1442

1442. Count Triplets That Can Form Two Arrays of Equal XOR (Medium)

Given an array of integers arr.

We want to select three indices i, j and k where (0 <= i < j <= k < arr.length).

Let's define a and b as follows:

• a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
• b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]

Note that ^ denotes the bitwise-xor operation.

Return the number of triplets (i, j and k) Where a == b.

 

Example 1:

Input: arr = [2,3,1,6,7]
Output: 4
Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)

Example 2:

Input: arr = [1,1,1,1,1]
Output: 10

Example 3:

Input: arr = [2,3]
Output: 0

Example 4:

Input: arr = [1,3,5,7,9]
Output: 3

Example 5:

Input: arr = [7,11,12,9,5,2,7,17,22]
Output: 8

 

Constraints:

• 1 <= arr.length <= 300
• 1 <= arr[i] <= 10^8

Related Topics:
Array, Math, Bit Manipulation

Solution 1.

// OJ: https://leetcode.com/problems/count-triplets-that-can-form-two-arrays-of-equal-xor/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int countTriplets(vector<int>& A) {
        unordered_map<int, vector<int>> m;
        m[0] = {-1};
        int x = 0, ans = 0;
        for (int k = 0; k < A.size(); ++k) {
            x ^= A[k];
            if (m.count(x)) {
                for (int i : m[x]) ans += k - i - 1;
            }
            m[x].push_back(k);
        }
        return ans;
    }
};