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  • + 0 comments

    Python: allocate extra space for linear runtime

    from collections import Counter
    
    def matchingStrings(stringList, queries):
        # Write your code here
        counts = Counter(stringList)
        return [counts[query] for query in queries]
    
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    Python

    def matchingStrings(stringList, queries):
        return [stringList.count(each) if each in stringList else 0 for each in queries]
    
  • + 0 comments

    Another simple but a bit less efficient python solution:

    def matchingStrings(stringList, queries):
        count = []
        for i in queries:
            occ = [j for j in stringList if i == j]
            count.append(len(occ))
        return count
    
  • + 0 comments

    A very simple python solution:

    def matchingStrings(stringList, queries):
        return [stringList.count(i) for i in queries]
    		
    
  • + 0 comments

    JavaScript Solution:

    let res=[];
    let count=0;
    queries.map(s=>{
        count=0;
        stringList.map(q=>{
            if(s===q) count++;
        })
        res.push(count);
    })
    return res;