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# XOR Queries of a Subarray

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`

```java
class Solution {
    public int[] xorQueries(int[] arr, int[][] queries) {
        int[] prefixXOR = new int[arr.length];
        prefixXOR[0] = arr[0];
        for (int i = 1; i < arr.length; i++)
            prefixXOR[i] = prefixXOR[i - 1] ^ arr[i];
        int[] ans = new int[queries.length];
        int index = 0;
        for (int[] query : queries)
            ans[index++] = prefixXOR[query[1]] ^ (query[0] > 0 ? prefixXOR[query[0] - 1] : 0);
        return ans;
    }
}
```