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Merge pull request #1 from devcoder007/CODER-001
CODER-001 Uploaded some more problems
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‎.ipynb_checkpoints/Untitled-checkpoint.ipynb‎

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{
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"cells": [],
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"metadata": {},
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"nbformat": 4,
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"nbformat_minor": 2
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}

‎aa.py‎

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‎compress_k_words.py‎

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# A function which will take two arguements, the string and number
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# of consecutive charecters to be compressed.
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def compressWord(word):
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#initializing a empty result string so that we can concatenate our resultant string.
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res = ""
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#initial value of i is 0 so that we can check the consecutiveness from starting.
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i = 0
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k=2
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while i < len(word):
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# it will run starting from the first element of string to the last element of string.
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if i < len(word) - (k-1) and [True for j in range(5,0,-1) if word[i]*(k+j) == word[i:i+k+j]]:
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#if stmt i <len(word) - (k-1) to prevent index out of range error.
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#and word[i]*k == word[i:i+k] will check whether we have k consecutive charecter
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#in string or not as word[i] ='c, word[i]*k ='ccc', word[i:i+k] = 'ccc', that means
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# we have k consecutive charecter. we will go for first recursion(it will work for i =3).
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i += k
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# if k consecutive charecter found then we will skip the i value upto k that is 3
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else:
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#if not then that means no need to compress, so we will concatenate that charecter
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#with our res variable
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res += word[i]
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#increment the i value to +1 to so that we can move to next charecter
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i += 1
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if len(res) == len(word):
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# if both strings have same length that means we have compressed succesfully
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#return the resulting string
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return res
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else:
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# recurssion (if the word again consist of group of k consecutive characters)
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return compressWord(res)
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word ="abbbbbcc"
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c=compressWord(word)
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print(c)

‎customer_island.py‎

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l=[1, 2, 3, 4, 5, 6, 7, 8]
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l1=[(1,2), (3,4), (6,5), (6,7), (2,8)]
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tt=[]
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count=0
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for i in l1:
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if i[0] in tt or i[1] in tt:
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continue
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else:
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if i[0] not in tt:
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tt.append(i[0])
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if i[1] not in tt:
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tt.append(i[1])
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count = count+1
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print(count)

‎email_scrap.py‎

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# from bs4 import BeautifulSoup
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# import requests
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# import requests.exceptions
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# from urllib.parse import urlsplit
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# from collections import deque
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# import re
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# # a queue of urls to be crawled
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# new_urls = deque(['https://mail.google.com/mail/u/1/#inbox'])
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# # a set of urls that we have already crawled
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# processed_urls = set()
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# # a set of crawled emails
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# emails = set()
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# # process urls one by one until we exhaust the queue
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# while len(new_urls):
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# # move next url from the queue to the set of processed urls
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# url = new_urls.popleft()
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# processed_urls.add(url)
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# # extract base url to resolve relative links
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# parts = urlsplit(url)
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# base_url = "{0.scheme}://{0.netloc}".format(parts)
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# path = url[:url.rfind('/')+1] if '/' in parts.path else url
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# # get url's content
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# print("Processing %s" % url)
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# try:
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# response = requests.get(url)
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# except (requests.exceptions.MissingSchema, requests.exceptions.ConnectionError):
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# # ignore pages with errors
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# continue
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# # extract all email addresses and add them into the resulting set
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# new_emails = set(re.findall(r"[a-z0-9\.\-+_]+@[a-z0-9\.\-+_]+\.[a-z]+", response.text, re.I))
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# emails.update(new_emails)
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# # create a beutiful soup for the html document
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# soup = BeautifulSoup(response.text)
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# # find and process all the anchors in the document
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# for anchor in soup.find_all("a"):
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# # extract link url from the anchor
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# link = anchor.attrs["href"] if "href" in anchor.attrs else ''
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# # resolve relative links
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# if link.startswith('/'):
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# link = base_url + link
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# elif not link.startswith('http'):
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# link = path + link
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# # add the new url to the queue if it was not enqueued nor processed yet
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# if not link in new_urls and not link in processed_urls:
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# new_urls.append(link)
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import imaplib
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import email
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from email.header import decode_header
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import HTMLParser
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# to unescape xml entities
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_parser = HTMLParser.HTMLParser()
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def decodeHeader(value):
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if value.startswith('"=?'):
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value = value.replace('"', '')
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value, encoding = decode_header(value)[0]
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if encoding:
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value = value.decode(encoding)
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return _parser.unescape(value)
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def listLastInbox(top = 10):
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mailbox = imaplib.IMAP4_SSL('imap.gmail.com')
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mailbox.login('[email protected]', 'Rish@1996')
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selected = mailbox.select('INBOX')
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assert selected[0] == 'OK'
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messageCount = int(selected[1][0])
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for i in range(messageCount, messageCount - top, -1):
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reponse = mailbox.fetch(str(i), '(RFC822)')[1]
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for part in reponse:
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if isinstance(part, tuple):
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message = email.message_from_string(part[1])
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# for h in ('subject','from','date'):
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# if "Firebase" in decodeHeader(message['from']):
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# print(decodeHeader(message['from']))
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yield {h: decodeHeader(message[h]) for h in ('subject', 'from', 'date') if "Firebase" in decodeHeader(message['from']) }
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mailbox.logout()
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if __name__ == '__main__':
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for message in listLastInbox():
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print ('-' * 40)
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for h, v in message.items():
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print (u'{0:8s}: {1}'.format(h.upper(), v))
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# def scramble(s1, s2):
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# # your code here
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# # print(list(s1),list(s2))
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# l = len(s2)
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# s1 = list(s1)
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# s2 = list(s2)
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# count = 0
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# for i in s2:
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# if i in s1:
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# count = count+1
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# # print(list(s1))
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# s1.remove(i)
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# # print(list(s1))
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# # if count == l:
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# # return True
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# # else:
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# # return False
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# print(count,l,len(s1))
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# scramble('scriptjava','javascript')

