2527.Query arrayXorBeautiful value Zhou Sai Third Question

6289.Query array Xor Beautiful value。

This question has been studied for a long time，Discovering a mathematical problem，But I didn’t want to be transparent，Because unfamiliar or valuable。So there is no better solution。

The first is to read the question and learn(nums[i]|nums[j])&nums[k]，becauseijkCan be any value in the array，So the tripleforcycle（It will definitely timeout），Then he was a little bit out of his own，Just a little bit，feeli，jThe value of the value is the same，So add an additional layer of judgment（Dictionary）。

class Solution:
    def xorBeauty(self, nums: List[int]) -> int:
        ans = []
        record = {}
        for i in range(len(nums)):
            for j in range(len(nums)):
                for k in range(len(nums)):
                    if k not in record:
                        if (i,j) not in record.values():
                            record[k] = (i,j)
                            flag = (nums[i]|nums[j])&nums[k]
                            if flag not in ans:
                                ans.append(flag)
                            else:
                                ans.remove(flag)
        if len(ans) == 0:
            return 0
        answer = ans[0]

        for i in range(1,len(ans)):
            answer ^= ans[i]
        return answer

I have been thinking about how to judge the same situation when I go out today，As a result@Lingcha Mountain AifuSolution，One by one，That’s okay。

class Solution:
    def xorBeauty(self, nums: List[int]) -> int:
        return reduce(xor, nums)

Then do，Search online，It turns out that if the two values ​​are the same，for exampleA^B^A = B，Soappend，removeTo implement this function。at lastforcycle求异或值。

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