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Decode the Coding Interview in Python: Real-World Examples

Feature #6: Most Common Token

Explore how to determine the most frequently used token in a program string, excluding language keywords and syntax. Understand the step-by-step process of normalizing code, tokenizing it, and using data structures to count and identify the top token efficiently while analyzing time and space complexity.

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Description

For this feature of the language compiler, the program statements are given to us as a string. We want to determine the variable or function that is most commonly referred to in a program. In this process, we want to ignore the language keywords. For example, a keyword may be used more frequently than any variable or function. The language keywords are given as an array of strings.

Let’s say you are given the following program as input:

C++
int main() {
  int value = getValue();  
  int sum = value + getRandom();
  int subs = value - getRandom();
  return 0;
}

The list of keywords given to you is ["int", "main", "return"]. In this example, your function will return "value". Note that, your functions should ignore syntax, such as parentheses, operators, semicolons, etc.

Solution

We can solve this problem by normalizing the code string and processing it step by step. The complete algorithm is as follows:

  • First, we replace all the syntax, including parentheses, operators, semicolons, etc., with spaces. Now, the string only contains alphanumeric tokens.

  • Then, we split the code obtained from the previous step into tokens.

  • We iterate through the tokens to count the appearance of each unique token as keys, excluding the keywords.

  • We create a HashMap count, which has tokens as keys and occurrences as values.

  • In the end, we check the tokens in count to find the token with the highest frequency.

Python 3.5
def most_common_token(code, keywords):
    # Replacing the syntax with spaces
    normalized_code = ''.join([c if c.isalnum() else ' ' for c in code])
    tokens = normalized_code.split()
    count = defaultdict(int)
    banned_words = set(keywords)
    # Count occurrence of each token, excluding the keywords 
    for token in tokens:
        if token not in banned_words:
            count[token] += 1
    
    return max(count.items(), key=operator.itemgetter(1))[0]
# Driver code
code = """int main() {
  int value = getValue();  
  int sum = value + getRandom();
  int subs = value - getRandom();
  return 0;
}"""
keywords = ["int", "main", "return"]
print(most_common_token(code, keywords))
Most Common Token

Complexity measures

Time complexity Space complexity
...