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Sparse Arrays
Sparse Arrays
+ 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]
+ 0 comments python 3
result1 = []
for i in queries: result = stringList.count(i) result1.append(result) return result1
+ 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;
+ 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; }
+ 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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