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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 * 10^41 <= arr[i] <= 10^91 <= queries.length <= 3 * 10^4queries[i].length == 20 <= 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;
}
};
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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; };