# Find sum of product of every number and its frequency in given range

Given an array arr[] of integers and an array of queries, the task is to find the sum of product of every number and its frequency in given range [L, R] where each ranges are given in the arrayt of queries.

Examples:

Input: arr[] = [1, 2, 1], Queries: [{1, 2}, {1, 3}]
Output: [3, 4]
Explanation:
For query [1, 2], freq = 1, freq = 1, ans = (1 * freq) + (2 * freq) => ans = (1 * 1 + 2 * 1) = 3
For query [1, 3], freq = 2, freq = 1; ans = (1 * freq) + (2 * freq) => ans = (1 * 2) + (2 * 1) = 4

Input: arr[] = [1, 1, 2, 2, 1, 3, 1, 1], Queries: [{2, 7}, {1, 6}]
Output: [10, 10]
Explanation:
For query (2, 7), freq = 3, freq = 2, freq = 3;
ans = (1 * freq) + (2 * freq ) + (3 * freq)
ans = (1 * 3) + (2 * 2) + (3 * 1) = 10

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach:

To solve the problem mentioned above the naive method is to iterate over the subarray given in the query. Maintain a map for the frequency of each number in the subarray and iterate over the map and compute the answer.

Below is the implementation of the above approach:

## C++

 // C++ implementation to find  // sum of product of every number  // and square of its frequency  // in the given range     #include  using namespace std;     // Function to solve queries  void answerQueries(      int arr[], int n,      vector >& queries)  {         for (int i = 0; i < queries.size(); i++) {             // Calculating answer          // for every query          int ans = 0;             // The end points          // of the ith query          int l = queries[i].first - 1;          int r = queries[i].second - 1;             // map for storing frequency          map<int, int> freq;          for (int j = l; j <= r; j++) {                 // Iterating over the given              // subarray and storing              // frequency in a map                 // Incrementing the frequency              freq[arr[j]]++;          }             // Iterating over map to find answer          for (auto& i : freq) {                 // adding the contribution              // of ith number              ans += (i.first                      * i.second);          }             // print answer          cout << ans << endl;      }  }     // Driver code  int main()  {         int arr[] = { 1, 2, 1 };      int n = sizeof(arr) / sizeof(arr);         vector > queries          = { { 1, 2 },              { 1, 3 } };      answerQueries(arr, n, queries);  }

## Java

 // Java program to find sum of  // product of every number and   // square of its frequency in   // the given range  import java.util.*;     class GFG{     // Function to solve queries       public static void answerQueries(int[] arr,                                   int n,                                    int[][] queries)   {      for(int i = 0; i < queries.length; i++)      {                     // Calculating answer           // for every query               int ans = 0;             // The end points           // of the ith query          int l = queries[i] - 1;          int r = queries[i] - 1;             // Hashmap for storing frequency          Map freq = new HashMap<>();             for(int j = l; j < r + 1; j++)          {                             // Iterating over the given               // subarray and storing               // frequency in a map                  // Incrementing the frequency              freq.put(arr[j],                        freq.getOrDefault(arr[j], 0) + 1);          }          for(int k: freq.keySet())          {                             // Adding the contribution               // of ith number              ans += k * freq.get(k);          }                  // Print answer      System.out.println(ans);      }  }     // Driver code  public static void main(String args[] )  {      int[] arr = { 1, 2, 1 };      int n = arr.length;      int[][] queries = { { 1, 2 },                          { 1, 3 } };                                 // Calling function      answerQueries(arr, n, queries);   }  }     // This code contributed by dadi madhav

## Python3

 # Python3 implementation to find  # sum of product of every number  # and square of its frequency  # in the given range     # Function to solve queries  def answerQueries(arr, n, queries):             for i in range(len(queries)):             # Calculating answer          # for every query          ans = 0            # The end points          # of the ith query          l = queries[i] - 1         r = queries[i] - 1            # Map for storing frequency          freq = dict()          for j in range(l, r + 1):                 # Iterating over the given              # subarray and storing              # frequency in a map                 # Incrementing the frequency              freq[arr[j]] = freq.get(arr[j], 0) + 1            # Iterating over map to find answer          for i in freq:                 # Adding the contribution              # of ith number              ans += (i * freq[i])             # Print answer          print(ans)     # Driver code  if __name__ == '__main__':             arr = [ 1, 2, 1 ]      n = len(arr)         queries = [ [ 1, 2 ],                  [ 1, 3 ] ]                         answerQueries(arr, n, queries)     # This code is contributed by mohit kumar 29