‎fizzbuzz.py‎

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# def fizzbuzz(n):
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# if n % 3 == 0 and n % 5 == 0:
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# return 'FizzBuzz'
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# elif n % 3 == 0:
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# return 'Fizz'
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# elif n % 5 == 0:
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# return 'Buzz'
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# else:
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# return str(n)
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# print("\n".join(fizzbuzz(n) for n in range(1, 21)))
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# Recursive FizzBuzz
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def fizzbuzz_recursive(n):
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if n == 0:
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return
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fizzbuzz_recursive(n - 1)
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if n % 3 == 0 and n % 5 == 0:
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print('FizzBuzz')
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elif n % 3 == 0:
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print('Fizz')
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elif n % 5 == 0:
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print('Buzz')
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else:
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print(n)
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n = [n for n in range(21)]
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fizzbuzz_recursive(n)

‎ideal_prime_factors.py‎

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import math
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def primeFactors(n):
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prime_factors = set()
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# Print the number of two's that divide n
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while n % 2 == 0:
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prime_factors.add(2)
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n = n / 2
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# n must be odd at this point
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# so a skip of 2 ( i = i + 2) can be used
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for i in range(3,int(math.sqrt(n)),2):
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# while i divides n , print i ad divide n
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while n % i== 0:
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prime_factors.add(i)
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n = n / i
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# Condition if n is a prime
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# number greater than 2
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if n > 2:
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prime_factors.add(n)
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return prime_factors
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def factor_of_three_and_five(first,last):
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#initializing value of count with 0 so that we can count ideal prime factors.
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count = 0
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# empty list for three and five div
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three_and_fives_div =[]
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#loop will run starting from the first arguement to last arguement
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for i in range(first,last+1):
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#it will check whether the i is a multiple of 3 or 5 or not.
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if (i%3==0) or (i%5==0):
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# if i is a multiple of 3 or 5 then pass it to the primeFactors function defined
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#above
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prime_factors = primeFactors(i)
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# there are two primeFactors, so we will check whether we have two or less than
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# two
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if len(set(prime_factors)) <= 2:
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# check whether we have 1 prime factor
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if len(set(prime_factors)) == 1:
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# if yes than just increment the count variable
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if (3 in prime_factors) or (5 in prime_factors):
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count += 1
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# if not equal to 1 then check whether that set has these elemnts or not
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# but to check we have a sorted list that's why we are converting the
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# returned set sorted list and again increment count if yes
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elif (sorted(list(set(prime_factors))) == [3,5]):
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count +=1
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# if 1 in range(first,last+1):
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# count += 1
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#just return value of count
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return count
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f = factor_of_three_and_five(200,405)
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print(f)
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# string="aaabbccaaee"
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# lst=list(string)
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# #print(lst)
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# count=1
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# ls=[]
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# for i in range(len(lst)):
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# if i+1 == len(lst):
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# # print(lst[i]," ",count," ",i)
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# ls.append((lst[i],count))#dict[lst[i]]=count
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# if i+1 < len(lst):
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# if lst[i] == lst[i+1]:
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# count=count+1
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# #print(i,count,lst[i])
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# continue
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# else:
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# ls.append((lst[i],count))#dict[lst[i]]=count
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# count=1
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# #print(i,lst[i],count,ls)
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# continue
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# # print(ls)
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# for i in ls:
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# print(i[0],i[1],sep="",end="")

‎least_common.py‎

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from collections import Counter
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line = [4,4,4,1,1,1,1,1,1,1,2,3,4,5,6,-1,1,2]
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temp = Counter(line)
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# # print(temp.keys())
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# print(temp)
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# t_l = [0]
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# ss = set(temp)
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# max = 0
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# for i in ss:
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# if temp[i]>max:
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# t_l.insert(0,i)
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# max = temp[i]
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# # print(max)
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# print(t_l[0])
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# print(t_l)
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print(temp)
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for k,v in temp:
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print(k,v)

‎nested_list_to_single.py‎

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import itertools
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a = [[1, 2], [3, 4], [5, 6]]
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print(list(itertools.chain.from_iterable(a)))

‎spiral_draw.py‎

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import pyautogui, time
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time.sleep(5)
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pyautogui.click() # click to put drawing program in focus
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distance = 200
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while distance > 0:
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pyautogui.dragRel(distance, 0, duration=0.2) # move right
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distance = distance - 2.5
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pyautogui.dragRel(0, distance, duration=0.2) # move down
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pyautogui.dragRel(-distance, 0, duration=0.2) # move left
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distance = distance - 2.5
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pyautogui.dragRel(0, -distance, duration=0.2) # move up

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