Question

1310. XOR Queries of a Subarray

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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

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]

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;
  }
};

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;
};