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

[leetcode] 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: 4Explanation: 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 <= 3001 <= arr[i] <= 10^8

分析

题目的意思是：给你一个数组，找出具有相同XOR值的三元组，如果需要a==b，则a^b=0，如果了解这个，则解法就出来了，找出所有异或为0的子数组就行了，如果一个子数组有N个元素，就能构成n-1个三元组。如果能够找到这个规律，解法就出来了。

代码

class Solution: def countTriplets(self, arr: List[int]) -> int: n=len(arr) res=0 for i in range(n-1): xor=0 for j in range(i,n): xor^=arr[j] if(xor==0): res+=j-i return res

参考文献

​​[LeetCode] Easy to understand Python solution, O(N^2) time, O(1) space​​

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