HR_Sherlock and Anagrams_TIMEOUT[UNDONE]

2019年1月10日15:39:23 去掉了所有不必要的循环区间 还是超时 本地运行大概3s

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the sherlockAndAnagrams function below.
def compareSubString(string1,string2):
    num_string1 = [0]*26
    num_string2 = [0]*26
    for index in range (len(string1)):
        temp = ord(string1[index]) - ord('a')
        num_string1[temp] +=1
        temp = ord(string2[index]) - ord('a')
        num_string2[temp] +=1

    if num_string1 == num_string2:
        return True
    else:
        return False

def sherlockAndAnagrams(s):     #s <= 100
    n = len(s)
    count = 0
    for index in range(n-1):                        #last one have no mean
        for length in range(1, n - index):       #only length between [1,n-1] passible have Anagrams
            tempstring1 = s[index : index + length]
            if (index + 1 <= n - length ):
                for index2 in range(index + 1, n - length +1):
                    tempstring2 = s[index2: index2 + length]
                    if compareSubString(tempstring1 , tempstring2):
                        count +=1
    return count

if __name__ == '__main__':
    # fptr = open(os.environ['OUTPUT_PATH'], 'w')

    q = 1

    for q_itr in range(q):
        s = 'zjekimenscyiamnwlpxytkndjsygifmqlqibxxqlauxamfviftquntvkwppxrzuncyenacfivtigvfsadtlytzymuwvpntngkyhw'

        result = sherlockAndAnagrams(s)

        print(result)

------

注意 anagrammatic  是拼字游戏的意思，任意调换顺序可以组成对方都成立，不是简单的回文数字。

------

1.第一种解法想省些时间 结果都比测试集少了20多个 应该是有些规则题目没有描述清楚。

5
ifailuhkqqhucpoltgtyovarjsnrbfpvmupwjjjfiwwhrlkpekxxnebfrwibylcvkfealgonjkzwlyfhhkefuvgndgdnbelgruel
gffryqktmwocejbxfidpjfgrrkpowoxwggxaknmltjcpazgtnakcfcogzatyskqjyorcftwxjrtgayvllutrjxpbzggjxbmxpnde
mqmtjwxaaaxklheghvqcyhaaegtlyntxmoluqlzvuzgkwhkkfpwarkckansgabfclzgnumdrojexnrdunivxqjzfbzsodycnsnmw
ofeqjnqnxwidhbuxxhfwargwkikjqwyghpsygjxyrarcoacwnhxyqlrviikfuiuotifznqmzpjrxycnqktkryutpqvbgbgthfges
zjekimenscyiamnwlpxytkndjsygifmqlqibxxqlauxamfviftquntvkwppxrzuncyenacfivtigvfsadtlytzymuwvpntngkyhw

399
471
370
403
428

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the sherlockAndAnagrams function below.
def sherlockAndAnagrams(s):
    n = len(s)
    count = 0
    # len = 1 && multi in string
    for i in range(n):
        for j in range(n):
            if j > i:
                if s[j] == s[i]:
                    count += 1
                    if j - i > 1:
                        count += 1

    # repeat substring && huiwen substring
    for start in range(n):
        for end in range(n):
            if end - start > 1:  # substring len > 1
                substring_num = 0
                reverse_substring_num = 0
                str1 = s[start:end]
                str2 = s[end:]
                str3 = str1[::-1]
                substring_num = str2.count(str1)
                reverse_substring_num = str2.count(str3)
                count += substring_num
                if str1 != str3:  # avoid kkk
                    count += reverse_substring_num

    return count

if __name__ == '__main__':

    s = 'zjekimenscyiamnwlpxytkndjsygifmqlqibxxqlauxamfviftquntvkwppxrzuncyenacfivtigvfsadtlytzymuwvpntngkyhw'

    result = sherlockAndAnagrams(s)
    # fptr = open(os.environ['OUTPUT_PATH'], 'w')
    #
    # q = int(input())
    #
    # for q_itr in range(q):
    #     s = 'abba'
    #
    #     result = sherlockAndAnagrams(s)
    #
    #     fptr.write(str(result) + '\n')
    #
    # fptr.close()

2.看s 最大也就100 直接列出所有子串排序求和就行 但是发现会超时，本地运行需要十几秒。。。：

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the sherlockAndAnagrams function below.
def sherlockAndAnagrams(s):
    n = len(s)
    count = 0
    #find all substring
    arr = []
    for i in range(n):
        for j in range(n+1):
            if j > i:
                arr.append(s[i:j])
    # count
    count = 0
    for i in range(len(arr)):
        for j in range(len(arr)):
            if j>i and len(arr[i] )== len(arr[j]):
                str1 = sorted(arr[i])
                str2 = sorted(arr[j])
                if str1 == str2 :
                    count += 1
    return count

if __name__ == '__main__':

    s = 'mqmtjwxaaaxklheghvqcyhaaegtlyntxmoluqlzvuzgkwhkkfpwarkckansgabfclzgnumdrojexnrdunivxqjzfbzsodycnsnmw'

    result = sherlockAndAnagrams(s)
    # fptr = open(os.environ['OUTPUT_PATH'], 'w')
    #
    # q = int(input())
    #
    # for q_itr in range(q):
    #     s = 'abba'
    #
    #     result = sherlockAndAnagrams(s)
    #
    #     fptr.write(str(result) + '\n')
    #
    # fptr.close()

3.优化了下 改成字符串 还是超时：

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the sherlockAndAnagrams function below.
def sherlockAndAnagrams(s):
    n = len(s)
    count = 0
    #find all substring
    arr = []
    for i in range(n):
        for j in range(n+1):
            if j > i:
                arr.append(s[i:j])
    # count
    count = 0
    # for i in range(len(arr)):
    #     for j in range(len(arr)):
    #         if j>i and len(arr[i] )== len(arr[j]):
    #             str1 = sorted(arr[i])
    #             str2 = sorted(arr[j])
    #             if str1 == str2 :
    #                 count += 1

    for i in range(len(arr)):
        temp = "".join(sorted(arr[i]))
        arr[i] = temp

    for i in range(len(arr)):
        for j in range(i,len(arr)):
            if j>i and len(arr[i] )== len(arr[j]) and arr[i] == arr[j]:
                count += 1
    return count

if __name__ == '__main__':

    s = 'ifailuhkqqhucpoltgtyovarjsnrbfpvmupwjjjfiwwhrlkpekxxnebfrwibylcvkfealgonjkzwlyfhhkefuvgndgdnbelgruel'

    result = sherlockAndAnagrams(s)

    print(result)
    # fptr = open(os.environ['OUTPUT_PATH'], 'w')
    #
    # q = int(input())
    #
    # for q_itr in range(q):
    #     s = 'abba'
    #
    #     result = sherlockAndAnagrams(s)
    #
    #     fptr.write(str(result) + '\n')
    #
    # fptr.close()