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  1. Sparse Arrays
  2. Discussions

Sparse Arrays

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400 Discussions

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  • derekcommongrou1
     + 0 comments

    JavaScript solution:

    function matchingStrings(strings, queries) {    
    const counts = new Map();
    
    strings.forEach(string => {
        const existingCount = counts.get(string);
        if (existingCount) {
            counts.set(string, existingCount + 1);
        } else {
            counts.set(string, 1);
        }
    })
        
    return queries.map(query => counts.get(query) || 0);
}

  • cooldilcool
     + 0 comments

    C# Code

    public static List matchingStrings(List strings, List queries) { var resultList = new List(); 

        foreach(string queryItem in queries)
    {
        resultList.Add(strings.Count(e => e == queryItem));
    }

    return resultList;
}

  • lo4251054
     + 0 comments

    Java solution using HashMap:

        public static List<Integer> matchingStrings(List<String> strings, List<String> queries) {

        Map<String, Integer> countMap = new HashMap<>();

        for(String query : queries){
            countMap.put(query, 0);
        }

        for(String str : strings){
            if(countMap.containsKey(str)){
                Integer count = countMap.get(str);
                countMap.put(str, count + 1);
            }
        }

        List<Integer> returnList = new ArrayList<>();
        for(String query : queries){
            returnList.add(countMap.get(query));
        }

        return returnList;
    }

  • bladimir_matos
     + 0 comments

    Java:

    public static List matchingStrings(List strings, List queries) { // Write your code here

    List result = new ArrayList<>();

    for (String string : queries) {

    int counter = 0;

for (String seek : strings) {

    if(seek.equals(string)){
        counter ++;
    }

}

result.add(counter);

    }

    return result;


}

    }

  • fagundes4181
     + 0 comments
    fun matchingStrings(strings: Array<String>, queries: Array<String>): Array<Int> {
        val resultArray = Array(queries.size){0}
        queries.forEachIndexed { index, element ->
            resultArray[index] = strings.count { it == element }
        }
        return resultArray
}