xor-queries-of-a-subarray

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Algorithms

You are given an array <code>arr</code> of positive integers. You are also given the array <code>queries</code> where <code>queries[i] = [lefti, righti]</code>.

For each query <code>i</code> compute the XOR of elements from <code>lefti</code> to <code>righti</code> (that is, <code>arr[lefti] XOR arr[lefti + 1] XOR ... XOR arr[righti]</code> ).

Return an array <code>answer</code> where <code>answer[i]</code> is the answer to the <code>ith</code> query.

&nbsp;
Example 1:

<pre>
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
</pre>

Example 2:

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

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= arr.length, queries.length &lt;= 3 * 104</code></li>
	<li><code>1 &lt;= arr[i] &lt;= 109</code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 &lt;= lefti &lt;= righti &lt; arr.length</code></li>
</ul>

xorQueries

XOR Queries of a Subarray

Bit Manipulation

Array

Prefix Sum

有一个正整数数组 <code>arr</code>，现给你一个对应的查询数组 <code>queries</code>，其中 <code>queries[i] = [Li, Ri]</code>。

对于每个查询 <code>i</code>，请你计算从 <code>Li</code> 到 <code>Ri</code> 的 XOR 值（即 <code>arr[Li] xor arr[Li+1] xor ... xor arr[Ri]</code>）作为本次查询的结果。

并返回一个包含给定查询 <code>queries</code> 所有结果的数组。

 

示例 1：

<pre>
输入：arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
输出：[2,7,14,8] 
解释：
数组中元素的二进制表示形式是：
1 = 0001 
3 = 0011 
4 = 0100 
8 = 1000 
查询的 XOR 值为：
[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
</pre>

示例 2：

<pre>
输入：arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
输出：[8,0,4,4]
</pre>

 

提示：

<ul>
	<li><code>1 <= arr.length <= 3 * 10^4</code></li>
	<li><code>1 <= arr[i] <= 10^9</code></li>
	<li><code>1 <= queries.length <= 3 * 10^4</code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 <= queries[i][0] <= queries[i][1] < arr.length</code></li>
</ul>