1310. X O R Queries of a Subarray

# 1310. XOR Queries of a Subarray#

## 题目 #

Given the array arr of positive integers and the array queries where queries[i] = [Li,Ri], for each query i compute the XOR of elements from Li to Ri (that is, arr[Li]xor arr[Li+1]xor ...xor arr[Ri]). Return an array containing the result for the given queries.

Example 1:

Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
Output: [2,7,14,8]
Explanation:
The binary representation of the elements in the array are:
1 = 0001
3 = 0011
4 = 0100
8 = 1000
The XOR values for queries are:
[0,1] = 1 xor 3 = 2
[1,2] = 3 xor 4 = 7
[0,3] = 1 xor 3 xor 4 xor 8 = 14
[3,3] = 8



Example 2:

Input: arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
Output: [8,0,4,4]



Constraints:

• 1 <= arr.length <= 3 * 10^4
• 1 <= arr[i] <= 10^9
• 1 <= queries.length <= 3 * 10^4
• queries[i].length == 2
• 0 <= queries[i][0] <= queries[i][1] < arr.length

## 解题思路 #

• 此题求区间异或，很容易让人联想到区间求和。区间求和利用前缀和，可以使得 query 从 O(n) 降为 O(1)。区间异或能否也用类似前缀和的思想呢？答案是肯定的。利用异或的两个性质，x ^ x = 0，x ^ 0 = x。那么有：（由于 LaTeX 中异或符号 ^ 是特殊字符，笔者用 $$\oplus$$ 代替异或）

\begin{aligned}Query(left,right) &=arr[left] \oplus \cdots \oplus arr[right]\\&=(arr[0] \oplus \cdots \oplus arr[left-1]) \oplus (arr[0] \oplus \cdots \oplus arr[left-1]) \oplus (arr[left] \oplus \cdots \oplus arr[right])\\ &=(arr[0] \oplus \cdots \oplus arr[left-1]) \oplus (arr[0] \oplus \cdots \oplus arr[right])\\ &=xors[left] \oplus xors[right+1]\\ \end{aligned}

按照这个思路解题，便可以将 query 从 O(n) 降为 O(1)，总的时间复杂度为 O(n)。

## 代码 #

package leetcode

func xorQueries(arr []int, queries [][]int) []int {
xors := make([]int, len(arr))
xors[0] = arr[0]
for i := 1; i < len(arr); i++ {
xors[i] = arr[i] ^ xors[i-1]
}
res := make([]int, len(queries))
for i, q := range queries {
res[i] = xors[q[1]]
if q[0] > 0 {
res[i] ^= xors[q[0]-1]
}
}
return res
}


Apr 8, 2023