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  1. Prepare
  2. Data Structures
  3. Arrays
  4. Sparse Arrays
  5. Discussions

Sparse Arrays

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  • __eskiimo
     + 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]

  • darylo1987
     + 0 comments

    python 3

    result1 = [] 

        for i in queries:

         result = stringList.count(i)

         result1.append(result)

return result1

  • himanshuaiml
     + 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;

  • prernakalwani34
     + 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;
}

  • Abhishek16012001
     + 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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