# cheatSheet.py

#https://leetcode.com/problems/minimum-ascii-delete-sum-for-two-strings/discuss/289435/Python-DP

#2d list

dp = [[0 for j in xrange(len(s2) + 1)] for i in xrange(len(s1) + 1)]

#get char num

ord(s1[i - 1])

total = 0

for c in s1:

total += ord(c)

#random list no duplicated

import random

random.sample(range(10), 10)

#findmiss

def findMiss(li):

return res

#assemble

s = 0

c = a[0]

for i in a[1:]:

c = c+i

s = s+c

#default dict

cnt = collections.defaultdict(int)

#819

words = re.split(r'\W+', paragraph)

print(words)

# 1380.2

logsSorted.insert(i, log)

logsSortedC + logsSortedN

# 000 maze

import collections

dq = collections.deque()

dq.append({'x':1, 'y':1, 's':1})

pos = dq.popleft()

# 001 init array

buckets = [0] * 100

# 002 random

random.randint(-10,10)

# list remove

a = [1,2,3,4]

a.remove(1)

# list index the first match element

a.index(3)

# smallest number

-float("inf")

# reverse range

range(5, 0, -1)

# set

vSet = set(['beginWord'])

wordSet = set(wordList)

mapNext[word] = mapNext[word] | graph[word[:i] + '_' + word[i + 1:]] & wordSet

mapNext[word] -= mapNext[word2]

# catch a spread list

a = []

a.append([1,2,3])

c,d,e = a.pop()

# set self.paramter in function

self.res=set()

# valid int

try:

n = int(sec, 16)

except ValueError:

return False

# check palindrome

s == s[::-1]

# pickle and PATH

pickle.dump(vecd, open(PATH/f'wiki.{lang}.pkl','wb'))

# binary operator

sign = [1,-1][x < 0]

# format binary padding 0

format(14, '#010b')

{}.format(14, '#010b')

#Fill the string with zeros until it is 10 characters long:

txt = "50"

x = txt.zfill(10)

# sort list

orgList

newList = sorted(orgList, key = lambda x: orgList[x])

# or

orgList.sort(orgList, key = lambda x: orgList[x])

# dict to list

l = list(dict.item())

# min heap: from min to max, maintain len

import heapq

a = [3,2,1]

heapq.heapify(a)

# a become [1,2,3]

if len(heap) < 3:

heapq.heappush(heap, num) # -num if max heap

else:

heapq.heappushpop(heap, num)

# !!!!!!!! heappop to get ordered item

#The traditional solution is to store (priority, task) tuples on the heap:

pq = [ ]

heappush(pq, (10, task1))

heappush(pq, (5, task2))

heappush(pq, (15, task3))

priority, task = heappop(pq)

# defaultdict

from collections import defaultdict

dint1 = defaultdict(lambda: 0)

dint2 = defaultdict(int)

dlist1 = defaultdict(lambda: [])

dlist2 = defaultdict(list)

for ch in 'abracadabra':

dint1[ch] += 1

dint2[ch] += 1

dlist1[ch].append(1)

dlist2[ch].append(1)

# init 2d matrix

m = [[0]*3]*3 no!!!!!

t = [ [0]*3 for i in range(3)] #instead

# slice last n number

nums[-n:]

# slice first n number

nums[:n] #nums[0] ~ nums[n-1]

# slice middle n number

nums[i:j] #nums[i] ~ nums[j-1]

# bit transfer

# xor

# string count char

bin(x ^ y).count('1')

# Standard Python dictionaries are unordered. Even if you sorted the (key,value) pairs, you wouldn't be able to store them in a dict in a way that would preserve the ordering.

# The easiest way is to use OrderedDict, which remembers the order in which the elements have been inserted:

In [1]: import collections

In [2]: d = {2:3, 1:89, 4:5, 3:0}

In [3]: od = collections.OrderedDict(sorted(d.items()))

In [4]: od

Out[4]: OrderedDict([(1, 89), (2, 3), (3, 0), (4, 5)])

# Never mind the way od is printed out; it'll work as expected:

In [11]: od[1]

Out[11]: 89

In [12]: od[3]

Out[12]: 0

In [13]: for k, v in od.iteritems(): print k, v

....:

1 89

2 3

3 0

4 5

Python 3

# For Python 3 users, one needs to use the .items() instead of .iteritems():

In [13]: for k, v in od.items(): print(k, v)

....:

1 89

2 3

3 0

4 5

# str to characters

list(str)