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You want to build a word cloud, an infographic where the size of a word corresponds to how often it appears in the body of text.
To do this, you’ll need data. Write code that takes a long string and builds its word cloud data in a dictionary ↴ , where the keys are words and the values are the number of times the words occurred.
Think about capitalized words. For example, look at these sentences:
'After beating the eggs, Dana read the next step:''Add milk and eggs, then add flour and sugar.'
What do we want to do with “After”, “Dana”, and “add”? In this example, your final dictionary should include one “Add” or “add” with a value of 2. Make reasonable (not necessarily perfect) decisions about cases like “After” and “Dana”.
Assume the input will only contain words and standard punctuation.
You could make a reasonable argument to use regex in your solution. We won’t, mainly because performance is difficult to measure and can get pretty bad.
Are you sure your code handles hyphenated words and standard punctuation?
Are you sure your code reasonably handles the same word with different capitalization?
Try these sentences:
"We came, we saw, we conquered...then we ate Bill's (Mille-Feuille) cake."
'The bill came to five dollars.'
We can do this in O(n) runtime and space.
The final dictionary we return should be the only data structure whose length is tied to n.
We should only iterate through our input string once.
We’ll have to go through the entire input string, and we’re returning a dictionary with every unique word. In the worst case every word is different, so our runtime and space cost will both be at least O(n).
This challenge has several parts. Let’s break them down.
How would you start the first part?
We could use the built-in split() function to separate our words, but if we just split on spaces we’d have to iterate over all the words before or after splitting to clean up the punctuation. And consider em dashes or ellipses, which aren’t surrounded by spaces but nonetheless separate words. Instead, we’ll make our own split_words() function, which will let us iterate over the input string only once.
We’ll check if each character is a letter with the string method isalpha().
Now how can we split each word? Let’s assume, for now, that our helper function will return a list of words.
We’ll iterate over all the characters in the input string. How can we identify when we’ve reached the end of a word?
Here’s a simple example. It doesn’t work perfectly yet—you’ll need to add code to handle the end of the input string, hyphenated words, punctuation, and edge cases.
def split_words(input_string): words = []
current_word_start_index = 0
current_word_length = 0
for i, char in enumerate(input_string):
if char.isalpha():
if current_word_length == 0:
current_word_start_index = i
current_word_length += 1
else:
word = input_string[current_word_start_index:
current_word_start_index + current_word_length]
words.append(word)
current_word_length = 0
return words
Careful—if you thought of building up the word character by character (using +=), you’d be doing a lot more work than you probably think. Every time we append a character to a string, Python makes a whole new string. If our input is one long word, then creating all these copies is O(n^2)O(n2) time.
Instead, we keep track of the index where our word starts and its current length. Once we hit a space, we can use string slicing to extract the current word to append to the list. That keeps our split function at O(n)O(n) time.
Now we’ve solved the first part of the challenge, splitting the words. The next part is populating our dictionary with unique words. What do we do with each word?
If the word is in the dictionary, we’ll increment its count. Otherwise, we’ll add it to the dictionary with a count of 1.
words_to_counts = {}
def add_word_to_dictionary(word):
if word in words_to_counts:
words_to_counts[word] += 1
else:
words_to_counts[word] = 1
Alright, last part! How should we handle words that are uppercase and lowercase?
Consider these sentences:
'We came, we saw, we ate cake.''Friends, Romans, countrymen! Let us eat cake.'
'New tourists in New York often wait in long lines for cronuts.'
Take some time to think of possible approaches. What are some other sentences you might run into. What are all your options?
When are words that should be lowercase not?
Why not?
What are the ideal cases we’d want in our dictionary?
Here are a few options:
What are the pros and cons for each one?
Pros and cons include:
Any of these could be considered reasonable. Importantly, none of them are perfect. They all have tradeoffs, and it is very difficult to write a highly accurate algorithm. Consider “cliff” and “bill” in these sentences:
'Cliff finished his cake and paid the bill.''Bill finished his cake at the edge of the cliff.'
You can choose whichever of the options you’d like, or another option you thought of. For this breakdown, we’re going to choose option (1).
Now, how do we update our add_word_to_dictionary() function to avoid duplicate words?
Think about the different possibilities:
Moving forward, we can either:
We’ll choose the second approach since it will save us a walk through our dictionary. How should we start?
If the word we’re adding is already in the dictionary in its current case, let’s increment its count. What if it’s not in the dictionary?
There are three possibilities:
Let’s start with the first possibility. What do we want to do?
Because we only include a word as uppercase if it is always uppercase, we simply increment the lowercase version’s count.
# Current dictionary# {'blue': 3}
# Adding
# 'Blue'
# Code
words_to_counts[word.lower()] += 1
# Updated dictionary
# {'blue': 4}
What about the second possibility?
This is a little more complicated. We need to remove the uppercase version from our dictionary if we encounter a lowercase version. But we still need the uppercase version’s count!
# Current dictionary# {'Yellow': 6}
# Adding
# 'yellow'
# Code (we will write our "capitalize()" method later)
words_to_counts[word] = 1
words_to_counts[word] += words_to_counts[word.capitalize()]
del words_to_counts[word.capitalize()]
# Updated dictionary
# {'yellow': 7}
Finally, what if the word is not in the dictionary at all?
