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  1. Sparse Arrays
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Sparse Arrays

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410 Discussions

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  • emailthegyro
     + 0 comments

    A O(strings+queries) solution in C++ using unordered_map where all operations used below are at O(1) according to to implementation document.

    vector<int> matchingStrings(vector<string> strings, vector<string> queries) {
    unordered_map<string, int> frequency;
    vector<int>results;
    
    //counts all occurances of str in strings and populates the frequency map
    for(string str: strings)
        frequency[str]++;
    //writes out the queries occurances in result
    for(string str: queries)
        results.push_back(frequency[str]);
    
     return results;
}

  • t7_muj
     + 0 comments

    Hi! This was my attempt using Javascript. I think its o(n) time or o(n+q) technically. I could probably avoid storing them in arrays in the groupedStrings array in hindsight 

    function matchingStrings(strings, queries) {
    // Write your code here
    
    const groupedStrings = {};
    
    strings.forEach((string) => {
        if (!groupedStrings[string]) {
            groupedStrings[string] = []
        }
        groupedStrings[string].push(string);
    })
    
    queries.forEach((query) => {
        if (groupedStrings[query] == undefined) { 
            console.log(0);
            return;
        }
        console.log(groupedStrings[query].length);
    })
    
}

  • douglasbuscam
     + 0 comments

    Javascript solution

    function matchingStrings(strings, queries) {

    let freqObj = {}
for(let i=0; i < strings.length; i++) {
    let str = strings[i];  
    freqObj[str] = (freqObj[str] || 0) + 1; 
    }
    return queries.map(querie => freqObj[querie]) 


return queries.map(querie => freqObj[querie] || 0);

    }

  • trupti29
     + 0 comments

    from collections import Counter

    def matchingStrings(strings, queries): count = Counter(strings) results = [] for i in queries: if i in count: results.append(count[i]) else: results.append(0) return results

  • juandadky2020
     + 0 comments

    Kotlin Solution

    fun matchingStrings(strings: Array<String>, queries: Array<String>): Array<Int> {
    val counts = strings.groupingBy { it }.eachCount()
    return queries.map {query -> counts[query] ?: 0 }.toTypedArray()
}