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

Description

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

 

Constraints:

  • 1 <= arr.length <= 300
  • 1 <= arr[i] <= 108

Solutions

  • class Solution {
    public int countTriplets(int[] arr) {
        int n = arr.length;
        int[] pre = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            pre[i + 1] = pre[i] ^ arr[i];
        }
        int ans = 0;
        for (int i = 0; i < n - 1; ++i) {
            for (int j = i + 1; j < n; ++j) {
                for (int k = j; k < n; ++k) {
                    int a = pre[j] ^ pre[i];
                    int b = pre[k + 1] ^ pre[j];
                    if (a == b) {
                        ++ans;
                    }
                }
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int countTriplets(vector<int>& arr) {
        int n = arr.size();
        vector<int> pre(n + 1);
        for (int i = 0; i < n; ++i) pre[i + 1] = pre[i] ^ arr[i];
        int ans = 0;
        for (int i = 0; i < n - 1; ++i) {
            for (int j = i + 1; j < n; ++j) {
                for (int k = j; k < n; ++k) {
                    int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
                    if (a == b) ++ans;
                }
            }
        }
        return ans;
    }
};

  • class Solution:
    def countTriplets(self, arr: List[int]) -> int:
        n = len(arr)
        pre = [0] * (n + 1)
        for i in range(n):
            pre[i + 1] = pre[i] ^ arr[i]
        ans = 0
        for i in range(n - 1):
            for j in range(i + 1, n):
                for k in range(j, n):
                    a, b = pre[j] ^ pre[i], pre[k + 1] ^ pre[j]
                    if a == b:
                        ans += 1
        return ans


  • func countTriplets(arr []int) int {
	n := len(arr)
	pre := make([]int, n+1)
	for i := 0; i < n; i++ {
		pre[i+1] = pre[i] ^ arr[i]
	}
	ans := 0
	for i := 0; i < n-1; i++ {
		for j := i + 1; j < n; j++ {
			for k := j; k < n; k++ {
				a, b := pre[j]^pre[i], pre[k+1]^pre[j]
				if a == b {
					ans++
				}
			}
		}
	}
	return ans
}

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