Output:

3
4


Time Complexity: O(Q * N)

Auxiliary Space Complexity: O(N)

Efficient Approach:

To optimize the above method we will try to implement the problem using Mo’s Algorithm.

• Sort the queries first according to their block of size using custom comparator and also store the indexes of each query in a map for printing in order.
• Now, we will maintain two-pointer L and R which we iterate over the array for answering the queries. As we move the pointers, if we are adding some number in our range, we’ll first remove the contribution of its previous freq from the answer, then increment the frequency and finally add the contribution of new frequency in answer.
• And if we remove some element from the range, we’ll do the same, remove the contribution of existing freq of this number, decrement the freq, add the contribution of its new frequency.

Below is the implementation of the above approach:

 // C++ implementation to find sum  // of product of every number  // and square of its frequency  // in the given range     #include  using namespace std;     // Stores frequency  const int N = 1e5 + 5;     // Frequnecy array  vector<int> freq(N);     int sq;     // Function for comparator  bool comparator(      pair<int, int>& a,      pair<int, int>& b)  {      // comparator for sorting      // accoring to the which query      // lies in the whcih block;      if (a.first / sq != b.first / sq)          return a.first < b.first;         // if same block,      // return which query end first      return a.second < b.second;  }     // Function to add numbers in range  void add(int x, int& ans, int arr[])  {      // removing contribution of      // old frequency from answer      ans -= arr[x]             * freq[arr[x]];         // incrementing the frequency      freq[arr[x]]++;         // adding contribution of      // new frequency to answer      ans += arr[x]             * freq[arr[x]];  }     void remove(int x, int& ans, int arr[])  {      // removing contribution of      // old frequency from answer      ans -= arr[x]             * freq[arr[x]];         // Decrement the frequency      freq[arr[x]]--;         // adding contribution of      // new frequency to answer      ans += arr[x]             * freq[arr[x]];  }     // Function to answer the queries  void answerQueries(      int arr[], int n,      vector >& queries)  {         sq = sqrt(n) + 1;         vector<int> answer(          int(queries.size()));         // map for storing the      // index of each query      map, int> idx;         // Store the index of queries      for (int i = 0; i < queries.size(); i++)          idx[queries[i]] = i;         // Sort the queries      sort(queries.begin(),           queries.end(),           comparator);         int ans = 0;         // pointers for iterating      // over the array      int x = 0, y = -1;      for (auto& i : queries) {             // iterating over all          // the queries          int l = i.first - 1, r = i.second - 1;          int id = idx[i];             while (x > l) {              // decrementing the left              // pointer and adding the              // xth number's contribution              x--;              add(x, ans, arr);          }          while (y < r) {                 // incrementing the right              // pointer and adding the              // yth number's contribution              y++;              add(y, ans, arr);          }          while (x < l) {                 // incrementing the left pointer              // and removing the              // xth number's contribution              remove(x, ans, arr);              x++;          }          while (y > r) {                 // decrementing the right              // pointer and removing the              // yth number's contribution              remove(y, ans, arr);              y--;          }          answer[id] = ans;      }         // printing the answer of queries      for (int i = 0; i < queries.size(); i++)          cout << answer[i] << endl;  }     // Driver Code  int main()  {         int arr[] = { 1, 1, 2, 2, 1, 3, 1, 1 };      int n = sizeof(arr) / sizeof(arr);         vector > queries          = { { 2, 7 },              { 1, 6 } };      answerQueries(arr, n, queries);  }

Output:

10
10


Time Complexity: O(N * sqrt{N})
Auxiliary Space Complexity: O(N)

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