Given a dictionary and a character array, print all valid words that are possible in Python

In this Python example we will discuss the following problem statement. Given a dictionary of words and a character array, write a function that prints all valid words that are possible using characters from the array. A word is considered valid if it can be formed using only the characters from the array, and if it is present in the dictionary.

Some of the topics which will be helpful for understanding the program implementation better are:

• List in Python
• Function in Python
• Dictionary in Python
• For loop in Python

1. Using a for loop

We can iterate through each word in the dictionary and check if it can be formed using only the characters in the char_array. To check if a word can be formed using the characters in the array, we can convert the word and the array into sets and use the issubset() method to check if that is the subset of the character array.

Now let’s implement a program for this problem –

def print_valid_words(dictionary, char_array):
  for word in dictionary:
    if set(word).issubset(set(char_array)):
      print(word)

dictionary = {'cat', 'dog', 'act', 'god', 'bat'}
char_array = ['a', 'c', 't', 'o', 'g', 'd']
print_valid_words(dictionary, char_array)

Output
cat
god
dog
act

2. Using list comprehension

This is exactly like the previous example but instead of using a for loop we will use list comprehension to achieve the same.

Now let’s implement a program for this problem –

def print_valid_words(dictionary, char_array):
  valid_words = [word for word in dictionary if set(word).issubset(set(char_array))]
  print(valid_words)

dictionary = {'cat', 'dog', 'act', 'god', 'bat'}
char_array = ['a', 'c', 't', 'o', 'g', 'd']
print_valid_words(dictionary, char_array)

Output
['cat', 'god', 'dog', 'act']

3. Conclusion

In this Python example, we discussed how to print all valid words that can be formed using characters from a given character array, using a dictionary in Python. We explored two methods to solve this problem

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