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1310. XOR Queries of a Subarray

Description

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

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

Return an array answer where answer[i] is the answer to the ith query.

 

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, queries.length <= 3 * 104
  • 1 <= arr[i] <= 109
  • queries[i].length == 2
  • 0 <= lefti <= righti < arr.length

Solutions

Solution 1: Prefix XOR

We can use a prefix XOR array \(s\) of length \(n+1\) to store the prefix XOR results of the array \(\textit{arr}\), where \(s[i] = s[i-1] \oplus \textit{arr}[i-1]\). That is, \(s[i]\) represents the XOR result of the first \(i\) elements of \(\textit{arr}\).

For a query \([l, r]\), we can obtain:

\[ \begin{aligned} \textit{arr}[l] \oplus \textit{arr}[l+1] \oplus \cdots \oplus \textit{arr}[r] &= (\textit{arr}[0] \oplus \textit{arr}[1] \oplus \cdots \oplus \textit{arr}[l-1]) \oplus (\textit{arr}[0] \oplus \textit{arr}[1] \oplus \cdots \oplus \textit{arr}[r]) \\ &= s[l] \oplus s[r+1] \end{aligned} \]

Time complexity is \(O(n+m)\), and space complexity is \(O(n)\). Here, \(n\) and \(m\) are the lengths of the array \(\textit{arr}\) and the query array \(\textit{queries}\), respectively.

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class Solution:
    def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]:
        s = list(accumulate(arr, xor, initial=0))
        return [s[r + 1] ^ s[l] for l, r in queries]

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

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class Solution {
public:
    vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& queries) {
        int n = arr.size();
        int s[n + 1];
        memset(s, 0, sizeof(s));
        for (int i = 1; i <= n; ++i) {
            s[i] = s[i - 1] ^ arr[i - 1];
        }
        vector<int> ans;
        for (auto& q : queries) {
            int l = q[0], r = q[1];
            ans.push_back(s[r + 1] ^ s[l]);
        }
        return ans;
    }
};

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

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

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/**
 * @param {number[]} arr
 * @param {number[][]} queries
 * @return {number[]}
 */
var xorQueries = function (arr, queries) {
    const n = arr.length;
    const s = Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] ^ arr[i];
    }
    return queries.map(([l, r]) => s[r + 1] ^ s[l]);
};

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