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.
// 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 prev = q[0] - 1;
prev = prev == -1 ? 0 : A[prev];
ans.push_back(A[q[1]] ^ prev);
}
return ans;
}
};
Java
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;
}
}