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1310. XOR Queries of a Subarray

中文文档

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

Python3

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]

Java

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;
    }
}

JavaScript

/**
 * @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;
};

C++

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

Go

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
}

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