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Hackerrank Python哈希表:赎金注-由于超时而终止:(

[英]Hackerrank Python Hash Tables: Ransom Note - Terminated due to timeout :(

令人讨厌的Hackerrank哈希表“因超时而终止”错误:21个测试案例中有6个赎金票据

实现开放地址哈希。 输入字符串的大小最多为30,000个字符串:已尝试将哈希表的大小从60,000更改为300,000,但没有成功。

CAPACITY = 300000
hashTable = [None] * CAPACITY

def checkMagazine(magazine, note):
    # Store Magazine into hashtable
    for element in magazine:
        # print("STORED " + element)
        position = calculateHash(element)
        # print(position)
        if hashTable[position] == None:
            hashTable[position] = element
            # print("Stored into " + str(position))
        else:
            i = 1
            # print("collided into " + str((position) % CAPACITY))
            while hashTable[(position + i) % CAPACITY] != None:
                # print("collided into " + str((position + i) % CAPACITY))
                i += 1
            hashTable[(position + i) % CAPACITY] = element


    # Check if all items in note is in hashtable
    included = True
    for item in note:
        position = calculateHash(item)
        if hashTable[position] != item:
            i = 1
            while hashTable[(position + i ) % CAPACITY] != item:
                if hashTable[(position + i ) % CAPACITY] == None:
                    included = False
                    print("No")
                    return
                else:
                    i += 1
            hashTable[(position + i ) % CAPACITY] = "DONED"
        else:
            hashTable[position] = "DONED"
        # print("Found " + item)            


    print("Yes") 

def calculateHash(string):
    return hash(string) % CAPACITY

假设哈希表是解决此问题的最佳方法(时间复杂度O(n)),那么发生超时的原因是因为开放地址哈希? 还是还有其他原因?

我认为问题与您的实施有关。 看看如果传递诸如["a", "a", "a", .... "a"]类的大型“ magazine”输入,代码会发生什么变化。

应该尝试过这样的事情:

magazine = 'two times three is not four'
note = 'two times two is four'

if set(magazine.split()) & set(note.split()) == set(note.split()) :
    print 'yes'
else :
    print 'no'

您可以通过预先计算set(note.split())节省一些时间,但是我怀疑它是否很大。

如果您关心字数,可以使用Counter

from collections import Counter

然后检查便笺中的每个单词,计数器是否比杂志中相同单词的计数器小。

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