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Sparse Arrays

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  • __eskiimo
    2 days ago+ 0 comments

    should've known that if a query string is repeated in queries array I had to count its occurance twice !! i.e stringList = ['a','a'], queries =['a','a','b'] outputs [2,2,0] not [2,0]

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  • darylo1987
    3 days ago+ 0 comments

    python 3

    result1 = []

        for i in queries:
    
             result = stringList.count(i)
    
             result1.append(result)
    
    return result1
    
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  • himanshuaiml
    4 days ago+ 0 comments

    unordered_mapm; vector< int>ans; for(auto i: stringList) { m[i]++; } for(auto i:queries) { int get=m[i]; ans.push_back(get); } return ans;

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  • prernakalwani34
    6 days ago+ 0 comments

    Easy C++ Solution:

    vector<int> matchingStrings(vector<string> stringList, vector<string> queries) {
        vector<int>result;
        int n = stringList.size();
        int m =queries.size();
        for(int i=0;i<m;i++){
            int count=0;
            for(int j=0;j<n;j++){
                if(queries[i]==stringList[j]){
                    count++;
                }
            }
            result.push_back(count);
        }
        return result;
    }
    
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  • Abhishek16012001
    6 days ago+ 0 comments
        public static List<Integer> matchingStrings(List<String> stringList, List<String> queries) {
        // Write your code here
         int count;
        ArrayList<Integer> ans= new ArrayList<>();
        for(int i=0;i<queries.size();i++){
             count=0;
            for(int j=0;j<stringList.size();j++){
                
                if((queries.get(i)).equals(stringList.get(j))){
                    count+=1;
                }
            }
            ans.add(count);
          
            
        }
          return ans;
        }
    
    }
    
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