Feature #6: Most Common Token
Implementing the "Most Common Token" feature for our "Language Compiler" project.
We'll cover the following...
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:
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.
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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.
class HelloWorld {public static String mostCommonToken(String code, String[] keywords){// Replacing the syntax with spacesString normalizedCode = code.replaceAll("[^a-zA-Z0-9]", " ");String[] tokens = normalizedCode.split("\\s+");HashMap<String, Integer> count = new HashMap<String, Integer>();Set<String> bannedWords = new HashSet<String>(Arrays.asList(keywords));// Count occurence of each token, excluding the keywordsfor(String token : tokens){if (!bannedWords.contains(token)){if(!count.containsKey(token)){count.put(token, 0);}count.put(token, count.get(token) + 1);}}// return the maximum valued keyreturn Collections.max(count.entrySet(), (entry1, entry2) -> entry1.getValue() - entry2.getValue()).getKey();}public static void main( String args[] ) {// Driver codeString code = "int main() {\n" +"int value = getValue();\n" +"int sum = value + getRandom();\n" +"int subs = value - getRandom();\n" +"return 0;\n" +"}";String[] keywords = {"int", "main", "return"};System.out.println(mostCommonToken(code, keywords));}}
Complexity measures
Time complexity | Space complexity |
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