Decode the Coding Interview in JavaScript: Real-World Examples/

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Feature #6: Most Common Token

Feature #6: Most Common Token

Implementing the "Most Common Token" feature for our "Language Compiler" project.

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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:

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

The array 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.

function mostCommonToken(code, keywords){
  // Replacing the syntax with spaces
  var normalizedCode = code.replace(/[^a-zA-Z0-9]/gi, function (x) {
    return " "
  });
  var tokens = normalizedCode.split(/(\s+)/).filter( function(e) { return e.trim().length > 0; } );
  var count = {}
  var bannedWords = new Set(keywords)
  // Count occurence of each token, excluding the keywords 
  tokens.forEach((token) => {
    if (!(bannedWords.has(token))){
      if(!(token in count)){
        count[token] = 0
      }
      count[token] = count[token] + 1
    }   
  })
  // return the maximum valued key
  // return Collections.max(count.entrySet(), (entry1, entry2) -> entry1.getValue() - entry2.getValue()).getKey();
  return Object.keys(count).reduce((a, b) => count[a] > count[b] ? a : b);
}
// Driver code
var code = "int main() {\n" +
  "int value = getValue();\n" + 
  "int sum = value + getRandom();\n" +
  "int subs = value - getRandom();\n" +
  "return 0;\n" +
"}"
var keywords = ["int", "main", "return"]
console.log(mostCommonToken(code, keywords))
Most Common Token
Time complexity Space complexity
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