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  1. Sparse Arrays
  2. Discussions

Sparse Arrays

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403 Discussions

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  • nozhana
     + 0 comments

    Swift solution:

    func matchingStrings(strings: [String], queries: [String]) -> [Int] {
    queries.reduce(into: [Int]()) { partialResult, query in
        let occurrences = strings
            .filter { $0 == query }
            .count
        partialResult.append(occurrences)
    }
}

  • adrianromanenco1
     + 0 comments
    def matchingStrings(strings, queries):
    query_results = []
    for i in range(len(queries)):
        occ = 0
        for j in range(len(strings)):
            if queries[i] == strings[j]:
                occ += 1
        query_results.append(occ) 
    return query_results

  • rtpardavila
     + 0 comments
    • Python 3

    One liner with list comprehensions and no aux/temp variables:

    def matchingStrings(strings, queries):

    return [
        sum(int(q == s) for s in strings)
        for q in queries
    ]

  • 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;
}