** denotes the **bitwise-xor** operation.
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Return *the number of triplets* (`i`, `j` and `k`) Where `a == b`.
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**Example 1:**
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```
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Input: arr = [2,3,1,6,7]
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Output: 4
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Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)
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```
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**Example 2:**
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```
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Input: arr = [1,1,1,1,1]
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Output: 10
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```
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**Example 3:**
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```
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Input: arr = [2,3]
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Output: 0
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```
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**Example 4:**
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```
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Input: arr = [1,3,5,7,9]
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Output: 3
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```
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**Example 5:**
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```
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Input: arr = [7,11,12,9,5,2,7,17,22]
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Output: 8
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```
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**Constraints:**
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- `1 <= arr.length <= 300`
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- `1 <= arr[i] <= 10^8`
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## 题目大意
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给你一个整数数组 arr 。现需要从数组中取三个下标 i、j 和 k ，其中 (0 <= i < j <= k < arr.length) 。a 和 b 定义如下：
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- a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
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- b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]
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注意：^ 表示 按位异或 操作。请返回能够令 a == b 成立的三元组 (i, j , k) 的数目。
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## 解题思路
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- 这一题需要用到 `x^x = 0` 这个异或特性。题目要求 `a == b`，可以等效转化为 `arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1] ^ arr[j] ^ arr[j + 1] ^ ... ^ arr[k] = 0`，这样 j 相当于可以“忽略”，专注找到所有元素异或结果为 0 的区间 [i,k] 即为答案。利用前缀和的思想，只不过此题非累加和，而是异或。又由 `x^x = 0` 这个异或特性，相同部分异或相当于消除，于是有 `prefix[i,k] = prefix[0,k] ^ prefix[0,i-1]`，找到每一个 `prefix[i,k] = 0` 的 i，k 组合，i < j <= k，那么满足条件的三元组 (i,j,k) 的个数完全取决于 j 的取值范围，(因为 i 和 k 已经固定了)，j 的取值范围为 k-i，所以累加所有满足条件的 k-i，输出即为最终答案。
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## 代码
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```go
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package leetcode
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func countTriplets(arr []int) int {
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prefix, num, count, total := make([]int, len(arr)), 0, 0, 0
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for i, v := range arr {
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num ^= v
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prefix[i] = num
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}
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for i := 0; i < len(prefix)-1; i++ {
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for k := i + 1; k < len(prefix); k++ {
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total = prefix[k]
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if i > 0 {
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total ^= prefix[i-1]
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}
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if total == 0 {
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count += k - i
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}
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}
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}
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return count
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}
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``` |