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: Prefix Sum

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} arr
 * @param {number[][]} queries
 * @return {number[]}
 */
const xorQueries = function (arr, queries) {
  const n = arr.length;
  const prefixXor = new Array(n).fill(0);

  prefixXor[-1] = 0;

  for (let index = 0; index < n; index++) {
    prefixXor[index] = arr[index] ^ prefixXor[index - 1];
  }

  return queries.map(([left, right]) => {
    return prefixXor[right] ^ prefixXor[left - 1];
  });
};