Easy—we add it and give it a count of 1.
# Current dictionary# {'purple': 2}
# Adding
# 'indigo'
# Code
words_to_counts[word] = 1
# Updated dictionary
# {'purple': 2, 'indigo': 1}
Now we have all our pieces! We can split words, add them to a dictionary, and track the number of times each word occurs without having duplicate words of the same case. Can we improve our solution?
Let’s look at our runtime and space cost. We iterate through every character in the input string once and then every word in our list once. That’s a runtime of O(n), which is the best we can achieve for this challenge (we have to look at the entire input string). The space we’re using includes a list for each word and a dictionary for every unique word. Our worst case is that every word is different, so our space cost is also O(n), which is also the best we can achieve for this challenge (we have to return a dictionary of words).
But we can still make some optimizations!
How can we make our space cost even smaller?
We’re storing all our split words in a separate list. That at least doubles the memory we use! How can we eliminate the need for that list?
Right now, we store each word in our list as we split them. Instead, let’s just immediately populate each word in our dictionary!
In our solution, we make three decisions:
To split the words in the input string and populate a dictionary of the unique words to the number of times they occurred, we:
If the input word is uppercase and there’s a lowercase version in the dictionary, we increment the lowercase version’s count. If the input word is lowercase and there’s an uppercase version in the dictionary, we “demote” the uppercase version by adding the lowercase version and giving it the uppercase version’s count.
class WordCloudData:
def __init__(self, input_string):
self.words_to_counts = {}
self.populate_words_to_counts(input_string)
def populate_words_to_counts(self, input_string):
# Iterates over each character in the input string, splitting
# words and passing them to add_word_to_dictionary()
current_word_start_index = 0
current_word_length = 0
for i, character in enumerate(input_string):
# If we reached the end of the string we check if the last
# character is a letter and add the last word to our dictionary
if i == len(input_string) - 1:
if character.isalpha():
current_word_length += 1
if current_word_length > 0:
current_word = input_string[current_word_start_index:
current_word_start_index + current_word_length]
self.add_word_to_dictionary(current_word)
# If we reach a space or emdash we know we're at the end of a word
# so we add it to our dictionary and reset our current word
elif character == ' ' or character == u'\u2014':
if current_word_length > 0:
current_word = input_string[current_word_start_index:
current_word_start_index + current_word_length]
self.add_word_to_dictionary(current_word)
current_word_length = 0
# We want to make sure we split on ellipses so if we get two periods in
# a row we add the current word to our dictionary and reset our current word
elif character == '.':
if i < len(input_string) - 1 and input_string[i + 1] == '.':
if current_word_length > 0:
current_word = input_string[current_word_start_index:
current_word_start_index + current_word_length]
self.add_word_to_dictionary(current_word)
current_word_length = 0
# If the character is a letter or an apostrophe, we add it to our current word
elif character.isalpha() or character == '\'':
if current_word_length == 0:
current_word_start_index = i
current_word_length += 1
# If the character is a hyphen, we want to check if it's surrounded by letters
# If it is, we add it to our current word
elif character == '-':
if i > 0 and input_string[i - 1].isalpha() and \
input_string[i + 1].isalpha():
current_word_length += 1
else:
if current_word_length > 0:
current_word = input_string[current_word_start_index:
current_word_start_index + current_word_length]
self.add_word_to_dictionary(current_word)
current_word_length = 0
def add_word_to_dictionary(self, word):
# If the word is already in the dictionary we increment its count
if word in self.words_to_counts:
self.words_to_counts[word] += 1
# If a lowercase version is in the dictionary, we know our input word must be uppercase
# but we only include uppercase words if they're always uppercase
# so we just increment the lowercase version's count
elif word.lower() in self.words_to_counts:
self.words_to_counts[word.lower()] += 1
# If an uppercase version is in the dictionary, we know our input word must be lowercase.
# since we only include uppercase words if they're always uppercase, we add the
# lowercase version and give it the uppercase version's count
elif word.capitalize() in self.words_to_counts:
self.words_to_counts[word] = 1
self.words_to_counts[word] += self.words_to_counts[word.capitalize()]
del self.words_to_counts[word.capitalize()]
# Otherwise, the word is not in the dictionary at all, lowercase or uppercase
# so we add it to the dictionary
else:
self.words_to_counts[word] = 1
Runtime and memory cost are both O(n). This is the best we can do because we have to look at every character in the input string and we have to return a dictionary of every unique word. We optimized to only make one pass over our input and have only one O(n) data structure.
To handle capitalized words, there were lots of heuristics and approaches we could have used, each with their own strengths and weaknesses. Open-ended questions like this can really separate good engineers from great engineers.
Good engineers will come up with a solution, but great engineers will come up with several solutions, weigh them carefully, and choose the best solution for the given context. So as you’re running practice questions, challenge yourself to keep thinking even after you have a first solution. See how many solutions you can come up with. This will grow your ability to quickly see multiple ways to solve a problem, so you can figure out the best solution. And use the hints and gotchas on each Interview Cake question—they’re designed to help you cultivate this skill.
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