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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

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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]
  
  

• 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
  }
  

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

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

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