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

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最新推荐文章于 2021-05-21 00:13:43 发布

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分类专栏： leetcode

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本文介绍了一种算法，用于在整数数组中找出所有可能的三元组(i, j, k)，使得arr[i]^arr[i+1]^...^arr[j-1]等于arr[j]^arr[j+1]^...^arr[k]。通过使用字典存储异或和，算法能高效地解决此问题。

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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

思路：找出异或和为0的区间

class Solution(object):
    def countTriplets(self, arr):
        """
        :type arr: List[int]
        :rtype: int
        """
        from collections import defaultdict
        d = defaultdict(list)
        d[0].append(-1)
        res = 0
        s = 0
        for j in range(len(arr)):
            s ^= arr[j]
            for i in d[s]:
                res += j-i-1
            d[s].append(j)
        return res