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  1. Sparse Arrays
  2. Discussions

Sparse Arrays

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230 Discussions

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  • rohelkhan1997
     + 0 comments

    Here my Python Solution:

    def matchingStrings(strings, queries): # Write your code here count_list = []

    for i in queries:
    count_list.append(strings.count(i))

return count_list

  • skillful_unleas1
     + 0 comments

    My Python 3 solution:

    def matchingStrings(strings, queries):
    for i in range(len(queries)):
        yield sum([1 for e in strings if queries[i] == e])

  • antoine_bourgeo1
     + 0 comments

    My Solution for C++ :

    vector<int> matchingStrings(vector<string> strings, vector<string> queries) {
    vector<int> ret;
    unordered_map<string, int> m;
    for (const string &s : strings) {
        m[s]++;
    }
    for (const string &q : queries) {
        if (m[q])
            ret.push_back(m[q]);
        else
            ret.push_back(0);
    }
    return (ret);
}

  • hiranyag_singh
     + 0 comments

    Python 3 Solution:

    def matchingStrings(strings, queries):
    res = []
    count = 0
    for i in range(len(queries)):
        for j in range(len(strings)):
            if strings[j] == queries[i]:
                count += 1
        res.append(count)
        count = 0
    return res

  • yassir_online_o1
     + 0 comments

    my version of code using JavaScript

    function matchingStrings(strings, queries) {
    // Write your code here
    const strCount = {};
    strings.forEach(str => {
        if(strCount[str])
        {
            strCount[str] += 1;
        } else {
            strCount[str] = 1;
        }
    });
    
    const result = queries.map(querie => {
        if(strCount[querie])
        {
            return strCount[querie];
        } else {
            return 0;
        }
    });

    return result;

    }