May 18, 2019
May 17, 2019 - Day 17 of 100!
100 days of Code! Reading Cracking the Coding Interview ! Preparing for Fall Internships!!
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May 17, 2019 - Day 17 of 100!
100 days of Code! Reading Cracking the Coding Interview ! Preparing for Fall Internships!!
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1955-56 Ford Thunderbird Roadsters
Issued by Brooklin Models in 1992. It is 1:43 scale and crafted in white metal. Released as the Classic Thunderbird Club International as the 1992 LE Club model for the CTCI Convention in Oklahoma. Issued as set, 1 of 500 (w/only 450 offered-50 sets were repurposed) The models are finished in Raven Black for the '55 and Goldenglow Yellow for the '56. BRK-13x
1956 Ford Thunderbird Roadster
Issued by Brooklin Models in 1982. It is 1:43 scale and crafted in white metal. The Classic Thunderbird Collector's International Club Edition (CTCI) Toronto Club Model. A Limited Edition, 1 of 300. Finished in Buckskin Tan BRK 13-003a
partition a linked list around a value x, such that all nodes less than x come before all nodes greater than or equal to x
# if x is contained within the list, the values of x only need to be after the elements less than x. # the partition element x can appear anywhere in the "right parition"; it does not need to appear between the left and right partitions # the original relative order should be preserved # can you do this in place? (without using extra memory) # x = 5 # 3 -> 5 -> 8 -> 5 -> 10 -> 2 -> 1 # 3 -> 2 -> 1 -> 5 -> 5 -> 8 -> 10 # create a seperate linked list for each partition # walk down the list starting at head # lt: 3 -> 2 -> 1 # et: 5 -> 5 # gt: 8 -> 10 # # 3 < 5 so it goes into the less than linked list # 5 == 5 so it goes into the == linked list # 8 > 5 so it goes into the > linked list # 5 == 5 so it goes into the == linked list # 10 > 5 so it goes into the > linked list # 2 < 5 # 1 < 5 # then combine linked lists in the following order: # lt -> et -> gt # 3 -> 2 -> 1 -> 5 -> 5 -> 8 -> 10 class Node: def __init__(self, value, n=None): self.val = value self.next = n class LinkedList: def __init__(self, h=None, t=None): self.head = h self.tail = t def add(self, value): if self.head is None: self.head = self.tail = Node(value) else: self.tail.next = Node(value) self.tail = self.tail.next return self.tail def deleteNode(self, value): current = self.head if current.val == value: return self.head.next while current.next is not None: if current.next.val == value: current.next = current.next.next return self.head current = current.next return self.head def partitionLinkedlist(lst, X): lessThanList = LinkedList() equalToList = LinkedList() greaterThanList = LinkedList() current = lst.head while current is not None: curVal = int(current.val) if curVal < X: lessThanList.add(curVal) elif curVal == X: equalToList.add(curVal) else: greaterThanList.add(curVal) current = current.next equalToList.tail.next = greaterThanList.head lessThanList.tail.next = equalToList.head return lessThanList lst = LinkedList() # 3 -> 5 -> 8 -> 5 -> 10 -> 2 -> 1 lst.add('3') lst.add('5') lst.add('8') lst.add('5') lst.add('10') lst.add('2') lst.add('1') lst2 = partitionLinkedlist(lst, 5) current = lst2.head while current.next is not None: print(current.val) current = current.next print(current.val)
Time: O(N), space: O(n)
Palindrome permutation: Given a string, write a function to check if it is a permutation of a palindrome.
# # A Palindrome is a word or phrase that is the same forwards and backwards. # A permutation is a rearrangemet of letters. The palindrome does not need to be limited to just dictionary words # A palindrome is a string that is the same forwards and backwards.Therefore, to decide if a string is a permu- tation of a palindrome, we need to know if it can be written such that it's the same forwards and backwards # # That is to say, if the string is a palindrome there is an even number of each unique character #edge case: Strings of an odd length must have exactly one character with an odd count. # That is to say to be a permutation of a palindrome, a string can have no more than one character that is odd.This will cover both the odd and the even cases. # Example # Input: Tact Coa # Output: # True (perumations: "taco cat", "atco cta", etc.) # brute force: # 1) sort the string in alphabetical order # 2) move a pointer along the string and keep a count of the current letter occurance # 3) if the count is not even (% 2 != 0), then return false # this solution is O(nlogn) but O(1) space def palindromePermutationSolution1(input): input = sorted(input.upper()) print(input) runningCount = 1 oddCount = 0 for letter in range(0, len(input)-1): #O(n) if input[letter] == input[letter+1]: runningCount += 1 if runningCount % 2 != 0: oddCount += 1 if input[letter] != input[letter+1]: runningCount = 1 if oddCount > 2: return False return True #aab b bcc #^^^^^^ print(palindromePermutationSolution1("taco cat")) # however a slightly more intuitive way may be done with a hashmap def palindromePermutationSolution2(input): input = input.upper() map = dict() for letter in input: if letter != " ": # edge case if letter in map: map[letter] += 1 else: map[letter] = 1 isOneOdd = False for value in map.values(): if value % 2 == 1: if isOneOdd == True: # we do this because the function loops return False # this will never run if the there is 0 or 1 odds! isOneOdd = True return True # this solution is O(n) time complexity however the hashmap causes O(n) space complexity print(palindromePermutationSolution2("taco cat"))
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Hey everyone, Â As promised here are some more photos of things I've been up to in the past several months. Â Enjoy!
CTCI India Freshers Walkin Recruitment in Noida On 10th November 2016
CTCI India Freshers Walkin Recruitment in Noida On 10th November 2016
CTCI India Freshers Walkin Recruitment in Noida On 27th October 2016. CTCI Corp job openings in Noida for the vacancies of administration assistant. Interested and eligible candidates can directly attend the walkin drive at below mentioned venue.  CTCI India Walkin Recruitment Details: Company Name CTCI India Pvt. Ltd. Qualification Any Graduate Experience 0 to 1 Years Job Location Noida Job…
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CTCI India Freshers Walkin Recruitment in Noida On 27th October 2016
CTCI India Freshers Walkin Recruitment in Noida On 27th October 2016
CTCI India Freshers Walkin Recruitment in Noida On 27th October 2016.
CTCI Corp job openings in Noida for the vacancies of administration assistant. Interested and eligible candidates can directly attend the walkin drive at below mentioned venue.
CTCI India Walkin Recruitment Details: Company Name CTCI India Pvt. Ltd. Qualification Any Graduate Experience 0 to 1 Years Job LocationNoida
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