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Formatted question description: https://leetcode.ca/all/1310.html

# 1310. XOR Queries of a Subarray (Medium)

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

Related Topics:
Bit Manipulation

## Solution 1.

• class Solution {
public int[] xorQueries(int[] arr, int[][] queries) {
int length = arr.length;
int[] xors = new int[length];
xors[0] = arr[0];
for (int i = 1; i < length; i++)
xors[i] = xors[i - 1] ^ arr[i];
int queriesCount = queries.length;
int[] result = new int[queriesCount];
for (int i = 0; i < queriesCount; i++) {
int[] query = queries[i];
int begin = query[0], end = query[1];
if (begin == end)
result[i] = arr[begin];
else {
if (begin == 0)
result[i] = xors[end];
else
result[i] = xors[end] ^ xors[begin - 1];
}
}
return result;
}
}

############

class Solution {
public int[] xorQueries(int[] arr, int[][] queries) {
int n = arr.length;
int[] xors = new int[n + 1];
for (int i = 0; i < n; ++i) {
xors[i + 1] = xors[i] ^ arr[i];
}
int m = queries.length;
int[] ans = new int[m];
for (int i = 0; i < m; ++i) {
int l = queries[i][0];
int r = queries[i][1];
ans[i] = xors[l] ^ xors[r + 1];
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/xor-queries-of-a-subarray/
// Time: O(N + Q)
// Space: O(1)
class Solution {
public:
vector<int> xorQueries(vector<int>& A, vector<vector<int>>& Q) {
for (int i = 1; i < A.size(); ++i) A[i] ^= A[i - 1];
vector<int> ans;
for (auto &q : Q) {
int from = q[0], to = q[1];
ans.push_back(A[to] ^ (from ? A[from - 1] : 0));
}
return ans;
}
};

• class Solution:
def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]:
xors = [0]
for v in arr:
xors.append(xors[-1] ^ v)
return [xors[l] ^ xors[r + 1] for l, r in queries]

############

# 1310. XOR Queries of a Subarray
# https://leetcode.com/problems/xor-queries-of-a-subarray/

class Solution:
def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]:
res = []
prefix = [0]
for x in arr:
prefix.append(x ^ prefix[-1])

for l,r in queries:
res.append(prefix[l] ^ prefix[r+1])

return res

• func xorQueries(arr []int, queries [][]int) []int {
xors := make([]int, len(arr)+1)
for i, v := range arr {
xors[i+1] = xors[i] ^ v
}
var ans []int
for _, q := range queries {
l, r := q[0], q[1]
ans = append(ans, xors[l]^xors[r+1])
}
return ans
}

• /**
* @param {number[]} arr
* @param {number[][]} queries
* @return {number[]}
*/
var xorQueries = function (arr, queries) {
let n = arr.length;
let xors = new Array(n + 1).fill(0);
for (let i = 0; i < n; i++) {
xors[i + 1] = xors[i] ^ arr[i];
}
let res = [];
for (let [l, r] of queries) {
res.push(xors[l] ^ xors[r + 1]);
}
return res;
};


• function xorQueries(arr: number[], queries: number[][]): number[] {
const n = arr.length;
const s: number[] = new Array(n + 1).fill(0);
for (let i = 0; i < n; ++i) {
s[i + 1] = s[i] ^ arr[i];
}
const ans: number[] = [];
for (const [l, r] of queries) {
ans.push(s[r + 1] ^ s[l]);
}
return ans;
}